Electromagnetics - Schaum - 2nd Edition - Joseph A. Ediminister by Blog da Engenharia de Produção

Second Edition

Co -rs a course fund~n nt uppf rllents any dass tc t

351 probl "' ot t p-b - t p

ld I for

IfÂˇ tud !

SCHAUM'S OUTLINE OF

THEORY AND PROBLEMS OJ!

ELECTROMAGNETICS Second Edition

â&#x20AC;¢

JOSEPHAED~STER

Professor Emeritus of Electrical Engirurering The University of Akron

SCHAUM'S OUTLINE SERIES McGRAW-HILL New York

Son Francisco Waahington, D.C. Auckland Bogota Corocaa London Madrid Mexico City Milan Montreal New Dehli Son Juan Singapore Sydney Tokyo Toronto

Lisbon

JOSEPH A. EDMINISTER is currently Director of Corporate Relations for the College of Engineering at Cornell University. In 1984 he held an IEEE Congressional Fellowship in the office of Congressman Dennis E. Eckart (D-OH). He received BEE, MSE and JD degrees from the University of Akron. He served as professor of electrical engineering, acting department head of electrical engineering, assistant dean and acting dean of engineering, all at the University of Akron. He is an attorney in the state of Ohio and a registered patent attorney. He taught electric circuit analysis and electromagnetic theory throughout his academic career. He is a Professor Emeritus of Electrical Engineering from The University of Akron.

Appendix B is jointly copyrighted© 1995 by McGraw-Hill, Inc. and MathSoft, Inc. Schaum's Outline or Theory and Problems or ELECfROMAGNETICS Copyright © 1993, 1979 by The McGraw-Hill Companies, Inc. All rlJbts reserved. Printed in the United States of America. Except as permitted under the Copyright Act of 1976, no put of this publication may be reproduced or distributed in any fonn or by any means, or stored in a data base or retrieval system, without the prior written pennission of the publisher.

567 8910 II 12131415161718 1920PRS PRS9 8

IS 8 N 0 - 0 7 - 0 212 3 4 - 1 (Formerly published under ISBN 0-07-018993-5.) Sponsoring Editor: David Beckwith Production Supervisor: AI Rihner Editing Supervisor: Patty Andrews Front Matter Editor: Maureen Walker

Libnry of Conpess CablogiaJ-in-hblialtion Dllbl

Edminister, Joseph. Schaum's outline or theory and problems or electromagnetics I by Joseph A. Edminister.-2nd ed. p. cm.-{Schaum's outline series) Includes index. ISBN 0-07-018993-5 I. Electromagnetism. 1. Title. II. Title: Theory and problems or electromagnetics. III. Series. OC760.E35 1993 S37'.02'02-dc20 91-43302

CIP

McGraw-Hill A Division of The M.:Graw·Hill Companies

Preface The second edition of Schaum's Outline of Electromagnetics offers three new chapters--in transmission lines, waveguides, and antennas. These have been included to make the book a more powerful tool for students and practitioners of electromagnetic field theory. I take pleasure here in thanking my colleagues M. L. Kult and K. F. Lee for their contribution of this valuable material. The basic approach of the first edition has been retained: "As in other Schaum's Outlines the emphasis is on how to solve problems. Each chapter consists of an ample set of problems with detailed solutions, and a further set of problems with answers, preceded by a simplified outline of the principles and facts needed to understand the problems and their solutions. Throughout the book the mathematics has been kept as simple as possible, and an abstract approach has been avoided. Concrete examples are liberally used and numerous graphs and sketches are given. I have found in many years of teaching that the solution of most problems begins with a carefully drawn sketch." Once again it is to my student~my former student~that I wish to dedicate this book. JosEPH A. EoMINISTER

iii

Contents

Chapter 1

VECTOR ANALYSIS ......................•...........•.............••.......•.....•.. 1.1 Introduction 1.2 Vector Notation 1.3 Vector Algebra Systems 1.5 Differential Volume, Surface, and Line Elements

Chapter 2

1.4 Coordinate

COULOMB FORCES AND ELECfRIC HELD INTENSITY ••••••••••••

2.1 Coulomb's Law 2.2 Electric Field Intensity 2.3 Charge Distributions 2.4 Standard Charge Configurations

Chapter 3

ELECfRIC FLUX AND GAUSS' LAW..........................................

3.1 Net Charge in a Region 3.2 Electric Flux and Flux Density 3.3 Gauss' Law 3.4 Relation Between Flux Density and Electric Field Intensity 3.5 Special Gaussian Surfaces

Olapter

DIVERGENCE AND THE DIVERGENCE THEOREM •••••••••••••••••••

4.1 Divergence 4.2 Divergence in Cartesian Coordinates 4.3 Divergence of D 4.4 The Del Operator 4.5 The Divergence Theorem

Chapter

THE ELECTROSTATIC HELD: WORK, ENERGY, AND POTENTI.AL •• . •. •. . .. •• •.. •. •. . •••. •. . •• •. . •• . . . •• •. . ••••. . .. •••. . . ••. ••• •. ••. . . . •. . •• ••.

5.1 Work Done in Moving a Point Charge 5.2 Conservative Property of the Electrostatic Field 5.3 Electric Potential Between Two Points 5.4 Potential of a Point Charge 5.5 Potential of a Charge Distribution 5.6 Gradient 5.7 Relationship Between E and V 5.8 Energy in Static Electric Fields

Chapter

CURRENT, CURRENT DENSITY, AND CONDUCTORS •••••••••••••••

6.1 Introduction 6.2 Charges in Motion 6.3 Convention Current Density J 6.4 Conduction Current Density J 6.5 Conductivity o 6.6 Current I 6. 7 Resistance R 6.8 Current Sheet Density K 6.9 Continuity of Current 6.10 Conductor-Dielectric Boundary Conditions

Chapter 7

CAPACITANCE AND DIELECTRIC MATERIALS........................

7.1 Polarization P and Relative Permittivity £, 7.2 Capacitance 7.3 MultipleDielectric Capacitors 7.4 Energy Stored in a Capacitor 7.5 Fixed-Voltage D and E 7.6 Fixed-ChargeD and E 7.7 Boundary Conditions at the Interface of Two Dielectrics

Chapter

LAPLACE'S EQUATION •••..••...••..•..••...••.••••••..•.••..•....•.........•..... 8.1 Introduction 8.2 Poisson's Equation and Laplace's Equation 8.3 Explicit 8.4 Uniqueness Theorem 8.5 Mean Value Forms of Laplace's Equation and Maximum Value Theorems 8.6 Cartesian Solution in One Variable 8.7 Cartesian Product Solution 8.8 Cylindrical Product Solution 8.9 Spherical Product Solution

114

CONTENTS

CUpter 9

AMPERE'S LAW AND THE MAGNETIC tlELD ........................... 135 9.1 Introduction 9.2 Biot-Savart Law 9.3 Am~re's Law 9.4 Curl 9.5 Relationship of J and H 9.6 Magnetic Flux Density B 9.7 Vector Magnetic Potential A 9.8 Stokes' Theorem

CUpter 10

FORCES AND TORQUES IN MAGNETIC tlELDS ........................ lS4 10.1 Magnetic Force on Particles 10.2 Electric and Magnetic Fields Combined 10.3 Magnetic Force on a Current Element 10.4 Work and Power 10.5 Torque 10.6 Magnetic Moment of a Planar Coil

CUpter

INDUCTANCE AND MAGNETIC CIRCUI1'S ................................ 169 11.1 Inductance 11.2 Standard Conductor Configurations 11.3 Faraday's Law andSelf-Inductance 11.4 lntemallnductance 11.5 Mutuallnductance 11.6 Magnetic Circuits 11.7 The B-H Curve 11.8 Am~re's Law for Magnetic Circuits 11.9 CoreswithAirGaps 11.10 MultipleCoils 11.11 ParallelMagneticCircuits

Chapter 12

DISPLACEMENT CURRENT AND INDUCED EMF ...................... 19l 12.1 Displacement Current 12.2 Ratio of lc to JD 12.3 Faraday's Law and Lenz's Law 12.4 Conductors in Motion Through Time-Independent Fields 12.5 Conductors in Motion Through Time-Dependent Fields

Chapter 13

MAXWELL'S EQUATIONS AND BOUNDARY CONDmONS ••••••• 205 13.1 Introduction 13.2 Boundary Relations for Magnetic Fields 13.3 Current Sheet at the Boundary 13.4 Summary of Boundary Conditions 13.5 Maxwell's Equations

Chapter 14

ELECTROMAGNETIC WAVES ................................................... Z16 14.1 Introduction 14.2 Wave Equations 14.3 Solutions in Cartesian Coordinates 14.4 Solutions for Partially Conducting Media 14.5 Solutions for Perfect Dielectrics 14.6 Solutions for Good Conductors; Skin Depth 14.7 Interface Conditions at Normal Incidence 14.8 Oblique Incidence and Snell's Laws 14.9 Perpendicular Polarization 14.10 Parallel Polarization 14.11 Standing Waves 14.12 Power and the Poynting Vector

Chapter

TRANSMISSION LINES •• ....... ••• •••••••• ••••• ••• ••• •••••••• •••••• ••••••••••• ••••••• '1:37 15.1 Introduction 15.2 Distributed Parameters 15.3 Incremental Model; Voltages and Currents 15.4 Sinusoidal Steady-State Excitation 15.5 The Smith Chart 15.6 Impedance Matching 15.7 Single-Stub Matching 15.8 DoubleStub Matching 15.9 Impedance Measurement 15.10 Transients in Lossless Lines

Chapter 16

WAVEGUIDES •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• 274 16.1 Introduction 16.2 Transverse and Axial Fields 16.3 TE and TM Modes; Wave Impedances 16.4 Determination of the Axial Fields 16.5 Mode Cutoff Frequencies 16.6 Dominant Mode 16.7 Power Transmitted in a Lossless Waveguide 16.8 Power Dissipation in a Lossy Waveguide

CONTENTS

Chapter 17

vii

ANTENNAS ••.• .•••• ••••• .•••. ••• .••••. •.. .•••• •.••••••••••••••• •••••••.•••••. ••••• •.••• •• Z93 17.1 Introduction 17.2 Current Source and the E and H Fields 17.3 Electric (Hertzian) Dipole Antenna 17.4 Antenna Parameters 17.5 Small CircularLoop Antenna 17.6 Finite-Length Dipole 17.7 Monopole Antenna 17.8 Selfand Mutual Impedances 17.9 The Receiving Antenna 17.10 Linear Arrays 17.11 Reflectors

Appendix A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315

AppendixB

317

INDEX

333

Chapter 1 Vector Analysis 1.1 INI'RODUCfiON Vectors are introduced in physics and mathematics courses, primarily in the cartesian coordinate system. Although cylindrical coordinates may be found in calculus texts, the spherical coordinate system is seldom presented. All three coordinate systems must be used in electromagnetics. As the notation, both for the vectors and the coordinate systems, differs from one text to another, a thorough understanding of the notation employed herein is essential for setting up the problems and obtaining solutions.

1.1 VECI'OR NOTATION In order to distinguish vectors (quantities having magnitude and direction) from scalars (quantities having magnitude only) the vectors are denoted by boldface symbols. A unit vector, one of absolute value (or magnitude or length) 1, will in this book always be indicated by a boldface, lowercase a. The unit vector in the direction of a vector A is determined by dividing A by its absolute value: A or A By use of the unit vectors a_.., ay, •z along the x, y, and z axes of a cartesian coordinate system, an arbitrary vector can be written in component form: A=A_..a_.. +Ay8y +Azaz

In terms of components, the absolute value of a vector is defined by

IAI =A= v'A;+A;+A~ 1.3 VECI'OR ALGEBRA 1.

Vectors may be added and subtracted. A± B = (A_..a_.. + Ay8y + Azaz) ± (B_..a_.. + Byay + Bzaz)

=(A... ± B_..)a_.. + (Ay ± By)ay + (Aze Bz)az 2.

The associative, distributive, and commutative laws apply. A+ (B +C) = (A+ B) + C k(A+B)=kA+kB A+B=B+A

The dot product of two vectors is, by definition, A• B=ABcos8

where 8 is the smaller angle between A and B.

(read "A dot B") In Example 1 it is shown that

A·B=A_..B... +AyBy +AzBz

which gives, in particular,

IAI =VA· A. 1

VECTOR ANALYSIS

EXAMPLE 1.

(CHAP. 1

The dot product obeys the distributive and scalar multiplication laws

A• (B+C) =A· B +A•C

A· kB=k(A· B)

This being the case,

A· B = (A..a.. +Ayay + Azaz)· (B..a.. + Byay +B.a.) = A_.B.. (a.. • a.. )+ A,.B,.(a,. • ay)

+ A.B.(a. • a.)

+ A .. B,.(a.. • ay) + · · ·+A. By(a.· a,.) However, a.. • a.. = a,. • a,. = a. • a. = 1 because the cos (J in the dot product is unity when the angle is zero. And when (J = 90", cos (J is zero; hence all other dot products of the unit vectors are zero. Thus

A· B =A,B.. +A,.By +A.B.

The cross product of two vectors is, by definition, A X B = (AB sin 8)an

(read "A cross B")

where 8 is the smaller angle between A and 8, and a" is a unit vector normal to the plane determined by A and B when they are drawn from a common point. There are two normals to the plane, so further specification is needed. The normal selected is the one in the direction of advance of a right-hand screw when A is turned toward B (Fig. 1-1). Because of this direction requirement, the commutative law does not apply to the cross product; instead, AXB=-BXA

Fig. 1-1

Expanding the cross product in component form, A X B = (A_..a... + Ayay + Azaz) X (B_..a_.. + Byay + Bzaz)

= (AyBz- AzBy)a.. + (AzB..-- A.. Bz)ay + (A ... By- AyB.. )az which is conveniently expressed as a determinant:

...

AXB= A ..

EXAMPLE 2.

VECfOR ANALYSIS

CHAP. 1)

Given A= 2a.. + 4ay - 3a, and B =a.. - a,., find A • B and A X B. A • B = (2)(1) + (4)( -1) + ( -3)(0) = -2

... .y •. AXB= 2

1 -1

= -3a.. -3ay -6a,

-3 0

1.4 COORDINATE SYSTEMS A problem which has cylindrical or spherical symmetry could be expressed and solved in the familiar cartesian coordinate system. However, the solution would fail to show the symmetry and in most cases would be needlessly complex. Therefore, throughout this book, in addition to the cartesian coordinate system, the circular cylindrical and the spherical coordinate systems will be used. All three will be examined together in order to illustrate the similarities and the differences. A point Pis described by three coordinates, in cartesian (x, y, z), in circular cylindrical (r, ¢, z), and in spherical {r, 0, ¢), as shown in Fig. 1-2. The order of specifying the coordinates is important and should be carefully followed. The angle ¢ is the same angle in both the cylindrical and spherical systems. But, in the order of the coordinates, ¢ appears in the second position in cylindrical, (r, q,, z ), and the third position in spherical, (r, 0, ¢ ). The same symbol, r, is used in both cylindrical and spherical for two quite different things. In cylindrical coordinates r measures the distance from the z axis in a plane normal to the z axis, while in the spherical system r measures the distance from the origin to the point. It should be clear from the context of the problem which r is intended. z

z ~ P(x,y,z)

~ P(r,~. z)

(a) Cartesian

(b) Cylindrical

Fig. 1-l

A point is also defined by the intersection of three orthogonal surfaces, as shown in Fig. 1-3. In cartesian coordinates the surfaces are the infinite planes x = const., y = const., and z = const. In cylindrical coordinates, z = const. is the same infinite plane as in cartesian; ¢ = const. is a half plane with its edge along the z axis; r = const. is a right circular cylinder. These three surfaces are orthogonal and their intersection locates point P. In spherical coordinates, ¢ = const. is the same half plane as in cylindrical; r = const. is a sphere with its center at the origin; 0 = const. is a right circular cone whose axis is the z axis and whose vertex is at the origin. Note that 0 is limited to the range 0 s 8 s 1r. Figure 1-4 shows the three unit vectors at point P. In the cartesian system the unit vectors have fixed directions, independent of the location of P. This is not true for the other two systems (except Each unit vector is normal to its coordinate surface and is in the direction in in the case of which the coordinate increases. Notice that all these systems are right-handed:

•z>·

a,Xae=a91

VECTOR ANALYSIS

(CHAP. 1 z

0 = const.

r= const. z = const.

1---.,._Y

X X

(b) Cylindrical

(a) Cartesian

= const.

Fig. 1-3

The component forms of a vector in the three systems are A= Ax&.r + Ay&y + Az&z

(cartesian)

= A,a, + A.,.a.,. + Azaz

(cylindrical)

= A,a, + A 9 a9 + A.,a.,.

(spherical)

It should be noted that the components Ax, A, A.,, etc., are not generally constants but more often are functions of the coordinates in that particular system. z

-y

(a) Cartesian

(b) Cylindrical

Fig. 1-4

1.5 DIFFERENTIAL VOLUME, SURFACE, AND LINE ELEMENTS There are relatively few problems in electromagnetics that can be solved without some sort of integration-along a curve, over a surface, or throughout a volume. Hence the corresponding differential elements must be clearly understood. When the coordinates of point P are expanded to (x + dx, y + dy, z + dz) or (r + dr, q, + dq,, z + dz) or (r + dr, 8 + d8, q, + dq, ), a differential volume dv is formed. To the first order in infinitesimal quantities the differential volume is, in all three coordinate systems, a rectangular box. The value of dv in each system is given in Fig. 1-5.

VECTOR ANALYSIS

CHAP. 1) z

dv=dxdydz

dv = rdrd,Pdz

(a) Cartesian

(b) Cylindrical

= r 2 sin tl drdtl d,P (c) Spherical

Fig. 1-5

From Fig. 1-5 may also be read the areas of the surface elements that bound the differential volume. For instance. in spherical coordinates. the differential suiface element perpendicular to a, is dS = (r dlJ)(r sin 8 d<J>) = r 2 sin 8 dO d<J>

The differential line element. dt is the diagonal through P. Thus dt2 = dx 2 + dy 2 + dz 2 2

dt = dr dt = dr

(cartesian)

+ r d<J> + dz + r 2 d82 + r 2 sin2 8 d<J> 2 2

(cylindrical) (spherical)

Solved Problems l.l.

Show that the vector directed from M(x .. y .. z1) to N(x 2 , y2 , z2 ) in Fig. 1-6 is given by (xz- x t)a.., + (Yz - Yt)a,. + (zz- Zt)az

Fig. 1-6 The coordinates of M and N are used to write the two position vectors A and B in Fig. 1-6. A =x 1a.., + y 1ay + z.a.

B=x2 a.., + y~Y + z~.

Then

B- A= (x2- x.)a.., + (y2 - y.)aY + (z2- z.)a.

1.2. ,.. 1

VECTOR ANALYSIS

(CHAP. 1

Find the vector A directed from (2, -4, 1) to (0, -2, 0) in ·cartesian coordinates and find the unit vector along A. A=(O- 2)a. + (-2- (-4))a,. + (0-l)a. = -2a. + 2a,. -a.

•

IAI 2 = (-2)2 + (2)2 + ( -1 )2 = 9 2 1 .A = IAIA = - 32 •• + 31ly - 3••

1.3.

Find the distance between (5, Jn/2, 0) and (5, n/2, 10) in cylindrical coordinates. First, obtain the cartesian position vectors A and B (see Fig. 1-7). A= -5a,.

B=5a,. +lOa.

z (5, n/2,10)

Fig. 1-7 Then B- A= lOlly+ lOa.

and the required distance between the points is IB-A1=10v'2

The cylindrical coordinates of the points cannot be used to obtain a vector between the points in the same manner as was employed in Problem 1.1 in cartesian coordinates.

1.4.

Show that A = 4a... - 2ay - &z that

and B = a_.. + 4&y

4az

are perpendicular.

Since the dot product contains cos 8, a dot product of zero from any two nonzero vectors implies 8 =90". A· 8=(4)(1) + (-2)(4) + (-1)(-4) =0

1.5.

ill

Given A= 2a_.. + 4ay and B = 6ay- 4az, the cross product, (b) the dot product .

.!II!! (a)

AXB=

a. a,. 2 4

find the smaller angle between them using (a) a. 0

=-16a.,+8a,.+l2a.

6 -4

IAI = V(2) 2 + (4) 2 + (0) 2 = 4.47 IBI = V(0) 2 + (6f + ( -4)2 = 7.21 lAX Bl = V( -16)2 + (8) 2 + (12)2 = 21.54

Then, since lA X Bl = IAIIBI sin 8, . SID

21.54

= (4.47)(7.21)

0.66S

8 =41.9"

A • B = (2)(0) + (4)(6) + (0)( -4) = 24

(b)

6= A·B

cos

1.6. •

VECTOR ANALYSIS

CHAP. 1)

!AliBI (4.47)(7.21)

Given F = (y - l)ax + 2xa,., where B =Sax -a,. + 2az.

=0.745

find the vector at (2, 2, 1) and its projection on 8, F(2, 2, 1) = (2- 1)a.. + (2)(2)a, =a.. + 4a,

As indicated in Fig. 1-8, the projection of one vector on a second vector is obtained by expressing the unit vector in the direction of the second vector and taking the dot product. Proj. A on B =A· a 8

A·B =Iii

Fig. l-8 Thus, at (2, 2, 1), Proj. F on B = F· B = (1)(5) + (4)(-1) + (0)(2) = 1

v'30

IBI L7.

•x

Given A= +a,., with A X (8 X C).

B =a..,+ 2az,

AXB=

Then

and· C = 2a,. + Bzo

v30 find (A X B) XC and compare it

a.. a, a. 1 1 0 =2a.. -2a,.-a. 1 0 2

(AXB)XC=

a.. a,. a. 2 -2 -1 =-2a,.+4a, 0 2 1

A similar calculation gives A X (B X C) = 2a.. - 2a,. + 3a,. Thus the parentheses that indicate which cross product is to be taken first are essential in the vector triple product.

1.8.

Using the vectors A, 8, and C of Problem 1.7, find A• B XC and compare it with A X B•C. From Problem 1.7,

B X C = -4a.. -a,. + 2a,.

Then

A· BXC = (1)(-4) + (1)(-1) + (0)(2) = -5

Also from Problem 1. 7,

A X B = 2a.. - 2ay -a..

Then

AX B • C = (2)(0)+ (-2)(2) + (-1)(1) = -5 Parentheses are not needed in the scalar triple product since it has meaning only when the cross product is taken first. In general, it can be shown that

A.. A,. A, A·BXC= B.. By B.

c.. c,. c.

VECfOR ANALYSIS

(CHAP. 1

As long as the vectors appear in the same cyclic order the result is the same. The scalar triple products not in this cyclic order have a change in sign.

1.9.

Express the unit vector which points from z = h coordinates. See Fig. 1-9.

on the z axis toward (r, tfl, 0) in cylindrical

z h

Fig. 1-9 The vector R is the difference of two vectors:

.-,R, -

R=ra,-ha. R

R ra,-haâ&#x20AC;˘ --y,2 +hi

The angle ,P does not appear explicitly in these expressions. through a,.

Nevertheless, both R and aR vary with

1.10. Express the unit vector which is directed toward the origin from an arbitrary point on the plane z = -5, as shown in Fig. 1-10. z

(x,y, -5 ..... 1-10

Since the problem is in cartesian coordinates, the two-point formula of Problem 1.1 applies.

R= -xa.,- ya,. + 5a, -xa.,- ya,. + 5a.

=-:=;;:==::;;:::::::;:::::""" 2 v'x + y 2 +25

1.11. Use the spherical coordinate system to find the area of the strip a :s 8 :s fJ on the spherical ..,. shell of radius a (Fig. 1-11 ). What results when a = 0 and fJ = n? 1 â&#x20AC;˘

The differential surface element is [see Fig. 1-5(c)] dS

= r 2 sin (J d(J d,P

CHAP. 1)

VECfOR ANALYSIS

Fig. 1-ll

Then

A= =

When

a= 0

and

f" f

a sin Od8df/J

2.tra 2 (cos a- cos {J)

fJ = .tr, A = 4.tra 2 , the surface area of the entire sphere.

1.12. Obtain the expression for the volume of a sphere of radius a from the differential volume. From Fig. 1-5(c ),

dv = r 2 sin 8 dr dO df/J. v =

Iz,I,Ia 0

Then 4 r 2 sin 0 dr d8 df/J =- .tra3 3

1.13. Use the cylindrical coordinate system to find the area of the curved surface of a right circular cylinder where r =2m. h = 5 m. and 30Â° :s; cp :s; 1200 (see Fig. 1-12).

i:l

,!!!!

Fig. 1-U

The differential surface element is dS = r df/J dz.

1.14. Transform

from cartesian to cylindrical coordinates.

J2x/3 2dlf>dz

1 5

Then

n/6

(CHAP. 1

VECfOR ANALYSIS

10 Referring to Fig. 1-2(b),

x=rcostJ> Hence

r= v'x2+ y2

y=rsintJ>

A = r sin tJ>a" + r cos t/>fly + r cos2 tJ>•~

Now the projections of the cartesian unit vectors on a,, a., and

•~

are obtained:

a"· a, =cos 4> fly • a, = sin 4>

a" ·a.= -sin 4>

8x•a~=O

fly· a. =cos 4>

fly.·~= 0

•~·a,=O

•~

·~··~=1

Therefore

·a.,=O

a" = cos tJ>a, - sin tJ>a. fly = sin tJ>a, + cos tJ>a.,

·~=·~

A = 2r sin 4> cos tJ>a, + (r cos2 4> - r sin2 4> )a., + r cos2 4>•~

and

1.1.5. A vector of magnitude 10 points from (5, 5n/4, 0) in cylindrical coordinates toward the origin • (Fig. 1-13). Express the vector in carte~an coordinates.

..... 1-13

In cylindrical coordinates, the vector may be expressed as lOa, where 1r 10 A =lOcos-=" 4 V2

• 1r 10 A =lOsm-=,. 4 V2

4> = n/4. Hence

A~=O

so that 10 10 A=-a +-fly V2"V2 Notice that the value of the radial coordinate, 5, is immaterial.

Supplementary Problems 1.16.

Given A= 4fly +lOa~ and B = 2a" + 3a,.,

1.17.

Given A= (10/V2)(a" + •~) and B = 3(fly + •~), the direction of A. Ans. 1.50(a" + •~)

1.18.

Find the angle between A = lOa,. + 2a~ cross product. Ans. l61S

1.D.

Find the angle between A= 5.8a,. + Ans. 135° and the cross product.

find the projection of A on B.

Ans.

12/vU

express the projection of B on A as a vector in

and B = -4fly + 0.5a~ using both the dot product and the

1.55•~

and B = -6.93a,. +

4.0a~

using both the dot product

VECfOR ANALYSIS

CHAP. 1]

1.20.

Given the plane 4x + 3y + 2z = 12, find the unit vector normal to the surface in the direction away from the origin. Ans. (4a_. + 3a,. + 2a.)/V29

1.Zl.

Find the relationship which the cartesian components of A and B must satisfy if the vector fields are everywhere parallel.

A_. Ay A. Ans. - = - = B.. By B.

1.12.

Express the unit vector directed toward the origin from an arbitrary point on the line described by X =0, y=3.

-3a,. -za• Ans. a= V9+z 2 1.13.

Express the unit vector directed toward the point (x., y., z,) from an arbitrary point in the plane y = -5.

Ans. a = ~<x;_;';:::-=x=)=a=·::;;:+:::::(=:Y:::•+=5=:)a~,.;=+~(z::::'=-=z~)-:;;:a• V(x,-xf+(y, +5f+(z,-zf 1.24.

Express the unit vector directed toward the point (0, 0, h) from an arbitrary point in the plane

Ans.

= -2.

a = _-_x-;::a..:;:-=y=a,.-;{=.+=(=h=+=2:::;),-=a• Vxz+ yz + (h + 2)2

1.25.

Given A= 5a_. and B = 4a_. + Byay, find B,. such that the angle between A and B is 45°. If B also has a term B.a., what relationship must exist between B,. and B.? Ans. By= ±4, VB!+ B! = 4

1.Z6.

Show that the absolute value of A • B X C is the volume of the parallelepiped with edges A, B, and C. (Hint: First show that the base has area IB X Cl.)

1.27.

Given

1.28.

Given A= a_. - ay, B = 2a., and C = -a.. + 3a,., this scalar triple product. Ans. -4, ±4

1.19.

Using the vectors of Problem 1.28 find (A X B) X C.

1.30.

Find the unit vector directed from (2, -5, -2) toward (14, -5, 3).

Ans.

A= 2a_.- a.,

B = 3a.. + ay,

and

C = -2a.. + 6a,.- 4a.,

show that Cis 1. to both A and B.

find A • B X C.

Ans.

Examine other variations of

-8a.

12 5 a_. + a. 13 13

1.31.

Find the vector directed from (10, 3;r/4, :rc/6) to (5, :rc/4, :rc), where the endpoints are given in spherical coordinates. Ans. -9.66a_.- 3.54a,. + 10.61a,

1.32.

Find the distance between (2, :rc/6, 0) and (1, :rr, 2), where the points are given in cylindrical coordinates. Ans. 3.53

1.33.

Find the distance between (1, :rr/4, 0) and (1, 3;r/4, :rc), where the points are given in spherical Ans. 2.0 coordinates.

1.34.

Use spherical coordinates and integrate to find the area of the region 0 s ,P s a of radius a. What is the result when a = 2:rr? Ans. 2aa 2 , A = 4;ra 2

1.35.

Use cylindrical coordinates to find the area of the curved surface of a right circular cylinder of radius a and height h. Ans. 2:rrah

on the spherical shell

VECfOR ANALYSIS

[CHAP. 1

1.36.

Use cylindrical coordinates and integrate to obtain the volume of the right circular cylinder of Problem 1.35. Ans. :rra 2h

1.37.

Use spherical coordinates to write the differential surface areas dS1 and d~ and then integrate to obtain the areas of the surfaces marked 1 and 2 in Fig. 1-14. Ans. :rr/4, :rr/6 z

y X

Fig. 1-14

1.38.

Use spherical coordinates to find the volume of a hemispherical shell of inner radius 2.00 m and outer radius 2.02 m. Ans. 0.162:rr m 3

1.39.

Using spherical coordinates to express the differential volume, integrate to obtain the volume defined 7:rr 3 by 1 ::S r ::S 2m, 0 ::S e ::S :rr/2, and 0 ::S 4> ::S :rr/2. Ans. 6m

Chapter 2 Coulomb Forces and Electric Field Intensity 2.1 COULOMB'S LAW There is a force between two charges which is directly proportional to the charge magnitudes and inversely proportional to the square of the separation distance. This is Coulomb's law, which was developed from work with small charged bodies and a delicate torsion balance. In vector form, it is stated thus,

QIQ28 2 4n£d

Rationalized SI units will be used throughout this book. The force is in newtons (N), the distance is in meters (m), and the (derived) unit of charge is the coulomb (C). The system is rationalized by the factor 4.tr, introduced in Coulomb's Jaw in order that it not appear later in Maxwell's equations. 2 2 £ is the permittivity of the medium, with the units C /N · m or, equivalently, farads per meter (F/m). For free space or vacuum, £=Eo=

8.854 X

w-9

w-!2 F/m =-- F/m 36.tr

For media other than free space, £ = £ 0£r, where £r is the relative permittivity or dielectric constant. Free space is to be assumed in all problems and examples, as well as the approximate value for £ 0 , unless there is a statement to the contrary. For point charges of like sign the Coulomb force is one of repulsion, while for unlike charges the force is attractive. To incorporate this information rewrite Coulomb's law as follows: _ QtQ2 _ QtQ2 R FI 21 2 8213 4n£oR2t

4n£oR2t

where F1 is the force on charge Q 1 due to a second charge Q2 , a21 is the unit vector directed from Q2 to Q 1 , and R21 = R 21a21 is the displacement vector from Q2 to Q 1 • Find the force on charge Q 1 , 20 #JC, due to charge Q 2 , -300 #JC, where Q 1 is at (0, 1, 2) m and Q2 at (2, 0, 0) m. Because 1 Cis a rather large unit, charges are often given in microcoulombs (IJC), nanocoulombs (nC), or picocoulombs (pC). (See Appendix for the SI prefix system.) Referring to Fig. 2-1, EXAMPLE 1.

R21

= -2a.. +By+ 2a

R 21

..... 2-l

= V(-2) 2 + 12 +

22 = 3

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

(CHAP. 2

1 tlza =3(-2a.. +a, +2az)

and

= (20 X 10-

1ben

300 X 104n(10- /36Jr)(3)2 )(-

) (-

=6(2a_. -

2a_. + ay + 2az) 3

j - 2az) N

1be force magnitude is 6 N and the direction is such that Q 1 is attracted to Q2 (unlike charges attract).

This force relationship is bilinear in the charges. Consequently, superposition applies, and the force on a charge Q 1 due ton- 1 other charges Q 2 , Q 3 , ... , Q" is the vector sum of the individual forces: F1 =

QlQ2 QlQ3 Q. ~ Q" 2 ~~ + 2 &31 + · · · = - - £J -2- akt U£oR21 4n£oR31 4n£o 1<-2 Rkl

This superposition extends in a natural way to the case where charge is continuously distributed through some spatial region: one simply replaces the above vector sum by a vector integral (see Section 2.3). The force field in the region of an isolated charge Q is spherically symmetric. This is made evident by locating Q at the origin of a spherical coordinate system, so that the position vector R, from Q to a small test charge Q, « Q, is simply ra,.. Then F,=4Q,Q 2a, Jl'£or

showing that on the spherical surface r =constant,

IF,I is constant, and F, is radial.

2.2 ELEcrRIC DELD INTENSITY

Suppose that the above-considered test charge Q, is sufficiently small so as not to disturb significantly the field of the fixed point charge Q. Then the electric field intensity, E, due to Q is defined to be the force per unit charge on Q,: E = F,/Q,. For Q at the origin of a spherical coordinate system [see Fig. 2-2(a)], the electric field intensity at an arbitrary point Pis, from Section 2.1,

/E P(r, 6, cf>)

(a) Spherical

(b) Cartesian

CHAP. 2]

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

In an arbitrary cartesian coordinate system [see Fig. 2-2(b)],

Q 4nE 0 R 2

The units of E are newtons per coulomb (N/C) or the equivalent, volts per meter (V/m). EXAMPLE 2.

Find E at (0, 3, 4) m in cartesian coordinates due to a point charge

Q = 0.5 J.lC

at the origin.

In this case R=3ay+4a,

R=5

aR=0.6ay+0.8a,

o.5 x w-" E = 41l(lo-9 /36n)(5)2 (0.6ay + 0.8a,) Thus

lEI= 180 V/m in the direction aR = 0.6a, + 0.8 a,.

2.3 CHARGE DISTRIBUTIONS Volume Charge When charge is distributed throughout a specified volume, each charge element contributes to the electric field at an external point. A summation or integration is then required to obtain the total electric field. Even though electric charge in its smallest division is found to be an electron or proton, it is useful to consider continuous (in fact, differentiable) charge distributions and to define a charge density by dQ dv

p=-

Note the units in parentheses, which is meant to signify that p will be in C/m3 provided that the variables are expressed in proper SI units (C for Q and m3 for v). This convention will be used throughout this book. With reference to volume v in Fig. 2-3, each differential charge dQ produces a differential electric field

IdE â&#x20AC;˘

Fig. 2-3

at the observation point P. Assuming that the only charge in the region is contained within the volume, the total electric field at Pis obtained by integration over the volume:

E=J

paR 2 dv

v41rEoR

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

Sheet Charge Charge may also be distributed over a surface or a sheet. the sheet results in a differential electric field dE=

(CHAP. 2

Then each differential charge dQ on

dQ aR 4.rcE0 R 2

at point P (see Fig. 2-4). If the surface charge density is Ps (C/m2 ) and if no other charge is present in the region, then the total electric field at P is E=

f PsaR dS Js4.TCE 0 R 2 P/dE

â&#x20AC;˘

s Fig. 2-4

Line Charge If charge is distributed over a (curved) line, each differential charge dQ along the line produces a differential electric field

at P (see Fig. 2-5). And if the line charge density is Pt (C/m), and no other charge is in the region, then the total electric field at P is

Fig. 2-5

It should be emphasized that in all three of the above charge distributions and corresponding integrals for E, the unit vector aR is variable, depending on the coordinates of the charge element dQ. Thus aR cannot be removed from the integrand. It should also be noticed that whenever the appropriate integral converges, it defines E at an internal point of the charge distribution.

CHAP. 2)

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

2.4 STANDARD CHARGE CONFIGURATIONS In three special cases the integration discussed in Section 2.3 is either unnecessary or easily carried out. In regard to these standard configurations (and to others which will be covered in this chapter) it should be noted that the charge is not "on a conductor." When a problem states that charge is distributed in the form of a disk, for example, it does not mean a disk-shaped conductor with charge on the surface. (In Chapter 6, conductors with surface charge will be examined.) Although it may now require a stretch of the imagination, these charges should be thought of as somehow suspended in space, fixed in the specified configuration.

Point Charge As previously determined, the field of a single point charge Q is given by E=

____Q_2 a 4JrE0 r

(spherical coordinates)

See Fig. 2-2(a). This is a spherically symmetric field that follows an inverse-square law (like gravitation). Infinite Line Charge If charge is distributed with uniform density p~ (C/m) along an infinite, straight line-which will be chosen as the z axis-then the field is given by

E=~a 2JrE0 r

(cylindrical coordinates)

See Fig. 2-6. This field has cylindrical symmetry and is inversely proportional to the first power of the distance from the line charge. For a derivation of E, see Problem 2.9.

-oo

Fig.l-6 EXAMPLE 3. A uniform line charge, infinite in extent, with p, = 20 nC/m, lies along the z axis. (6,8,3)m. In cylindrical coordinates r = V62 + 82 = 10m. The field is constant with z. Thus 20x 10-9 E = 2Jr(l0 9 /36n)(l0) a,= 36a, V /m

Find E at

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

(CHAP. 2

Infinite Plane CllarJe If charge is distributed with uniform density Ps (C/m2 ) over an infinite plane, then the field is

given by

E=-.&a 2Eo " See Fig. 2-7. This field is of constant magnitude and has mirror symmetry about the plane charge. For a derivation of this expression, see Problem 2.12.

Fig.l-7 EXAMPLE 4. Charge is distributed (1/3n) nC/m 2 • Find E.

uniformly over the

(1/3n)10- 9 lEI = 2£o = 2(10 9 /36n) p.

Above the sheet (z > 10 em),

plane z

= 10 em

with a density p, =

V /m

E = 68. V /m; and for z < 10 em,

E = -68. V /m.

Solved Problems Z.l. • 1 111

Two point charges, Q 1 =501JC and Q 2 =10~JC, are located at (-1,1,-J)m and (3, 1, 0) m, respectively (Fig. 2-8). Find the force on Q 1 • R21

= -4a. - 3a.

al, =

-4a -3a z 5 :K

Q.Q2

F, =4.7r£o R221 a2t 6

=(SO X 10- )(10- ) 411"(10 "/36n)(S)2

= (0.18)( -O.Ba. -

(-4a. -3a.) 5

0.68.) N

Ql (-1, 1, -3)

Fig.l-8

CHAP. 2]

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

The force has a magnitude of 0.18N and a direction given by the unit vector -0.8a_. -o.&a... component form,

2.2.

19 In

Refer to Fig. 2-9. Find the force on a 100 f.'C charge at (0, 0, 3) m if four like charges of 20 14C are located on the x and y axes at ±4 m. z

Fig. 2.-9 Consider the force due to the charge at y = 4, (10- 4 )(20 x 10-") ( -4ay + 3a..) 4n(10-11/36n)(5)2 5

The y component will be canceled by the charge at y = -4. other two charges will cancel. Hence

Similarly, the x components due to the

F=4(~)a•. )= 1.73•. N 1.3.

Refer to Fig. 2-10. force

Point charge Q 1 = 300 14C, located at (1, -1, -3) m, experiences a

Fig. 2.-10

due to point charge Q 2 at (3, -3, -2) m. Determine Q 2 • Note that, because

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

(CHAP. 2

the given force is along R 21 (see Problem 1.21), as it must be. Q.Q2 F.=4-R2aR 1r£o

Solving,

1.4.

Q 2 = -40 !JC.

Find the force on a point charge of 50 pC at (0, 0, 5) m due to a charge of S001r pC that is uniformly distributed over the circular disk r ::=;; 5 m, z = 0 m (see Fig. 2-11). :z

(0,0, 5)

Fig.l-11 The charge density is Q

=A=

SOOn x 10-6 n( 5)2

0.2 x 10- C/m

In cylindrical coordinates, R= -ra,+5a, Then each differential charge results in a differential force 6

dF= (50x 10- )(p.rdrdrp) (-ra,+5a.) 4n(10 9 /36n)(r2 + 25) Vr 2 + 25 Before integrating, note that the radial components will cancel and that a. is constant. Hence F = i:z.. is (50 x 10- )(0.2 x 10- )5r dr dtp 8 4n(l0- 9 /36n)(r 2 + 25)312 • 0 0 6

= 90n

z.s .

•

rdr [ -1 (r2 + 25 )312 a,= 90n y,2 + a. = 16.568, N 25 0

Repeat Problem 2.4 for a disk of radius 2 m. Reducing the radius has two effects: the charge density is increased by a factor P2=(5)2=625 (2) 2 0

while the integral over r becomes 2

rdr (r2 + 25 ) 312 = 0.0143

1 0

instead of

s rdr o (rz + 25)312 = 0.0586

The resulting force is 0.0143) (16.568, N) = 25.27a. N F = (6.25) ( _ 0 0586

2.6.

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

CHAP. 2)

Find the expression for the electric field at P due to a point charge Qat (xt> y 1 , z.). with the charge placed at the origin.

Repeat

As shown in Fig. 2-12, R = (x- x 1)a., + (y - y 1)a,. + (z- z 1)a. z P(x,y,z)

Fig. 2-12

Then

Q E=4-R2.R 1r£o _ Q (x -x 1)a., + (y- y 1)ay + (z- z1)a, - 4:rr£o [(x - x 1) 2 + (y - y1) 2 + (z - z 1) 2)312

When the charge is at the origin, E=

Q xa_. + ya,. + za.

4:rr£o (x2 + y2 + z2)312

but this expression fails to show the symmetry of the field. origin,

In spherical coordinates with Q at the

and now the symmetry is apparent.

2.7.

Find E at the origin due to a point charge of 64.4 nC located at ( -4, 3, 2) m in cartesian coordinates. The electric field intensity due to a point charge Q at the origin in spherical coordinates is E=___g__a 4:rr£oT2 r

In this problem the distance is evaluated, is R = 4a., - 3a,. -

v'29 m and

2a•.

the vector from the charge to the origin, where E is to be

2a•) = (20.0) (4a_. -v'2§ 3ay- 2a,) V/m

64.4 X 10(4a_. - 3ay E = 4:rr(l0-9/J6:rr)(Z9) v'2§

2.8.

Find E at (0, 0, 5) m due to Q 1 = 0.35 pC at (0, 4, 0) m and (3, 0, 0) m (see Fig. 2-13). R 1 = -4ay + 5a. R2 = -3a., + 5a,

E _

0.35 x 10-

9 I - 4:rr(l0- /36:rr)(41)

= -48.0ay + 60.0..

(-4a,. + 5a.) V4i

V/m

Q 2 = -0.55 pC at

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

[CHAP. 2

y X

Fia-2-13 ~

-0.55 X 10-

4n(10- /36n)(34) =74.911_. -124.911,

and

2.9.

(-3a~

+ 5az)

v'34

V/m

E = E 1 + ~ = 74.911~- 48.0ay- 64.911. V /m

Charge is distributed uniformly along an infinite straight line with constant density p~. Develop the expression for E at the general point P. Cylindrical coordinates will be used, with the line charge as the z axis (see Fig. 2-14).

At P,

za.)

dE = _.5?.._ ('•· 4n£oR2 Vr2 + z2

~-Fig.l-14

Since for every dQ at z there is another charge dQ at -z, the z components cancel.

Then

p,rdz

~ 4n£o(r2 + z 2) 312 11, p,r [ z ]~ p, = 4.7r£o r2Vr2 + z2 -~a,= 2n£or a,

E =[

2.10. On the line described by x = 2 m, y = -4 m there is a uniform charge distribution of density p~:::::: 20 nC/m. Determine the electric field E at ( -2. -1. 4) m. With some modification for cartesian coordinates the expression obtained in Problem 2. 9 can be used with this uniform line charge. Since the line is parallel to a. , the field has no z

CHAP. 2]

COULOMB FORCES AND ELECfRIC FIELD INTENSITY

component. Referring to Fig. 2-15,

R = -4a... + 3a,. and

20 X 10E = 2.1l'£o(5)

9 (

-4a.. + 3a,.) = -57.611_. + 43.28,. V/m 5

Fia-1-15

2.11. As shown in Fig. 2-16, two uniform line charges of density x = 0 plane at y = ±4 m. Find E at (4, 0, 10) m.

p~ =

4 nC/m lie in the

y P~

(0,4,z)

(0, -4,z)

Fig. 1-16

The line charges are both parallel to •.; their fields are radial and parallel to the xy plane. either line charge the magnitude of the field at P would be E

For

= .J!.!_ =]! V /m 2.7r£or

The field due to both line charges is, by superposition,

E=2(~cos45°)a.. = 18a.. V/m

z.U. •

Develop an expression for E due to charge uniformly distributed over an infinite plane with density Ps·

COULOMB FORCES AND ELECTRIC FIELD INTENSITY 1be c:ylindrical coordinate system will be used, with the charge in the z Fig. 2-17.

[CHAP. 2 plane as shown in

P(O••• z)

Fig. Z..l7 Symmetry about the z axis results in cancellation of the radial components.

1" [ 0

p.rz dr drp

41&'£o(r + Z 2 ) 312 ••

'Ibis result is for points above the xy plane. Below the xy plane the unit vector changes to generalized form may be written using a, , the unit normal vector: E

-a..

The

=-· p.

2£o "

1be electric field is everywhere normal to the plane of the charge and its magnitude is independent of the distance from the plane.

2.13. As shown in Fig. 2-18, the plane y = 3 m contains a uniform charge distribution of density Ps = (10-8/6n) C/m2 • Determine Eat all points.

-:%~~.1,\';!S•~~~~:::,.r) Fig. Z..l8 For y >3m, E=.E!..a, 2£o

=308,. V/m andfor y<3m, E= -30ay V/m

CHAP. 2]

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

2.14. Two infinite uniform sheets of charge, each with density Ps , are located at x = ± 1 (Fig. 2-19). Determine E in all regions.

Fig. 2-19 Only parts of the two sheets of charge are shown in Fig. 2-19. are directed along x, independent of the distance. Then -(p8/£o)a., E,+~== 0 { (p8/£o)a.,

2.15. Repeat Problem 2.14 with Ps on x = -1

Both sheets result in E fields that

x<-1 -1<x<l x>1

and - Ps on x = 1.

0 E1 + E2 = (ps/€o)ax { 0

x<-1 -1 <x < 1 x>1

2.16. A uniform sheet charge with Ps = (1/3n) nC/m2 is located at z = 5 m and a uniform line charge with p, = (- 25/9) nC/m at z = -3 m, y = 3 m. Find E at (x, -1, 0) m. The two charge configurations are parallel to the x axis. Hence the view in Fig. 2-20 is taken looking at the yz plane from positive x. Due to the sheet charge, P2 E.=2-a, £o z

Fig. 2-20 At P,

a, = -az and

E8=-6a. V/m Due to the line charge,

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

[CHAP. 2

and at P

= 88,. - 6a. V /m E = E~ +E. =88,. - 12a. V /m. E~

The total electric field is the sum,

2.17. Determine E at (2, 0, 2) m due to three standard charge distributions as follows: a uniform sheet at x = 0 m with Psi = (1/3n) nC/m2 , a uniform sheet at x = 4 m with p,2 = ( -1/3n) nC/m 2 , and a uniform line at x = 6 m, y = 0 m with p, = -2 nC/m. Since the three charge configurations are parallel with a. , there will be no z component of the field. Point (2, 0, 2) will have the same field as any point (2, 0, z). In Fig. 2-21, Pis located between the two sheet charges, where the fields add due to the difference in sign.

p,, P.2 P~ E =2-a, +2- 8 " +2--a, £o

£o

1f£oT

= 6a~ + 6a~ + 9a~ =21a_. V/m PrJ

Ps2

--.- ....... E

x=O

x=4

Fig.l-21

2.18. As shown in Fig. 2-22, charge is distributed along the z axis between z = ±5 m with a . uniform density p, = 20 nC/m. Determine E at (2, 0, 0) m in cartesian coordinates. Also express the answer in cylindrical coordinates. 9

dE 20 X 10- dz - 4H(10- 9 /36H)(4 + z2 )

(2aV4+? . - za.) (VI m)

dQ= p1 dz

(2, O , O ) i t - - - - - J ' X

-S

Symmetry with respect to the

z= 0

plane eliminates any z component in the result. 5

2dz

E = 180 -s ( 4 + z 2) 312 a~ = 167•~ V/m In cylindrical coordinates, E = 167a, V /m.

CHAP. 2]

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

2.19. Charge is distributed along the z axis from z = 5 m to oo and from z = -5 m to -oo (see Fig. 2-23) with the same density as in Problem 2.18, 20 nC/m. Find Eat (2, 0, 0) m. E d

20 x 10- dz = 4n(10- 9 /36n)(4+z 2 )

{2a. - za,) v'4+zz

I (V m)

-5

t-eo Fig.l-13

Again the z component vanishes.

I-s

2dz 2dz ] E = 180 [[ s (4 + z2)312 + -~ (4 + z~J/2 a_. = 13a_. V/m

In cylindrical coordinates, E = 13a, V/m. When the charge configurations of Problems 2.18 and 2.19 are superimposed, the result is a uniform line charge.

p, E=--a,= 180a, V/m 21fÂŁ0 r

2.20. Find the electric field intensity E at (0, q,, h) in cylindrical coordinates due to the uniformly charged disk r sa, z = 0 (see Fig. 2-24).

Fig.l-24 If the constant charge density is p.,

dE=

p,rdrdt/) {-ra,+ha,) 4HÂŁo(r2 + hz) v'r2 + hz

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

(CHAP. 2

The radial components cancel. Therefore

l2A l"o (rr2dr+ hdq,2)312 a• p,h( -1 1) = 2£o Va2+h2 +h a.

E -p.h -- 4Jr£o o

Note that as a--.rx',

E-+(p,/2£0 )a.,

the field due to a uniform plane sheet.

1.21. Charge lies on the circular disk r ::sa, at (0, tJ>, h) .

•

with density Ps = p 0 sin2 t/J.

Determine E

dE= Po(sin .P )r dr dq, { -ra, + ha,)

4Jr£o(r2 + h2)

Vr2 + h2

The charge distribution, though not uniform, still is symmetrical such that all radial components cancel.

l2nl"(sin 2.P)rdrd.p =p h{ h2)312 •• •

E = Poh 4Jr£o o

0 -1 rr::-;-;2 + 4£o va+ h h ••

(r +

r ::s 4 m, z = 0 with density Ps = (10. . . 4 /r) (C/m2 ).

2.2Z. Charge lies on the circular disk • Determine E at r = 0, z = 3m. 4

dE= (10- /r)r dr d.P {-ra. + 3a.) 4Jr£0 (r2 + 9) y,.z + 9

(V /m)

As in Problems 2.20 and 2.21 the radial component vanishes by symmetry.

"'i L 2

E = (2.7 X Hf')

( 2drd.p312 a.= 1.51 r +9)

Hf'a. V /m

1.51a. MV/m

2.23. Charge lies in the z = -3 m plane in the form of a square sheet defined by -2 ::s x ::s 2m, -2 ::s y ::s 2m with charge density p.. = 2(x 2 + y 2 + 9)312 nC/m2 • Find E at the origin. From Fig. 2-25, R= -xa.- yay+ 3a. dQ

= p, dx dy =2(x + y 2

+ 9)

312

(m) 10- 9 dx dy

z dE

(-2,-2,-3)

....Ic------y

>Jf-,.lll,..;......,.,.....,-.-,(-2, 2, -3)

(2, -2, -3)

(2, 2, -3)

Jix. z-zs and so

(C)

CHAP. 2)

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

Due to symmetry, only the z component of E exists. 2 2 9 6 x w- cU- dy E= a. = 864a, V/m 41f€o -2 -2

2.24. A charge of uniform density Ps = 0.3 nC/m2 covers the plane 2x - 3y + z = 6 m. on the side of the plane containing the origin. Since this charge configuration is a uniform sheet, unit normal vectors for a plane Ax + By + Cz = D are •

= Ps/2€0

and

Find E

E = (17.0)a, V/m.

The

Aa_. + Ba, + Ca, ± ~=:::;;::==::=.=::::::::~ VA2+B2+C2

Therefore, the unit normal vectors for this plane are

2a_. -

3a, +

... = ± --"----<.---"

Vt4

From Fig. 2-26 it is evident that the unit vector on the side of the plane containing the origin is produced by the negative sign. The electric field at the origin is

Supplementary Problems 2.25.

Two point charges, respectively.

2.26.

Find the force on Q2

Two point charges, respectively.

Q, = 250 pC and

Q 2 = -300 pC,

Ans.

are located at (5, 0, 0) m and (0, 0, -5) m,

8_. + 8z) N = (13.5) ( '1/2

Q, = 30pC and Q 2 = -100pC, are located at (2,0, 5) m and (-1, 0, -2) m,

Find the force on Q,.

Ans.

F, = (0.465)( -

~ ?az) N

2.'1.7.

In Problem 2.26 find the force on Q2 •

2.28.

Four point charges, each 20 pC, are on the x and y axes at ±4 m. charge at (0, 0, 3) m. Ans. 1. 73az N

Ans.

-F, Find the force on a 100-pC point

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

(CHAP. 2

2.19.

Ten identical charges of 500 pC each are spaced equally around a circle of radius 2 m. Find the force on a charge of -20 pC located on the axis, 2m from the plane of the circle. Ans. (79.5)( -a.,) N

2.30.

Determine the force on a point charge of 50 pC at (0, 0, 5) m due to a point charge of 500n pC at the origin. Compare the answer with Problems 2.4 and 2.5, where this same total charge is distributed over a circular disk. Ans. 28.3a. N

2.31.

Find the force on a point charge of 30 pC at (0, 0, 5) m due to a 4 m square in the z = 0 plane between x =±2m and y =±2m with a total charge of 500 pC, distributed uniformly. Ans. 4.66a. N

2.32.

Two identical point charges of Q(C) each are separated by a distance d(m). Express the electric field E for points along the line joining the two charges. Ans. If the charges are at x = 0 and x = d, then, for 0 < x < d, E=

4;EJ~- (d ~x)2

]•.

(V/m)

2.33.

Identical charges of Q(C) are located at the eight corners of a cube with a side f (m). coulomb force on each charge has magnitude (3.29Q 2 /4nE 0 f 2 ) N.

2.34.

Show that the electric field E outside a spherical shell of uniform charge density p. is the same as E due to the total charge on the shell located at the center.

2.35.

Develop the expression in cartesian coordinates for E due to an infinitely long, straight charge E = p, xa. +yay Ans. configuration of uniform density P~2 2 2HEo x

Show that the

= 0,

y = ±4 m.

z = 0,

y = -2m

2.36.

Two uniform line charges of p~ = 4 nC/m each are parallel to the z axis at x Ans. ± 18a, V /m Determine the electric field E at ( ±4, 0, z) m.

2.37.

Two uniform line charges of p, = 5 nC/m each are parallel to the x axis, one at andtheotherat z=O, y=4m. FindEat(4,1,3)m. Ans. 30a.V/m

2.38.

Determine E at the origin due to a uniform line charge distribution with at x=3m, y=4m. Ans. -7.13a.-9.50ay V/m

2.39.

Referring to Problem 2.38, at what other points will the value of E be the same?

2.40.

Two meters from the z axis, lEI due to a uniform line charge along the z axis is known to be 1.80 X UJ' V /m. Find the uniform charge density p,. Ans. 2.0 pC/m

2.41.

The plane

-x + 3y - 6z = 6 m contains a uniform charge distribution · Ans. 30(•· - :~ ,. + 68•) V/m the sid e containing the ongin. V'Kl

p, = 3.30 nC/m

Ans.

p. = 0.53 nC/m2 •

located

(0, 0, z)

Find E on

2.42.

Two infinite sheets of uniform charge density p. = (10- 9 /6n) C/m2 are located at z = -5 m and y = -5 m. Determine the uniform line charge density p, necessary to produce the same value of Eat (4, 2, 2) m, if the line charge is located at z = 0, y = 0. Ans. 0.667 nC/m

2.43.

Two uniform charge distributions are as follows: a sheet of uniform charge density p, = -50 nC/m2 at y =2m and a uniform line of p~ = 0.2 pC/m at z =2m, y =-1m. At what Ans. (x, -2.273, 2.0) m points in the region will E be zero?

2.44.

A uniform sheet of charge with p. = ( -l/3n) nC/m2 is located at z = 5 m and a uniform line of charge with p, = (- 25/9) nC/m is located at z = -3m, y =3m. Find the electric field E at (0,-l,O)m. Ans. Bay V/m

CHAP. 2)

COULOMB FORCES AND ELECTRIC FIELD INTENSITY

2.45.

A uniform line charge of p, = (Vl x 10-8 /6) C/m lies along the x axis and a uniform sheet of charge is located at y = 5 m. Along the line y = 3 m, z = 3 m the electric field E has only a z component. What is p. for the sheet? Ans. 125 pC/m 2

2.46.

A uniform line charge of p, = 3.30 nC/m is located at x =3m, y = 4 m. A point charge Q is 2m from the origin. Find the charge Q and its location such that the electric field is zero at the origin. Ans. 5.28 nC at ( -1.2, -1.6, 0) m

2.47.

A circular ring of charge with radius 2 m lies in the z = 0 plane, with center at the origin. If the uniform charge density is p, = 10 nC/m, find the point charge Q at the origin which would produce the same electric field E at (0, 0, 5) m. Ans. 100.5 nC

2.48.

The circular disk r :S 2 m in the

z = 0 plane has a charge density p 3 = 10-8 /r 3

Determine the electric field E for the point (0,

q,, h).

Ans.

1.13Xl0 ( I ) â&#x20AC;˘ ~a,Vm hv.. -t-n-

2.49.

Examine the result in Problem 2.48 as h becomes much greater than 2 m and compare it to the field at h which results when the total charge on the disk is concentrated at the origin.

2.50.

A finite sheet of charge, of density p. = 2x(x 2 + y 2 + 4) 312 (C/m2 ), lies in the 0 s x s 2m and 0 s y :S 2m. Determine E at (0, 0, 2) m.

Ans.

(18 x 10

)(-

a.. -4ay

z =0

plane for

+Sa.) V/m = 1s(- ~ a, -4ay +Sa.) GV/m

2.51.

Determine the electric field E at (8, 0, 0) m due to a charge of 10 nC distributed uniformly along the x axis between x = -5 m and x = 5 m. Repeat for the same total charge distributed between x =-1m and x =1m. Ans. 2.31a.., V/m, 1.43a.. V/m

2.52.

The circular disk r :S 1m, z = 0 has a charge density p. = 2(r2 + 25) 312e-â&#x20AC;˘o. (C/m2 ). (0, 0, 5) m. Ans. 5.66a.., GV /m

2.53.

Show that the electric field is zero everywhere inside a uniformly charged spherical shell.

2.54.

Charge is distributed with constant density p throughout a spherical volume of radius a. results of Problems 2.34 and 2.53, show that r:Sa r~a

where r is the distance from the center of the sphere.

Find E at

By using the

Chapter 3 Electric Flux and Gauss' Law 3.1 NET CHARGE IN A REGION With charge density defined as in Section 2.3, it is possible to obtain the net charge contained in a specified volume by integration. From dQ=pdv

(C)

it follows that Q= Lpdv (C)

In general, p will not be constant throughout the volume v. EXAMPLE 1. Find the charge in the volume defined by

5 cos

p=--

1 :S r :S 2 m in spherical coordinates, if (C/m3)

By integration,

3.1 ELECTRIC FLUX AND FLUX DENSITY

_.. 1

â&#x20AC;˘

Electric flux 'II, a scalar field, and its density D, a vector field, are useful quantities in solving certain problems, as will be seen in this and subsequent chapters. Unlike E, these fields are not directly measurable; their existence was inferred from nineteenth-century experiments in electrostatics. EXAMPLE 2. Referring to Fig. 3-1, a charge +Q is first fixed in place and a spherical, concentric, conducting shell is then closed around it. Initially the shell has no net charge on its surface. Now if a conducting path to ground is momentarily completed by closing a switch, a charge -Q, equal in magnitude but of opposite sign, is discovered on the shell. This charge -Q might be accounted for by a transient ftow of negative charge from the ground, through the switch, and onto the shell. But what could provoke such a ftow? The early experimenters suggested that a flux from the +Q to the conductor surface induced, or displaced, the charge -Q onto the surface. Consequently, it has also been called displacement flux, and the use of the symbol D is a reminder of this early concept.

By definition, electrix flux 'I' originates on positive charge and terminates on negative charge. In the absence of negative charge, the flux 'I' terminates at infinity. Also by definition, one coulomb of electric charge gives rise to one coulomb of electric flux. Hence 'I'= Q (C)

In Fig. 3-2(a) the flux lines leave +Q and terminate on -Q. This assumes that the two charges are of equal magnitude. The case of positive charge with no negative charge in the region is illustrated in Fig. 3-2(b ). Here the flux lines are equally spaced throughout the solid angle, and reach out toward infinity. 32

ELECfRIC FLUX AND GAUSS' LAW

CHAP. 3)

Fig.3-l

(a)

(b)

+Q.

;Âˇ\

-....:. -Q

Fig.3-2 If in the neighborhood of point P the lines of flux have the direction of the unit vector a (see Fig. 3-3) and if an amount of flux d'll crosses the differential area dS, which is a normal to a, then the electric flux density at P is

Fig.3-3

A volume charge distribution of density p (C/m3 ) is shown enclosed by surface S in Fig. 3-4. Since each coulomb of charge Q has, by definition, one coulomb of flux 'I', it follows that the net flux crossing the closed surfaceS is an exact measure of the net charge enclosed. However, the

ELECTRIC FLUX AND GAUSS' LAW

[CHAP. 3

density D may vary in magnitude and direction from point to point of S; in general, D will not be along the normal to S. If, at the surface element dS, D makes an angle 8 with the normal, then the differential flux crossing dS is given by d'P = D dS cos 8 = D · dSan = D • dS

where dS is the vector surface element, of magnitude dS and direction an. The unit vector an is always taken to point out of S, so that d'P is the amount of flux passing from the interior of S to the exterior of S through dS.

Fig. 3-4

3.3 GAUSS' LAW Gauss' law states that The total flux out of a closed surface is equal to the net charge within the surface. This can be written in integral form as

A great deal of valuable information can be obtained from Gauss' law through clever choice of the surface of integration; see Section 3.5.

3.4 RELATION BETWEEN FLUX DENSITY AND ELECTRIC FIELD INTENSITY Consider a point charge Q (assumed positive, for simplicity) at the origin (Fig. 3-5). If this is enclosed by a spherical surface of radius r, then, by symmetry, D due to Q is of constant magnitude over the surface and is everywhere normal to the surface. Gauss' law then gives Q=

D • dS = D

dS = D ( 4.rrr

Fig. 3-5

)

from which D = Q/4~rr 2•

ELECTRIC FLUX AND GAUSS' LAW

CHAP. 3)

Therefore

But, from Section 2.2, the electric field intensity due to Q is

It follows that D = £oE. More generally, for any electric field in an isotropic medium of permittivity

£,

D=£E Thus, D and E fields will have exactly the same form, since they differ only by a factor which is a constant of the medium. While the electric field E due to a charge configuration is a function of the permittivity £, the electric fiux density Dis not. In problems involving multiple dielectrics a distinct advantage will be found in first obtaining D, then converting toE within each dielectric.

3.5 SPECIAL GAUSSIAN SURFACES The surface over which Gauss' law is applied must be closed. but it can be made up of several surface elements. If these surface elements can be selected so that D is either normal or tangential, and if IDI is constant over any element to which D is normal, then the integration becomes very simple. Thus the defining conditions of a special gaussian surface are 1.

The surface is closed.

2. 3.

At each point of the surfaceD is either normal or tangential to the surface. D is sectionally constant over that part of the surface where D is normal.

EXAMPLE 3 Use a special gaussian surface to find D due to a uniform line change p, (C/m). Take the line charge as the z axis of cylindrical coordinates (Fig. 3-6). By cylindrical symmetry, D can

-oo

Fig.U

ELECTRIC FLUX AND GAUSS' LAW

[CHAP. 3

only have an r component, and this component can only depend on r. Thus, the special gaussian surface for this problem is a closed right circular cylinder whose axis is the z axis (Fig. 3-7). Applying Gauss' law, Q

D . dS +

dS +

LD. dS

D dS 00

Fig. 3-7 Over surfaces 1 and 3, D and dS are orthogonal, and so the integrals vanish. Over 2, D and dS are parallel (or antiparallel, if p~ is negative), and D is constant because r is constant. Thus, Q

where Lis the length of the cylinder.

= D(2nrL)

But the enclosed charge is D =1!.!_

p~L.

Hence,

and

2nr

Observe the simplicity of the above derivation as compared to Problem 2.9.

The one serious limitation of the method of special gaussian surfaces is that it can be utilized only for highly symmetrical charge configurations. However, for other configurations, the method can still provide quick approximations to the field at locations very close to or very far from the charges. See Problem 3.36.

Solved Problems 3.1.

Find the charge in the volume defined by 0::::; x ::::; 1 m, 0 ::::; y ::::; 1 m, and 0::::; z ::::; 1 m if p = 30x 2y (pC/m3 ). What change occurs for the limits -1 ::=;y :::;Om? Since

p dv,

= (' Jo

f' i' o

30x 2y dx dy dz

= 5 J.tC

For the change in limits on y,

i'J" i' (J

-1

30x 2y dx dy dz = -5 J.tC

CHAP. 3)

3.2.

â&#x20AC;˘ _

3.3.

Ia _

ELECTRIC FLUX AND GAUSS' LAW

Three point charges, Q 1 = 30 nC, surface S. What net flux crosses S?

Q 2 = 150 nC,

and

Q3 = -70 nC, are enclosed by

Since electric flux was defined as originating on positive charge and terminating on negative charge, part of the flux from the positive charges terminates on the negative charge. 'l'ne, = Qneâ&#x20AC;˘ = 30 + 150- 70 = 110 nC

What net flux crosses the closed surface S shown in Fig. 3-8, which contains a charge distribution in the form of a plane disk of radius 4 m with a density Ps = (sin 2 t/J )/2r (C/m2 )?

11 4

'I'= Q =

(sin 4') rdrd4' =2n-C ~

2Jr

Fig.3-8

3.4.

A circular disk of radius 4 m with a charge density Ps = 12 sin tJ> p.C/m2 is enclosed by surface S. What net flux crosses S? 'I'= Q =

L2n

(12 sin t/')rdr dt/' = 0 llC

Since the disk contains equal amounts of positive and negative charge (sin (4' + .n) =-sin 4'), flux crosses S.

3.5.

no net

Charge in the form of a plane sheet with density Ps = 40 p.C/m2 is located at z = -0.5 m. A uniform line charge of p, = -6 p.C/m lies along the y axis. What net flux crosses the surface of a cube 2m on an edge, centered at the origin, as shown in Fig. 3-9? 'I'= Qen<: z

Fig.3-9

ELECfRIC FLUX AND GAUSS' LAW

(CHAP. 3

The charge enclosed from the plane is Q

= (4 m2 )(40 tJC/m 2 ) = 160 tJC

= (2 m )( -6 tJC/m) = -12 tJC

and from the line

Thus,

3.6.

Qenc = 'P = 160- 12 = 1481JC.

A point charge Q is at the origin of a spherical coordinate system. Find the flux which crosses the portion of a spherical shell described by a::; fJ::; f3 (Fig. 3-10). What is the result if a = 0 and f3 = 1r /2? z

Fig. 3-10

The total flux given by

'P = Q

crosses a complete spherical shell of area 4.rrr 2 •

= 2.n-r 2 ( cos

The area of the strip is

a - cos fJ)

Then the flux through the strip is 'Pner

For

3.7.

a= 0,

= - -2 Q =-(cos a- cosfJ) 4nr 2

fJ = .n-/2 (a hemisphere), this becomes 'Pne• = Q/2.

A uniform line charge with p£= 50 f.J.Cim lies along the x axis. What flux per unit length, 'I' I L, crosses the portion of the z = -3m plane bounded by y = ±2m? The flux is uniformly distributed around the line charge. Thus the amount crossing the strip is obtained from the angle subtended compared to 2.rr. In Fig. 3-11, a= 2 arctan

Then

(j) = 1.176 rad

'P (1.176) -=50 - - = 9.36tJC/m L 2.rr

CHAP. 3)

ELECTRIC FLUX AND GAUSS' LAW

Fig. 3-11

3.8.

A point charge, Q = 30 nC, is located at the origin in cartesian coordinates. electric flux density D at (1, 3, -4) m.

Find the

Referring to Fig. 3-12,

Q D= 4nRz•R

=30 x 10-

9 ( •..

4n(26)

+ 3ay - 4a.) v26

= (9.18 x 10- 11) ( •.. + 38Y -

48 •) C/m2

v26 or, more conveniently, D

= 91.8pC/m2 • z

(1,3,-4)

\n Fig. 3-12

3.9.

Two identical uniform line charges lie along the x and y axes with charge densities p, = 20 pC/m. Obtain D at (3, 3, 3) m. The distance from the observation point to either line charge is charge on the x axis, p~

(lly +_•·)

D ·=2-a..

20 p.C/m 2n(3V2 m)

Dz = 2P~ 8rz nrz

20 p.C/m (•.. +_•·) 2n(3V2m) V2

nr1

v'2

and now the y axis line charge,

3v2 m.

Considering first the line

ELECTRIC FLUX AND GAUSS' LAW

(CHAP. 3

The total flux density is the vector sum, D

20 (•.. + ay + 2az) = (2.25)(11.. +a,. + 2az) 1JC/m2 2.n-(3v'2) v'2 v'6

3.10. Given that D = lOxax (C/m 2), determine the flux crossing a 1-m2 area that is normal to the x axis at x =3m. Since D is constant over the area and perpendicular to it, 'I' = DA

= (30 C/m2)(1 m2) = 30 C

3.11. Determine the flux crossing a 1 mm by 1 mm area on the surface of a cylindrical shell ifl. at r =10m, z =2m, tJ> = 53.2° if

.!II!!

D = 2xax + 2(1- y)ay + 4z&z

(C/m 2 )

At point P (see Fig. 3-13),

10cos53.2°= 6

y = 10 sin 53.2° = 8 z

Fig. 3-13 Then, at P, D

= 12az -14a,. +Sa,

C/m2

Now, on a cylinder of radius lOrn, a 1-mm2 patch is essentially planar, with directed area dS = 10- 6 (0.6a..

Then

+ O.Sa,.) m2

d'P = D • dS = (12a.. -14ay + 8a,) • 10- 6 (0.6a.. + O.Sa,.) = -4.0 1JC

The negative sign indicates that flux crosses this differential surface in a direction toward the z axis rather than outward in the direction of dS.

3.U. A uniform line charge of Pt = 3pc;/m lies along the z axis, and a concentric circular cylinder of radius 2m has p.. = ( -1.5/4.n") pC/m2 • Both distributions are infinite in extent with z. Use Gauss' law to find D in all regions. Using the special gaussian surface A in Fig. 3-14 and processing as in Example 3, p,

D =-a,.

2nr

O<r<2

CHAP. 3]

ELECfRIC FLUX AND GAUSS' LAW

Fig. 3-14

Using the special gaussian surface B, Qenc=fDâ&#x20AC;˘dS (p, + 4:rrp,)L = D(2:rrrL)

from which D=

p,+4:rrp. a 2:rrr '

r>2

For the numerical data, 0.477 - a , (J1.CI m2)

D-

O<r<2m

r {

0.~39 a,

(p.C/mz)

r>2m

3.13. Use Gauss' law to show that D and E are zero at all points in the plane of a uniformly charged circular ring that are inside the ring. Consider, instead of one ring, the charge configuration shown in Fig. 3-15, where the uniformly charged cylinder is infinite in extent, made up of many rings. For gaussian surface 1, Qcnc= 0= Dfds

_j_dz

T X

Fig. 3-15

ELECTRIC FLUX AND GAUSS' LAW

(CHAP. 3

Hence D,.. 0 for r < R. Since 'I' is completely in the radial direction, a slice dz can be taken from the cylinder of charge and the result found above will still apply to this ring. For all points within the ring, in the plane of the ring, D and E are zero.

3.14. A charge configuration in cylindrical coordinates is given by p = 5re-2r (C/m3). Use Gauss' law to find D. Since p is not a function of ,P or z, the flux 'I' is completely radial. It is also true that, for r constant, the flux density D must be of constant magnitude. Then a proper special gaussian surface is a closed right circular cylinder. The integrals over the plane ends vanish, so that Gauss' law becomes Qenc: =

lateral

D • dS

LL L,.,. f 5re-2rrdrdt/'dz = D(21UL) aurfacc

5nL[e-2r( -,Z- r -l> + l] = D(2nrL) Hence

25 D = · (!- e--zr(r2 + r + nJa,

(C/m2 )

3.8. The volume in cylindrical coordinates between r = 2 m and r = 4 m contains a uniform charge density p (C/m3). Use Gauss' law to find Din all regions. From Fig. 3-16, for O<r<2m,

Q•..., = D(2ML) D=O

..,._ 3-16 For 2srs4m, npL(,Z- 4) = D(2ML) p

D = 2r (,Z- 4)a,

(C/m2 )

For r>4m, 12npL = D(2ML)

6p D =-a, r

(C/m2 )

ELECfRIC FLUX AND GAUSS' LAW

CHAP. 3)

3.16. The volume in spherical coordinates described by r ~a contains a uniform charge density p. Use Gauss' law to determine D anj compare your results with those for the corresponding E field, found in Problem 2.5<::,. What point charge at the origin will result in the same D field for r >a? For a gaussian surface such as l: in Fig. 3-17,

D=pra, 3

and

r~a

r=a

Fig. 3-17

For points outside the charge distribution,

r>a

whence

If a point charge Q = ~n-a 3p is placed at the origin, the D field for same. This point charge is the same as the total charge contained in the volume.

r >a will be the

3.17. A parallel-plate capacitor has a surface charge on the lower side of the upper plate of +Ps (C/m2 ). The upper surface of the lower plate contains -ps (C/m 2 ). Neglect fringing and use Gauss' law to find D and E in the region between the plates. • e«< All flux leaving the positive charge on the upper plate terminates on the equal negative charge on the lower plate. The statement neglect fringing insures that all flux is normal to the plates. For the special gaussian surface shown in Fig. 3-18, Qenc =

lop

=0+

D • dS +

bouom

I..

D • dS

&~de

Lom D·dS+O p,A = D

or where A is the area.

D • dS +

JdS =DA

Consequently, and

E=P·a.. (V/m) Eo

Both are directed from the positive to the negative plate.

ELECTRIC FLUX AND GAUSS' LAW

(CHAP. 3

Fig. 3-18

Supplementary Problems 3.18.

Find the net charge enclosed in a cube 2m on an edge, parallel to the axes and centered at the origin, if the charge density is

3.19.

Find the charge enclosed in the volume 1 s r s 3m, 0 s tp s :rr/3, density p = 2z sin 2 4' (C/m3 ). Ans. 4.91 C

3.20.

Given a charge density in spherical coordinates p = ___!!!2__ e _,,, cos2

(r/ro)z

0 s z s 2m given the charge

find the amounts of charge in the spherical volumes enclosed by Ans. 3.fJ7poro. 6.24poro. 6.28poro 3.21.

A closed surface S contains a finite line charge distribution, 0 s â&#x201A;Ź s :rr m,

What net flux crosses the surface S?

3.22.

p,=

-posin~

Ans.

-2po

r = r0 ,

r = 5r0 ,

and r = oo.

with charge density

(C/m) (C)

Charge is distributed in the spherical region r s 2m with density -200

p = --

What net flux crosses the surfaces r =1m, Ans. -~:rr~C.-l~:rr~C.-l~:rr~C

(~C/m )

r = 4 m,

and r = 500 m?

3.23.

A point charge Q is at the origin of spherical coordinates and a spherical shell charge distribution at r =a has a total charge of Q'- Q, uniformly distributed. What flux crosses the surfaces r = k for k <a and k >a? Ans. Q, Q'

3.24.

A uniform line charge with p, = 31JC/m lies along the x axis. centered at the origin with r = 3 m? Ans. l81JC

3.25.

If a point charge Q is at the origin, find an expression for the flux which crosses the portion of a sphere, (3-a centered at the origin, described by as 4' s f3. Ans. ~ Q

What flux crosses a spherical surface

ELECTRIC FLUX AND GAUSS' LAW

CHAP. 3)

3.26.

A point charge of Q (C) is at the center of a spherical coordinate system. Find the flux 'P which crosses Ans. Q/9 (C) an area of 4.7r m2 on a concentric spherical shell of radius 3m.

3.27.

An area of 40.2 m2 on the surface of a spherical shell of radius 4 m is crossed by 10 pC of flux in an inward direction. What point charge at the origin is indicated? Ans. -50 pC

3.28.

A uniform line charge p~ lies along the x axis. What percent of the flux from the line crosses the strip of the y = 6 plane having -1 s z s 1? Ans. 5.26%

3.19.

A point charge, Q = 3 nC, is located at the origin of a cartesian coordinate system. What flux 'P crosses the portion of the z =2m plane for which -4 s x s 4 m and -4 s y s 4 m? Ans. 0.5nC

3.30.

A uniform line charge with

Ans. 3.31.

(0.356)(2aVi az)

p~=

pC/m

5 fJC/m

lies along the x axis.

Find D at (3, 2, 1) m.

A point charge of +Q is at the origin of a spherical coordinate system, surrounded by a concentric uniform distribution of charge on a spherical shell at r =a for which the total charge is -Q. Find the flux 'P crossing spherical surfaces at r < a and r > a. Obtain D in all regions.

Ans.

'P = 4.7rr 2D =

+Q {0

r<a r>a

3.31.

Given that D = 500e-o_txa. (1JC/m2), find the flux 'P crossing surfaces of area 1m2 normal to the x axis and located at x = 1 m, x = 5 m, and x = 10m. Ans. 452 fJC, 303 fJC, 184 pC

3.33.

Given that D = 5x 2a. + lOzaz (C/m2 ), find the net outward flux crossing the surface of a cube 2m on an edge centered at the origin. The edges of the cube are parallel to the axes. Ans. 80 C

3.34.

Given that

in cylindrical coordinates, find the outward flux crossing the right circular cylinder described Ans. 129b2 (C) by r = 2b, z = 0, and z = 5b (m).

3.35.

Given that sin cp D= 2rcos cpa~ -~az in cylindrical coordinates, find the flux crossing the portion of the z = 0 plane defined by r sa, 0 s cp s .1r /2. Repeat for 3Jr /2 s cp s 2Jr. Assume flux positive in the a, direction.

Ans. 3.36.

a a 3'3

In cyclindrical coordinates, the disk r sa, z = 0 carries charge with nonuniform density p,(r, cp). Use appropriate special gaussian surfaces to find approximate values of D on the z axis (a) very close to the disk (0 < z ~a), (b) very far from the disk (z :!!>a). p (0 cp) Q Ans. (a) â&#x20AC;˘ ; ; (b) .7rz 2 where Q= Jo p.(r, cp)rdrdcp 4 0

(2"[

3.37.

A point charge, Q = 2000 pC, is at the ongtn of spherical coordinates. A concentric spherical distribution of charge at r =1m has a charge density p. = 40Jr pC/m 2 â&#x20AC;˘ What surface charge 'Ca. density on a concentric shell at r = 2m would result in D = 0 for r > 2 m? ~ Ans. -71.2 pC/m2 ,..1+

ELECfRIC FLUX AND GAUSS' LAW

=5r

(C/m3 )

[CHAP. 3

3.38.

Given a charge distribution with density p find D. Ans. (5r2/4)a,. (C/m2)

in spherical coordinates, use Gauss' law to

3.39.

A uniform charge density of 2C/m3 exists in the volume 2sx s4m (cartesian coordinates). Gauss' law to find Din all regions. Ans. -2a.. C/m 2 , 2(x- 3)a.. (C/m2 ), 2a.. C/m 2

3.40.

Use Gauss' law to find D and E in the region between the concentric conductors of a cylindrical capacitor. The inner cylinder is of radius a. Neglect fringing. Ans. p."(a/r), p.. (a/ÂŁ 0 r)

3.41.

A conductor of substantial thickness has a surface charge of density p.. Assuming that 'P = 0 within the conductor, show that D = Âąp. just outside the conductor, by constructing a small special gaussian surface.

Use

Chapter 4 Divergence and the Divergence Theorem 4.1 DIVERGENCE There are two main indicators of the manner in which a vector field changes from point to point throughout space. The first of these is divergence, which will be examined here. It is a scalar and bears a similarity to the derivative of a function. The second is curl, a vector which will be examined when magnetic fields are discussed in Chapter 9. When the divergence of a vector field is nonzero, that region is said to contain sources or sinks, sources when the divergence is positive, sinks when negative. In static electric fields there is a correspondence between positive divergence, sources, and positive electric charge Q. Electric flux 'I' by definition originates on positive charge. Thus, a region which contains positive charges contains the sources of '1'. The divergence of the electric flux density D will be positive in this region. A similar correspondence exists between negative divergence, sinks, and negative electric charge. Divergence of the vector field A at the point P is defined by

fAÂˇdS divA= l i m - - A...-.o

Here the integration is over the surface of an infinitesimal volume

that shrinks to point P.

4.2 DIVERGENCE IN CARTESIAN COORDINATES The divergence can be expressed for any vector field in any coordinate system. For the development in cartesian coordinates a cube is selected with edges~. ~y. and ~z parallel to the x, y, and z axes, as shown in Fig. 4-1. Then the vector field A is defined at P, the corner of the cube with the lowest values of the coordinates x, y, and z. A= A_..a... +Ayay + Azaz z

Fig.4-l In order to express pA â&#x20AC;˘ dS for the cube, all six faces must be covered. On each face, the direction of dS is outward. Since the faces are normal to the three axes, only one component of A will cross any two parallel faces.

DIVERGENCE AND THE DIVERGENCE THEOREM

[CHAP. 4

In Fig. 4-2 the cube is turned such that face 1 is in full view; the x components of A over the faces to the left and right of 1 are indicated. Since the faces are small,

A•dS=-A.. (x)~y~z

{ Jleft face

A • dS =A.. (x

+ ~) ~y !u

right face

[ A .. (x)+

aA.. ] ax~

~y ~z

t.x

so that the total for these two faces is

aA .. ax ~~y~z The same procedure is applied to the remaining two pairs of faces and the results combined.

f Dividing by

~ ~y ~z = ~v

(- + - + aA.. ax

A·dS=

and letting . A dIV

aA.., ax

aAy ay

aAz) az

~v-+0,

aAy ay

~~y~z

one obtains

aAz az

=-+-+-

(cartesian)

The same approach may be used in cylindrical (Problem 4.1) and in spherical coordinates. t a divA=--(rA

r ar

. drv A

1 aA., r aq,

aAZ az

)+--+-

(cylindrical)

1 a . 1 aA., =r21 -aar (r2A,) +-------:--a (A sm B ) + - . - r sm 8 (J 6 r sm (J aq,

EXAMPLE 1. Given the vector field

•

.7[%)

A= 5x 2 ( sin 2(

';)a..

.7[%) :tr

(spherical)

find divA at x = 1. •

.7[% s

.7[%

dtvA=- 5x sm- =5x cos- -+10xsm-=-.7tX cos-+lOxsinax 2 22 22 2 2 and div Al.-- 1 = 10.

EXAMPLE 2. In cylindrical coordinates a vector field is given by A = r sin cPa, + r 2 cos cPa., + 2re - 5·a.. divA at {!, tr/2, 0). div A=!~(~sin cP)+!~(r2 cos cP) +~(2re- 5') =2sin cP -rsin cP -lOre-5%

r ar

and

r acP

Find

DIVERGENCE AND TilE DIVERGENCE niEOREM

CHAP. 4)

EXAMPLE 3. In spherical coordinates a vector field is given by A= (5/r1 sin 6a, + r cot 6a, + r sin (J cos </Ia.. Find div A. . 6 ) +-:-1 - a<rsm . 6 cot(}) +-.--(rsiDlJcos<i')=-1-sin</1 1 a . - 5 SID IV = 2ta( d.A

r ar

r SID 6 alJ

r SID (J a</1

4.3 DIVERGENCE OF D From Gauss' law (Section 3.3),

fD·dS

Q.,nc ~v

In the limit,

. I1m Av-O

fD·dS

Q,nc = d"IV D = 1"1m -= p

A...-.o ~v

This important result is one of MaxweU's equations for static fields: divD=p

and

if£ is constant throughout the region under examination (if not, div £E = p). fields will have divergence of zero in any isotropic charge-free region.

Thus both E and D

EXAMPLE 4. In spherical coordinates the region r s a contains a uniform charge density p, while for r >a the charge density is zero. From Problem 2.54, E = E,a,, where E, = (pr /3£0 ) for r sa and E, =(pa 3 /3£0 r 2 ) for r >a. Then, for r sa,

1a( P') r1( zP) P

2 dJVE=-z- r - =-z 3 r - =-

r ar

3£o

£o

and, for r >a,

( z pal2 ) =0 . E =1- a dIV 2 - r -r ar

3£or

4.4 mE DEL OPERATOR Vector analysis has its own shorthand, which the reader must note with care. vector operator, symbolized V, is defined in cartesian coordinates by

At this point a

V=~a +~•+~a dx ay -y az z >:

In the calculus a differential operator Dis sometimes used to represent d/dx. The symbols V and J are also operators; standing alone, without any indication of what they are to operate on, they look strange. And so V, standing alone, simply suggests the taking of certain partial derivatives, each foUowed by a unit vector. However, when V is dotted with a vector A, the result is the divergence of A.

a a a ) aAx aAy aAz . V•A= ( axa +-a ay y +-a az z ·(A a +A ya y +A z a z )=-+-+-=dtvA ax ay az X

Hereafter, the divergence of a vector field will be written V• A.

DIVERGENCE AND THE DIVERGENCE THEOREM

50 Warning!

(CHAP. 4

The del operator is defined only in cartesian coordinates. When V ·A is written for the divergence of A in other coordinate systems, it does not mean that a del operator can be defined for these systems. For example, the divergence in cylindrical coordinates will be written as

V•A=!~(rA )+! aA.,+ aAz rar

(see Section 4.2).

r aq,

This does not imply that

1a Ia() a() V=;ar(r )a,+;--ac5a41 +~az in cylindrical coordinates. In fact, the expression would give false results when used in VV (the gradient, Chapter 5) or V X A (the curl, Chapter 9).

4.5 THE DIVERGENCE THEOREM Gauss' law states that the closed surface integral of D · dS is equal to the charge enclosed. If the charge density function p is known throughout the volume, then the charge enclosed may be obtained from an integration of p throughout the volume. Thus,

fD· J dS=

But p = V • D,

pdv = Qenc

and so

D • dS

= J(V • D) dv

This is the divergence theorem, also known as Gauss' divergence theorem. It is a three-dimensional analog of Green's theorem for the plane. While it was arrived at from known relationships among D, Q, and p, the theorem is applicable to any sufficiently regular vector field.

divergence theorem

fA • dS = [ (V · A) dv

Of course, the volume vis that which is enclosed by the surfaceS. EXAMPLE 5.

The region

r s a in spherical coordinates has an electric field intensity pr

E=-a

3£ ,

Examine both sides of the divergence theorem for this vector field.

For S, choose the spherical surface r =

bsa.

f =

1 L" 2,..

4npb =--

3t:

pb3

-sin 6d6dtf> 3E'

then

(V · E)dv

L2"L"[p- r sin 2

6 dr d(J dtf>

o E

4npb 3 =--

The divergence theorem applies to time-varying as well as static fields m any coordinate

DIVERGENCE AND THE DIVERGENCE THEOREM

CHAP. 4)

system. The theorem is used most often in derivations where it becomes necessary to change from a closed surface integration to a volume integration. But it may also be used to convert the volume integral of a function that can be expressed as the divergence of a vector field into a closed surface integral.

Solved Problems 4.1.

Develop the expression for divergence in cylindrical coordinates. A delta-volume is shown in Fig. 4-3 with edges ll.r, r ll.t/), and ll.z. The vector field A is defined at P, the corner with the lowest values of the coordinates r, 1/), and z, as A= A,a, +A.a.+ Azaz z

Fig.4-3

By definition, fA·dS divA= l i m - - .&...-- ll.v To express ~A • dS all six faces of the volume must be covered. to Fig. 4-4.

(1)

For the radial component of A refer

~+ll.r)

A,(r)

Fig.44

Over the left face,

JA • dS = - A,r

fl. I/) ll.z

and over the right face,

JA • dS = A,(r + ll.r)(r + ll.r) ll.f'/» ll.z BA, ll.r) (r + ll.r) ll.f'/» ll.z = ( A, +a; =A,r ll.t/) ll.z

(

aA,) ll.r ll.t/) ll.z

+ A, +ra;

DIVERGENCE AND THE DIVERGENCE THEOREM where the term in (l'lr) 2 has been neglected.

(CHAP. 4

The net contribution of this pair of faces is then

aA,) ( A,+ ra; llr I'll/) l'lz = ar (rA,) l'lr I'll/) l'lz

=-;:1 ara (rA,) l'lv

(2)

since l'lv = r l'lr I'll/) l'lz. Similarly, the faces normal to -. yield and for a net contribution of (3)

and the faces normal to

yield and

for a net contribution of

aAz u..v

(4)

When (2), ( 3) and ( 4) are combined to give ~ A • dS, ( 1) yields divA=! a(rA,) +! aA'~' + aA, r

4.2.

•

r a,p

Show that V • E is zero for the field of a uniform line charge. For a line charge, in cylindrical coordinates, Pl'

E=--a, 2:trf' 0 r

Then The divergence of E for this charge configuration is zero everywhere except at r = 0, expression is indeterminate.

4.3.

Show that the D field due to a point charge has a divergence of zero. For a point charge. in spherical coordinates,

Then, for r > 0,

4.4.

Given

A=e-Y(cosxa.. -sinxay),

a ax

find V·A.

V ·A=- (e-r cosx) +-( -e-r sinx) = e-r( -sinx) + e-r(sinx) = 0

where the

CHAP. 4)

4.5.

Given

DIVERGENCE AND TilE DIVERGENCE THEOREM

A= x 2ax + yzay

+ xyaz , find V • A. a ax

a ay

a az

V •A=-(x 2 ) +-(yz) +-(xy) =2x +z

4.6.

Given

11 4.7.

A= (x 2 + y 2 ) - 112ax,

find V • A at (2. 2, 0).

v • Al 12.2.o) =

and Given

A= r sin q,a, + 2r cos f/Ja.p

-8.84 x

w-z

+ 2z 2az, find V • A.

V •A=-- (r2 sm tJ>) + --(2r cos tJ>) +-(2z 2 )

r ar

=2 sin tJ> 4.8.

Given

2 sin

r atJ> tJ> + 4z =

A= 10sin2 q,a, +ra.p + [(z 2 /r)cos2 q,]az, 2

find V • A at (2, and

Given

A= (5/r2 )a, + (10/sin B)a8 ta(

q,, 5).

V . A = 10 sin tJ> + 2z cos tJ>

4.9.

r 2 q, sin Ba.p.

find V • A.

•

V•A=-5)+---(10)+---(-r2 tJ>sm 6)= -r 2 r

4.10. Given

A= 5 sin Ba8

r sin 6 a6

r sin 6 atJ>

+ 5 sin f/Ja.p, find V ·A at (0.5. n/4, n/4).

. 2 6) +1 -a (5smtJ> . ) =10--+5-cos6 costJ> V · A = -1 - -a(5 SID

r sin 6 a6

r sin 6 atJ>

and

V • Al(o.S.nl4.n/4) = 24.14

4.11. Given that D = p 0 zaz in the region -1 ~ z (p0 z/lzl)az elsewhere, find the charge density.

1 in cartesian coordinates and

V·D=p

For -1 s

z s 1,

P = az (PoZ) =Po and for

r sin 6

z < -1

z > 1,

The charge distribution is shown in Fig. 4-5.

:o __ -1

Fig.4-5

DIVERGENCE AND THE DIVERGENCE THEOREM

[CHAP. 4

4.U. Given that D = (10r3 /4)a, (C/m2 ) in the region 0< r ~3m in cylindrical coordinates and D = (810/4r)a, (C/m2 ) elsewhere, find the charge density. For O<r s3 m,

and for

r > 3 m, 1a p=--(810/4}=0 rar

4.13. Given that D= -

nr 2

(1 -cos 3r)a,

in spherical coordinates, find the charge density. Q ] =-sm3r 3Q . p =12-a - [ r 2 -(l-cos3r) 2 2

r ar

1tr

Jtr

4.14. In the region O<r~lm, D=(-2x 10-4 /r)a, (C/m2 ) and for r>lm, D=(-4x I0- 4 /r2 )a, (C/m 2 ), in spherical coordinates. Find the charge density in both regions. For O<rs 1m,

and for

r > 1 m,

4.15. In the region r ~ 2, D = (5r 2 /4)a, and coordinates. Find the charge density.

for r > 2,

D = (20/r 2)a,

spherical

For rs2, 1 a r ar

4 p =--(5r /4) = 5r 2

and for

r>2,

1 a p=--(20)=0 r 2 ar

4.16. Given that D = (10x 3 /3)ax (C/m2 ), evaluate both sides of the divergence theorem for the volume of a cube, 2m on an edge, centered at the origin and with edges parallel to the axes.

rJ. D • dS =

(V • D) dv

vol

Since D has only an x component, D • dS is zero on all but the faces at x =1m and x =-1m Fig. 4-6).

II It

:rD·dS= _

II it

10(1) -ax·dydza.+ _ 3 1

40 40 80 =-+-=-C 3 3 3

10(-1) -a.·dydz(-a.) 3

(see

CHAP. 4)

DIVERGENCE AND THE DIVERGENCE THEOREM

Now for the right side of the divergence theorem.

111_11 11 1

""'(V·D)dv= _

Since V • D = 10x2•

80 1 1 [ x]]z_dydz=3C _ 103 1111

(10x )dtdydz= _

4.17. Given that A= 30e-'a,- 2zaz in cylindrical coordinates, evaluate both sides of the • divergence theorem for the volume enclosed by r = 2. z = 0, and z = 5 (Fig. 4-7). fA· dS=

(V • A)dv

It is noted that Az

for z

fA·dS=

and hence A • dS is zero over that part of the surface.

f f" 30e- a,•2d~dza,+ f 2

L'Z>I

-2(S)a.·rdrd4Jaz

=60e- (2Jr)(S) -10(2n)(2) = 129.4 2

For the right side of the divergence theorem: 1a a 30e-' V ·A""; ar (30re-') + az (-2z) =-,-- 30e-'- 2

and

4.18 ~

J(V · A)dv =f 121< f (~-, -30e-' -z)rdrdq,dz = 129.4

Given that D = (10r3 /4)a, (C/m2 ) in cylindrical coordinates, evaluate both sides of the divergence theorem for the volume enclosed by r = 1 m, r = 2 m, z = 0 and z = 10m (see Fig. 4-8). fn· dS=

(V· D)dv

DIVERGENCE AND THE DIVERGENCE THEOREM

(CHAP. 4

Since D has no z component, D • dS is zero for the top and bottom. is in the direction -a,.

D • dS""

fo 12>1 ~O +

On the inner cylindrical surface dS

(1)3a, · {1) dq, dz( -a,)

1012'"10 1 4 (2) a, • (2) dq, dza, 3

-200.11'

200.11'

=--+ 16--=750nC From the right side of the divergence theorem:

1 {10r

a V·D=-rar 4

4 )

=10r2

and

4.19. Given that D = (5r 2 /4)a, (C/m2 ) in spherical coordinates, evaluate both sides of the ~ divergence theorem for the volume enclosed by r = 4 m and 8 = n/4 (see Fig. 4-9).

fo·dS=

J(V·D)dv

FIJ.4-9

Since D has only a radial component, D • dS has a nonzero value only on the surface

r =4 m.

CHAP. 4)

DIVERGENCE AND THE DIVERGENCE THEOREM

For the right side of the divergence theorem: 4

1 a (5r V·D=-r2 ar 4 and

(V ·D) dv =

:z,..l,.-1414 (5r)r 0

)

=5r

sin B dr dB d4J

= 589.1 C

Supplementary Problems 4.20.

Develop the divergence in spherical coordinates. rsin B A4J.

4.21.

Show that V • E is zero for the field of a uniform sheet charge.

4.22.

The field of an electric dipole with the charges at ±d/2 on the z axis is E=

Use the delta-volume with edges Ar, r AB, and

Qd (2 cos Ba, +sin Ba11) 41CE0 r 3

Show that the divergence of this field is zero.

A= e5xax + 2 cos yay+ 2 sin za. , find V • A at the origin.

4.23.

Given

4.24.

Given A= (3x + y 2 )ax + (x- y 2 )ay ,

4.25.

Given A= 2xyax + zay + yz 2a.,

4.26.

Given A= 4xyax- xy 2 ay + 5 sin za. ,

4.27.

Given

4.28.

Given A= (10/r2 )a, + 5e- 2•a.,

4.29.

Given

A= 5cos ra, + (3ze-:z,/r)a.,

4.30.

Given

A = lOa, + 5 sin Ba11 , find V ·A.

Ans.

(2 +cos B)(IO/r)

4.31.

Given

A = ra, - r 2 cot Ba11

Ans.

3-r

4.32.

Given

A= [(10 sin2 B)/r)a, (N/m), find V • A at (2m, 1C/4 rad, n/2 rad).

4.33.

Given A = r2 sin Ba, + 131/lall + 2ra41

4.34.

Show that the divergence of E is zero if E = (100/r)a41 + 40a•.

4.35.

In the region as r s b

Ans.

3-2y

find V ·A at (2, -1, 3).

Ans.

find V • A.

find V • A at (2, 2, 0).

A= 2r cos2 1Jla, + 3r2 sin z~ + 4z sin 2 1/1 a.,

Ans.

find V ·A at (2, IJI, 1). find V ·A at

find V• A.

•

find V· A.

(cylindrical coordinates),

(1C,

Ans.

IJI, z).

Ans.

-8.0

Ans.

find V ·A.

1.0

5.0 4.0

Ans.

-2.60 Ans.

4r sin B +

-1.59

Ans.

c~IP) cot B

1.25N/m2 •

DIVERGENCE AND THE DIVERGENCE THEOREM

[CHAP. 4

and for r> b,

For r <a,

D = 0.

Find p in all three regions.

0, Po. 0

Ans.

4.36.

In the region O<r:s2 (cylindrical coordinates), D=(4r- 1 +2e- 0 · 5 '+4r-•e-0 · 5')a, 2, D = (2.057/r)a, _ Find p in both regions. Ans. -e-05', 0

4.37.

In the region r :s 2 (cylindrical [3/{128r)]a,. Find pin both regions.

4.38.

Given

4.39.

Given

D = 10 sin Ba, + 2 cos Baa ,

coordinates), D =[lOr+ (r2 /3))a., Ans. 20+r, 0

find the charge density.

Ans.

and

and for r>

for

r>2,

sin B - -(18+ 2coe B) 5

3r D=--a ,Z + 1 r in spherical coordinates, find the charge density.

4.40.

Given

in spherical coordinates, find the charge density-

4.41.

Ans.

40e-:z..

In the region r :s 1 (spherical coordinates), D= and for r > 1,

(~-~)a.

D = [5/(63r2 ))a, _ Find the charge density in both regions.

Ans.

4- r 2 , 0

4.42.

The region r:s2m (spherical coordinates) has a field E=(5rxl0- 5/Eo)a. (V/m). Ans. 5.03 X 10- 3 C charge enclosed by the shell r =2m.

4.43.

Given that D = (5r 2 /4)a, in spherical coordinates, evaluate both sides of the divergence theorem for Ans. 15:tr the volume enclosed between r = 1 and r = 2.

4.44.

Given that D = (10r1 /4)a, in cylindrical coordinates, evaluate both sides of the divergence theorem for the volume enclosed by r = 2, z = 0, and z = 10. Ans. BOO:tr

4.45.

Given that D = 10 sin Ba, + 2 cos Baa , evaluate both sides of the divergence theorem for the volume enclosed by the shell r = 2. Ans. 40:tr 2

Find the net

Chapter 5 The Electrostatic Field: Work, Energy, and Potential 5.1 WORK DONE IN MOVING A POINT CHARGE A charge Q experiences a force F in an electric field E. In order to maintain the charge in equilibrium a force Fa must be applied in opposition (Fig. 5-l): Fa= -QE

F=QE

F. - • - F

_______ Q_________

Fig. 5-1

Work is defined as a force acting over a distance. Therefore, a differential amount of work dW is done when the applied force Fa produces a differential displacement dl of the charge; i.e. moves the charge through the distance dt = ldll. Quantitatively, dW =Fa· dl= -QE • dl

Note that when Q is positive and dl is in the direction of E, dW = - QE dt < 0, indicating that work was done by the electric field. [Analogously, the gravitational field of the earth performs work on a (positive) mass Mas it is moved from a higher elevation to a lower one.) On the other hand, a positive dW indicates work done against the electric field ( cf. lifting the mass M). Component forms of the differential displacement vector are as follows: dl = dxa" + dyay + dzaz

(cartesian)

dl = dra, + r d¢a4>

(cylindrical)

+ dzaz

dl = dra, + r d8a 8 + r sin 8 d¢a4>

(spherical)

The corresponding expressions for dt were displayed in Section 1.5. An electrostatic field is given by E = (x/2 + 2y)ax + 2xay (V /m). Find the work done in moving a point charge Q = -20 llC (a) from the origin to (4, 0, 0) m, and (b) from (4, 0, 0) m to (4, 2, 0) m.

EXAMPLE 1.

(a) The first path is along the x axis, so that dl = dx dW

•x·

= -QE • dl = (20 X 10- 6)(~ + 2y) dx

w= (20 x w-

6 )

f (~

+ 2y) dx = so 111

(b) The second path is in the ay direction, so that dl = dya_..

w = (20 x w-

)

2x dy = 320 111

THE ELECfROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

[CHAP. 5

5.2 CONSERVATIVE PROPERTY OF THE ELECfROSTATIC FIELD The work done in moving a point charge from one location, B, to another, A, in a static electric field is independent of the path taken. Thus, in terms of Fig. 5-2,

E. dl =

-L

E. dl

E·di=O

where the last integral is over the closed contour formed by Q) described positively and ~ described negatively. Conversely, if a vector field F has the property that ~ F • dl = 0 over every closed contour, then the value of any tine integral of F is determined solely by the endpoints of the path. Such a field F is called conservative; it can be shown that a criterion for the conservative property is that the curl ofF vanish identically (see Section 9.4).

Fig.S-Z

EXAMPLE 2. For theE field of Example 1, find the work done in moving the same charge from (4, 2, 0) back to (0, 0, 0) along a straight-line path. W = (20 X w-~ fO,O,O) (4,2,0)

= (20 x

w-(>>

(11,11,0)

(4,2.0)

The equation of the path is y = x/2;

[ (~ + 2y )a.. + 2xay] • (dx 8.., + dy 8y)

(X + 2y ) dx + 2x dy 2

therefore, dy = ~ dx

w = (20 x w-

)

and

dx = -400 "'J

From Example 1, 80 + 320 = 400 Ill of work was spent against the field along the outgoing, right-angled path. Exactly this much work was returned by the field along the incoming, straight-line path, for a round-trip total of zero (conservative field).

5.3 ELECfRIC POTENTIAL BETWEEN TWO POINTS The potential of point A with respect to point B is defined as the work done in moving a unit positive charge, Q", from B to A.

VA 8= Qu W = - J.A E•dl B

(J/C or V)

It should be observed that the initial, or reference, point is the lower limit of the line

integral. Then, too, the minus sign must not be omitted. This sign came into the expression by way of the force Fa= -QE, which had to be applied to put the charge in equilibrium. Because E is a conservative field,

CHAP. 5)

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

whence VA 8 may be considered as the potential difference between points A and B. When VA 8 is positive, work must be done to move the unit positive charge from B to A, and point A is said to be at a higher potential than point B.

5.4 POTENTIAL OF A POINT CHARGE Since the electric field due to a point charge Q is completely in the radial direction, VA 8 = -

EÂˇdl=-

L'A E â&#x20AC;˘n

dr= - -Q4.7rEo

L'A -dr= -Q- ( -1 - -1 r8

4.7rEo rA

rB)

For a positive charge Q, point A is at a higher potential than point B when rA is smaller than r8. If the reference point B is now allowed to move out to infinity,

VA~= 4;E

c: -~)

or Considerable use will be made of this equation in the materials that follow. The greatest danger lies in forgetting where the reference is and attempting to apply the equation to charge distributions which themselves extend to infinity.

5.5

POTENTIAL OF A CHARGE DISTRIBUTION If charge is distributed throughout some finite volume with a known charge density p (C/m 3 ),

then the potential at some external point can be determined. To do so, a differential charge at a general point within the volume is identified, as shown in Fig. 5-3. Then at P, dV

=__!!Q_ 4nE0R

Integration over the volume gives the total potential at P:

v- J -pdv vol4nE0R

where dQ is replaced by p dv. Now R must not be confused with r of the spherical coordinate system. And R is not a vector but the distance from dQ to the fixed point P. Finally, R almost always varies from place to place throughout the volume and so cannot be removed from the integrand. dV

'"P R

Fig. 5-3

If charge is distributed over a surface or a line, the above expression for V holds, provided that the integration is over the surface or the line and that p_. or pf' is used in place of p. It must be emphasized that all these expressions for the potential at an external point are based upon a zero reference at infinity.

THE ELECfROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

[CHAP. 5

EXAMPLE 3. A total charge of ~ nC is uniformly distributed in the form of a circular disk of radius 2m. Find the potential due to this charge at a point on the axis, 2m from the disk. Compare this potential with that which results if all of the charge is at the center of the disk. Using Fig. 5-4,

w-M

Ps =A= ).n-

C/m

R=v'4+r2

(m)

V=30 (2"I2rdrdcp=49.1V

and

.7r

o v'4+r 2

With the total charge at the center of the disk, the expression for the potential of a point charge applies:

Q V=-4.n-E 0 Z

9 â&#x20AC;˘"xl03

4.n-(10 9 /36.n-)2

=60V

10. 0. 21

Fig. 5-4

5.6 GRADIENT At this point another operation of vector analysis is introduced. Figure 5-5(a) shows two neighboring points, M and N, of the region in which a scalar function V is defined. The vector separation of the two points is

M(x,y,z) N(x

+ dx, y + dy, z + dz)

~V(x,y,z)=c 2

~ V(x, y, z)

(a)

(b)

Fig. 5-5

=c 1

CHAP. 5)

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

From the calculus, the change in V from M toN is given by

av ax

av ay

av az

dV = - dx + - dy +- dz Now, the del operator, introduced in Section 4.4, operating on V gives

av ax

av ay

av az

VV=-a +-a +-a X

It follows that

dV=VV·dr The vector field VV (also written grad V) is called the gradient of the scalar function V. It is seen that, for fixed ldrl, the change in V in a given direction dr is proportional to the projection of VV in that direction. Thus VV lies in the direction of maximum increase of the function V. Another view of the gradient is obtained by allowing the points M and N to lie on the same equipotential (if V is a potential) surface, V(x, y, z) = c 1 [see Fig. 5-5( b)). Then dV = 0, which implies that VV is perpendicular to dr. But dr is tangent to the equipotential surface; indeed, for a suitable location of N, it represents any tangent through M. Therefore, VV must be along the surface normal at M. Since VV is in the direction of increasing V, it points from V(x, y, z) = c 1 to V(x, y, z) = c2 , where c2 > c 1 • The gradient of a potential function is a vector field that is everywhere normal to the equipotential surfaces. The gradient in the cylindrical and spherical coordinate systems follows directly from that in the cartesian system. It is noted that each term contains the partial derivative of V with respect to distance in the direction of that particular unit vector.

av av •x +x ay ay +-az az

VV =-a

VV =-a a,+ -a a.,

+-a az z

(cartesian) (cylindrical)

VV = - a , + - a 6 + . a (spherical) ar r aB r sm 8 a<f> "' While VV is written for grad V in any coordinate system, it must be remembered that the del operator is defined only in cartesian coordinates.

5.7 RELA TIONSmP BETWEEN E AND V From the integral expression for the potential of A with respect to B, the differential of V may be written dV=-E·dl On the other hand, dV=VV •dr Since dl = dr is an arbitrary small displacement, it follows that E=-VV The electric field intensity E may be obtained when the potential function V is known by simply taking the negative of the gradient of V. The gradient was found to be a vector normal to the equipotential surfaces, directed to a positive change in V. With the negative sign here, the E field is found to be directed from higher to lower levels of potential V.

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

EXAMPLE 4. In spherical coordinates and relative to infinity, the potential in the region a point charge Q is V = Q/41Ct::0 r. Hence,

[CHAP. 5

r > 0 surrounding

-vv = -~ (_g__) a =_g_a ar 41CEoT , 41Ct:: r 2 , 0

in agreement with Coulomb's law. differentiation of V gives back E.)

(V is obtained in principle by integrating E; so it is not surprising that

5.8 ENERGY IN STATIC ELECTRIC FIELDS Consider the work required to assemble, charge by charge, a distribution of n = 3 point charges. The region is assumed initially to be charge-free and with E = 0 throughout. Referring to Fig. 5-6, the work required to place the first charge, Q 1 , into position 1 is zero. Then, when Q2 is moved toward the region, work equal to the product of this charge and the potential due to Q 1 is required. The total work to position the three charges is WE=

w. + W2+ WJ

= o + (Q2 v2.•> + (Q3 v3.1 + Q3 V3.2) The potential V2.1 must be read "the potential at point 2 due to charge Q 1 at position l." (This rather unusual notation will not appear again in this book.) The work WE is the energy stored in the electric field of the charge distribition. (See Problem 5.17 for a comment on this identification.) Now if the three charges were brought into place in reverse order, the total work would be wE=

w3 + W2 + w.

= o + (Q2 V2.3) + (Q. v •. 3 + Q. v ..2> When the two expressions above are added, the result is twice the stored energy: 2WE = Q,(VI.2 + Vu) + Q2(V2.1 + V2.3) + Q3(V3.1

+ VJ.2)

The term Q 1(V1 •2 + Vu) was the work done against the fields of Q 2 and Q 3 , the only other charges in the region. Hence, V1. 2 + V1.3 = V1 , the potential at position 1. Then 2WE = Q. V1

+ Q2V2 + Q3V3 n

and

WE=2m~l QmVm

for a region containing n point charges. becomes an integration,

For a region with a charge density p (C/m3 ) the summation

Other forms (see Problem 5.12) of the expression for stored energy are WE=21

EE dv

Fig. 5-6

D2 WE=21 -;-dv

CHAP. 5)

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

In an electric circuit, the energy stored in the field of a capacitor is given by WE = !QV = !CV 2

where C is the capacitance (in farads), V is the voltage difference between the two conductors making up the capacitor, and Q is the magnitude of the total charge on one of the conductors. EXAMPLE 5. A parallel-plate capacitor, for which C = £A/d, plates (Fig. 5-7). Find the stored energy in the electric field.

has a constant voltage V applied across the

Fig. S-7

With fringing neglected, the field is E

= (V /d)a,

between the plates and

E = 0 elsewhere.

WE=iJ ££2dv

2 =-CV 2

As an alternate approach, the total charge on one conductor may be found from D at the surface via Gauss' law (Section 3.3).

£V

D=-a d "

£VA

Q=IDIA=d 2

1 1 (£AV ) 1 2 W=zQV=z -d- =zCV

Then

Solved Problems 5.1.

a .rJ+

Mathcacl

Given the electric field E = 2xax - 4yay (V /m), find the work done in moving a point charge +2 C (a) from (2, 0, 0) m to (0, 0, 0) and then from (0, 0, 0) to (0, 2, 0); (b) from (2, 0, 0) to (0, 2, 0) along the straight-line path joining the two points. (See Fig. 5-8.) (a)

Along the x axis, y

= dy =dz = 0, dW

and

= -2(2xa.. ) • (d.x a..)= -4x dx

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

(CHAP. 5

(ll. 2.0)

(/>) (a)

-+---------.\ tal

(0. 0. ())

(2. ll. ll)

Fig. 5-8 Along the y axis,

x = dx = dz = 0, dW

Thus

and

= -2( -4ya,) · (dyay) = 8y dy

W = -4

xdx

y dy = 24J

(b) The straight-line path has the parametric equations

x=2-2t where

0~ t

y=2t

z=O

Hence, dW = -2[2(2- 2t)ax- 4(2t)ay]• ((-2 dt)a,

+ (2 dt)a,]

=16(1+t)dt

w = 16 Jl)(' (1 + t) dt = 24 J

and

5.2.

a -rf+

Find the work done in moving a point charge spherical coordinates, in the field

Q = 5 11C

10 E=5e-'14a + - - a r r sin f) <P

Mat head

from the origin to (2m, :rr/4, :rr/2),

(V/m)

In spherical coordinates, dl = dra, + r dfJaH + r Sin fJ d¢a.p Choose the path shown in Fig. 5-9.

Along segment/,

dfJ = d¢ = 0,

dW = -QE · dl = ( -5 X 10

Along segment II,

dr = dfJ = 0,

)(5e __ ,, dr) 4

and dW = -QE · dl= (-5

Fig. 5-9

10 ")(lOd¢)

and

CHAP. Sf

TIIE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

Along segment Ill,

dr = d,P

= 0,

and dW = -QE·di=O

Therefore,

W = {-25 X 10-6 )

e-'14 dr+ (-50+ 10-~

dt/J

= -117.9 1Jl

In this case, the field does 117.9 1Jl of work on the moving charge. 5.3.

Given the field E = (k/r)a, in cylindrical coordinates, show that the work needed to move a point charge Q from any radial distance r to a point at twice that radial distance is independent of r. Since the field has only a radial component, -kQ dW= -QE·dl= -QE,dr=--dr r

For the limits of integration use r1 and 2r1 • z,,

W = -kQ

J.,,

-= -kQ ln2

independent of r1 •

5.4.

For a line charge p~= (10- 9 /2) C/m B is ( 4 m, 1r, 5 m). VAB

on the z axis, find VAB• where A is (2m, 1£/2, 0) and

= - L" E • dl

where

Since the field due to the line charge is completely in the radial direction, the dot product with dl results in E,dr. VAB =-

" f. 8

5.5.

10 -~~

( .1fEor) dr = -9[ln rJ: = 6.24 V 22

In the field of Problem 5.4, find V8 c, where compare with the sum of VAs and V8 c.

r8 = 4 m

and

= -9[ln r)~~ =

VAB

+ VBc = 6.24 V + 8.25 V = 14.49 V =

Then find VAc and

-9(ln 4 -In 10) =8.25 V VAc = -9[ln rJ~: = -9(ln 2 - In 10) = 14.49 V

5.6.

VBc

rc =10m.

Given the field E = ( -16/r2)a, (V/m) (2m, 1r, 1£/2) with respect to (4 m, 0,1r). B.

i!:i

.!I!!

in spherical coordinates, find the potential of point

The equipotential surfaces are concentric spherical shells. Let r = 2m be A and r = 4 m, Then VAB

5.7.

VAc

f (~! ) 6

dr = -4 V

A line charge p~ = 400 pC/m lies along the x axis and the surface of zero potential passes through the point (0, 5, 12) m in cartesian coordinates (see Fig. 5-10). Find the potential ~t

(2, 3, -4) m .

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

[CHAP. 5

Line charge

Fig. S-10

With the line charge along the x axis, the x coordinates of the two points may be ignored. '" = v'9 + 16 = 5 m

Then V,.. 8 = -

I'" ,

5.8.

= V25 + 144 = 13m

Pt Pt r" --dr= ---ln-=6.88V 21rtE0 r 2JrÂŁ0 Te

Find the potential at rA = 5 m with respect to r 8 =15m 500 pC at the origin and zero reference at infinity.

due to a point charge

Due to a point charge,

To find the potential difference, the zero reference is not needed. -

VAB-

soo x to-'

(I 1)

- - - =060V 4Jr{l0 /36Jr) 5 15 . 9

The zero reference at infinity may be used to find

y, and

~s-

(..!.) =0.30 V

v.=_g_ (-1)= o.90v 4JrEo 5

V15 = __Q_ 4JrEo 15

Then 5.9.

Forty nanocoulombs of charge is uniformly distributed around a circular ring of radius 2m. Find the potential at a point on the axis 5 m from the plane of the ring. Compare with the result where all the charge is at the origin in the form of a point charge. With the charge in a line,

v=J 4JrE R Ptdf 0

40 x

Here and (see Fig. 5-11) R

Pt=

= v'29 m,

2Jr(2)

df =(2m) df/>.

w-s

=-C/m

CHAP. 5)

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

F1g. 5-ll

If the charge is concentrated at the origin, V

5.10. Five equal point charges, tial at the origin.

Milt head

40x

4JrE 0

(5)

= 72.0V

Q =20 nC, are located at x = 2, 3, 4, 5, 6 m.

1 " Qm 20 V=-=

L 4JrEom=tRm

Find the poten-

10- (1

4JrEo

1 1 1 1) -+-+-+-+=261 v 2

5 6

5.11. Charge is distributed uniformly along a straight line of finite length 2L (Fig. 5-12). Show that for two external points near the midpoint, such that r 1 and r2 are small compared to the length, the potential V12 is the same as for an infinite line charge.

-L Fig. 5-U

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

(CHAP. 5

The potential at point 1 with zero reference at infinity is V, =2

p, dz

, , (P u 4.7r£n(z- + rl) -

2p, [In (z + v'-,-, ]/. z- + rl) n

= --

4.7r£n

=__!!_!_(In (L + v'L 2 + ri}- In r.] 2JrE 0

Similarly, the potential at point 2 is

Now if

L~r,

and

L~r2 ,

p, V, =--(In 2L -In r,) 2JrE 11

V, = - - (In 2L - In r 2 ) 2Jr£n

Then

V12 = V,- V2 =--ln2JrE11 r,

which agrees with the expression found in Problem 5.7 for the infinite line.

5.U. Charge distributed throughout a volume v with density p gives rise to an electric field with energy content

Show that an equivalent expression for the stored energy is

WE=21

J EE

Figure 5-13 shows the charge-containing volume v enclosed within a large sphere of radius Since p vanishes outside v.

Wt.-=. 2

, 1

pdv=2

~phcn(..al

pVdv=(V·D)Vdv 2 ..phcru:::.tl

v~,IURlC

voltunc

Sphere

Fig. 5-13

CHAP. 5)

THE ELECfROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

The vector identity V • VA= A • VV + V(V • A),

WE =!J 2

applied to the integrand, gives

(V • VD)dv opherical volume

_!J 2

(D • VV)dv opheric:al wolume

This expression holds for an arbitrarily large radius R; the plan is to let The first integral on the right equals, by the divergence theorem,

!1 2

R-+ co.

VD·dS

J'apheric:al surface

Now, as the enclosing sphere becomes very large, the enclosed volume charge looks like a point charge. Thus, at the surface, D appears as k 1/R 2 and V appears as k 2 /R. So the integrand is decreasing as 1/ R 3 • Since the surface area increases only as R 2 , it follows that lim

VD · dS = 0

R - Tspherical aurfacc

The remaining integral gives, in the limit,

And since D = eE,

the stored energy is also given by

WE=~J eE

5.13. Given the potential function V = 2x + 4y (V) in free space, find the stored energy in a 1-m3 volume centered at the origin. Examine other 1-m3 volumes.

E= -VV =- ( ax a.. +

av ay a,.+ av az a. ) =

-2a. -4a,.

(V/m)

This field is constant in magnitude (E = v'20 V/m) and direction over all space, and so the total stored energy is infinite. {The field could be that within an infinite parallel-plate capacitor. It would take an infinite amount of work to charge such a capacitor.) Nevertheless, it is possible to speak of an energy density for this and other fields. The expression

WE=~J eedv suggests that each tiny volume dv be assigned the energy content w dv, where w=~ee

For the present field, the energy density is constant: 1

w-s

36n

w =- E 0 (20) = - J/m 3

and so every 1-m3 volume contains (10- 8 /36n) J of energy.

5.14. Two thin conducting half planes; at t1> = 0 and t1> = n/6, are insulated from each other along the z axis. Given that the potential function for 0 ~ tJ> ~ n/6 is V = ( -OO,P/n) V, find the energy stored between the half planes for 0.1 ~ r ~ 0. 6 m and 0 ~ z ~ 1 m. Assume free space. To find the energy, w;;, stored in a limited region of space, one must integrate the energy density (see Problem 5.13) through the region. Between the half planes, E=

-vv = _!~ r aq,

(-ooq,)a .1r

00 = nr a .,

(V/m)

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

[CHAP. 5

and so £ ° 2

W:0=

Llln/6 r-6 (60)2rdrdq>dz=~ln6=1.51nJ 300£ 0

_ 0 1

trr

5.15. The electric field between two concentric cylindrical conductors at r = 0.01 m and r = 0.05 m is given by E = (lW/r)ar (V /m), fringing neglected. Find the energy stored in a 0.5-m length. Assume free space.

w;, = 2

£ 0£

dv = 2°

J.""

+O.S

12" LO.OS (71Cf)2 rdr dq> dz = 0.224 J o.ol

5.16. Find the stored energy in a system of four identical point charges, Q = 4 nC, at the corners of a square 1 m on a side. What is the stored energy in the system when only two charges at opposite corners are in place? 2WE = Ql \'; + Qz"i+ Q3\--3+ Q4V4=4Q 1 V1 where the last equality follows from the symmetry of the system. VI=

41f£oR 12

9 Q 1) J + Q4 = 4 x w- - +-1 +-:::: = 97.5 v 4Jr£oR 13 41f£oR 14 4Jr£o 1 1 v'2

W,r;; = 2Q 1 V1 = 2(4 X 10- 9 ){97.5) =780nJ

Then

For only two charges in place, 2W.r;=Q 1 V1 =(4x 10 _ 9 )

(4 x w-") =102nJ 41f£ 0 V2

5.17. What energy is stored in the system of two point charges, -3 nC, separated by a distance of d = 0.2 m? 2W,r;; = Q 1 VI+ Q 2V2 =

whence

W: E

= QIQz =4n£ 0 d

Q 1 = 3 nC

and

Q2 =

Q1( 4Jr~: d) + Q 2 ( 4Jr~: d)

(3 X 10-9)2 = -405 nJ 4n(l0 "/36.n-)(0.2)

It may seem paradoxical that the stored energy turns out to be negative here. whereas ~££ 2 , and

hence

is necessarily positive. The reason for the discrepancy is that in equating the work done in assembling a system of point charges to the energy stored in the field. one neglects the infinite energy already in the field when the charges were at infinity. (It took an infinite amount of work to create the separate charges at infinity.) Thus, the above result, W,r;; = -405 nJ, may be taken to mean that the energy is 405 nJ below the (infinite) reference level at infinity. Since only energy differences have physical significance, the reference level may properly be disregarded.

5.18. A spherical conducting shell of radius a, centered at the origin, has a potential field

v-{VoV a/r 0

with the zero reference at infinity.

r:s,a r>a

Find an expression for the stored energy that this

CHAP. 5]

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

potential represents r<a r>a

W,r;;=z

€ £ 0 edv=0+;

Jot"L"r(v.a) 1a r~

2 2

r sin0drd0dtj>=2n£0 V:,C,

Note that the total charge on the shell is, from Gauss' law,

while the potential at the shell is V = V0 • Thus, WE= ~QV, the familiar result for the energy stored in a capacitor (in this case, a spherical capacitor with the other plate of infinite radius).

Supplementary Problems 5.19.

Find the work done in moving a point charge Q = -20 pC from the origin to (4, 2, 0) min the field E = 2(x 2

along the path x =By.

Ans.

+ 4y )ax + 8xa,. (V /m}

1.60mJ

5.10.

Repeat Problem 5.2 using the direct radial path. Ans. -39.35 pJ (the nature of the singularity along the z axis makes the field nonconservative)

5.21.

Repeat Problem 5.2 using the path shown in Fig. 5-14.

Ans.

-117.9 pJ

Fig. 5-14

5.22.

Find the work done in moving a point charge Q = 3 pC from (4 m, rc, 0) to (2m, rc/2, 2m), cylindrical coordinates, in the field E = (Uf/r)a, + l(fza, (V /m). Ans. -0.392 J

5.23.

Find the difference in the amounts of work required to bring a point charge Q = 2 nC from infinity to r =2m and from infinity to r = 4 m, in the field E = (Hf/r)a, (V/m). Ans. 1.39 X 10-4 J

5.24.

A uniform line charge of density p~ = 1 nC/m is arranged in the form of a square 6 m on a side, as Ans. 35.6 V shown in Fig. 5-15. Find the potential at {0, 0, 5) m.

5.25.

Develop an expression for the potential at a point d meters radially outward from the midpoint of a finite line charge L meters long and of uniform density p, (C/m). Apply this result to Problem 5.24 as a check. p~ L/2 + Vd 2 + L2 /4 Ans. 21f£o In d (V)

THE ELECfROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

[CHAP. 5

(0, 0, S)

Fig. 5-l.S

S.26.

Show that the potential at the origin due to a uniform surface charge density p. over the ring z = 0, R :5 r :5 R + 1 is independent of R.

S.27.

A total charge of 160 nC is first separated into four equal point charges spaced at 90" intervals around a circle of 3 m radius. Find the potential at a point on the axis. 5 m from the plane of the circle. Separate the total charge into eight equal parts and repeat with the charges at 45° intervals. What would be the answer in the limit p, = (l60/6.1r) nC/m? Ans. 247 V

S.l8.

In spherical coordinates, point A is at a radius 2m while B is at 4 m. Given the field E = (-16/r2 )a. (V/m), find the potential of point A, zero reference at infinity. Repeat for point B. Now express the potential difference V,.- V8 and compare the result with Problem 5.6. Ans. V,. = 2V8 = -8 V

5.19.

If the zero potential reference is at r =10m and a point charge Q = 0.5 nC is at the origin, find the potentials at r = 5 m and r = 15 m. At what radius is the potential the same in magnitude as that at r = 5 m but opposite in sign? Ans. 0.45 V, -0.15 V, oo

5.30.

A point charge Q = 0.4 nC is located at {2, 3, 3) m in cartesian coordinates. Find the potential Ans. 2. 70 V difference VAn• where point A is (2, 2, 3) m and B is ( -2, 3, 3) m.

5.31.

Find the potential in spherical coordinates due to two equal but opposite point charges on the y axis Ans. (Qdsin6)/(4Jr£0 r 2 ) at y=±d/2. Assume r~d.

5.32.

Repeat Problem 5.31 with the charges on the z axis.

5.33.

Find the charge densities on the conductors in Problem 5.14. +60£o -60£o 2 .1r Ans. - - (C/m 2 ) on f/> = 0, - - (C/m ) on f/> =nr

.1rr

5.34.

A uniform line charge p, = 2 nC/m lies in the z = 0 plane parallel to the x axis at y =3m. the potential difference V,. 8 for the points A(2 m, 0, 4 m) and B(O, 0, 0) Ans. -18.4 V

S.35.

A uniform sheet of charge, p, = {1/6rr) nC/m2 , is at x = 0 and a second sheet, p. = (-1/6Jr)nC/m\ is at x=lOm. Find V" 8 , V8 c, and VAc for A(lOm,O,O), B(4m,O,O), and C(O,O,O). Ans. -36V,-24V,-60V

5.36.

Given the cylindrical coordinate electric fields E = (5/r)a. (V /m) for 0:5 r :52 m and 2.5a. V /m for r >2m, find the potential difference V" 8 for A(1 m, 0, 0) and B(4 m, 0, 0). Ans. 8.47V

5.37.

A parallel-plate capacitor 0.5 m by 1.0 m, has a separation distance of 2 em and a voltage difference of 10 V. Find the stored energy, assuming that £ = £0 • Ans. 11.1 nJ

Find

CHAP. 5) 5.38.

THE ELECTROSTATIC FIELD: WORK, ENERGY, AND POTENTIAL

The capacitor described in Problem 5.37 has an applied voltage of 200 V. (a) Find the stored energy. Hold d 1 (Fig. 5-16) at 2 em and the voltage difference at 200 V, while increasing d 2 to 2.2cm. Find the final stored energy. [Hint: .6WE = l{.6C)V 2 ) Ans. (a) 4.4 pJ; (b) 4.2 pJ (b)

I•

:=-- .•-.-:-.--.-·

-=... --.

••·

-.-

O.Sm-----1

Fig. 5-16

Q = 2 nC, arranged in a line with

5.39.

Find the energy stored in a system of three equal point charges, 0.5 m separation between them. Ans. 180 nJ

5.40.

Repeat Problem 5.39 if the charge in the center is -2 nC.

5.41.

Four equal point charges, Q = 2 nC, are to be placed at the comers of a square ~ m on a side, one at a time. Find the energy in the system after each charge is positioned. Ans. 0, 108 nJ, 292 nJ, 585 nJ

5.42.

Given the electric field E = -5e-'10&, in cylindrical coordinates, find the energy stored in the volume Ans. 7.89x10- 10a 3 describedby r$.2a and O$.z$.5a.

5.43.

Given a potential V = 3x 2 + 4y 2 (V), 1m, O$.y $.1m, and O$.z $.1m.

Ans.

-108nJ

find the energy stored in the volume described by 0 $, x $. Ans. 147pJ

Chapter 6 Current, Current Density, and Conductors 6.1 INTRODUCfiON Electric current is the rate of transport of electric charge past a specified point or across a specified surface. The symbol I is generally used for constant currents and i for time-variable currents. The unit of current is the ampere (1 A= 1 C/s; in the Sl, the ampere is the basic unit and the coulomb is the derived unit). Ohm's law relates current to voltage and resistance. For simple de circuits, I= VIR. However, when charges are suspended in a liquid or a gas, or where both positive and negative charge carriers are present with different characteristics, the simple form of Ohm's law is insufficient. Consequently, the current density J (A/m 2) receives more attention in electromagnetics than does current /.

6.2 CHARGES IN MOTION Consider the force on a positively charged particle in an electric field in vacuum, as shown in Fig. 6-1(a). This force, F= +QE, is unopposed and results in constant acceleration. Thus the charge moves in the direction of E with a velocity U that increases as long as the particle is in the E field. When the charge is in a liquid or gas, as shown in Fig. 6-l(b), it collides repeatedly with particles in the medium, resulting in random changes in direction. But, for constant E and a homogeneous medium, the random velocity components cancel out, leaving a constant average velocity, known as the drift velocity U, along the direction of E. Conduction in metals takes place by movement of the electrons in the outermost shells of the atoms making up the crystalline structure. According to the electron-gas theory, these electrons reach an average drift velocity in much the same way as a charged particle moving through a liquid or gas. The drift velocity is directly proportional to the electric field intensity, U=11E

where llÂˇ the mobility, has the units m 2/V Âˇ s. Each cubic meter of a conductor contains on the order of IOU atoms. Good conductors have one or two electrons from each atom free to move upon application of the field. The mobility 11 varies with temperature and the crystalline structure of the solid. The particles in the solid have a vibratory motion which increases with temperature. This makes it more difficult for the charges to move. Thus, at higher temperatures the mobility 11 is reduced, resulting in a smaller drift velocity (or current) for a given E. In circuit analysis this phenomenon is accounted for by stating a resistivity for each material and specifying an increase in this resistivity with increasing temperature.

u ~+~Q~Âˇ==~~------~E

(a)

(b) Liquid or gas

Vacuum

Fig. 6-1

CHAP. 6]

CURRENT, CURRENT DENSITY, AND CONDUCTORS

6.3 CONVECTION CURRENT DENSITY J A set of charged particles giving rise to a charge density p in a volume v is shown in Fig. 6-2 to have a velocity U to the right. The particles arc assumed to maintain their relative positions within the volume. As this charge configuration passes a surfaceS it constitutes a convection current, with density J = pU

(A/m 2 )

Of course, if the cross section of v varies or if the density pis not constant throughout v, then J will not be constant with time. Further, J will be zero when the last portion of the volume crosses S. Nevertheless, the concept of a current density caused by a cloud of charged particles in motion is at times useful in the study of electromagnetic field theory.

Fig. 6-2

6.4 CONDUCTION CURRENT DENSITY J Of more interest is the conduction current that occurs in the presence of an electric field within a conductor of fixed cross section. The current density is again given by

J = pU (A/m 2 ) which, in view of the relation

U = 11E,

can be written J=aE

where a= P/1 is the conductivity of the material, in siemens per meter (S/m). In metallic conductors the charge carriers are electrons, which drift in a direction opposite to that of the electric field (Fig. 6-3). Hence, for electrons, both p and 11 are negative, which results in a positive conductivity a, just as in the case of positive charge carriers. It follows that J and E have the same direction regardless of the sign of the charge carriers. It is conventional to treat electrons moving to the left as positive charges moving to the right. and always to report p and 11 as positive. The relation J = aE is often referred to as the point form of Ohm's law. The factor a takes into account the density of the electrons free to move (p) and the relative ease with which they move through the crystalline structure (11)- As might be expected, a is a function of temperature.

s Fig. 6-3

CURRENT, CURRENT DENSITY, AND CONDUCTORS

[CHAP. 6

What electric field intensity and current density correspond to a drift velocity of 6.0 x 10- 4 m/s in a silver conductor? For silver a= 61.7 MS/m and I'= 5.6 x 10 3 m2 /V · s. EXAMPLE 1.

U 6.0 X 10 4 E =-= _ = 1.07 X 10 1J 5.6X 10 -3 1 = aE

= 6.61 x 10

V/m

A/m2

6.5 CONDUCfiVITY a In a liquid or gas there are generally present both posatave and negative ions, some singly charged and others doubly charged, and possibly of different masses. A conductivity expression would include all such factors. However. if it is assumed that all the negative ions are alike and so too the positive ions, then the conductivity contains two terms as shown in Fig. 6-4(a). In a metallic conductor, only the valence electrons are free to move. In Fig. 6-4(b) they are shown in motion to the left. The conductivity then contains only one term, the product of the charge density of the electrons free to move, p, , and their mobility, Jle-

-0 ®---

---- -B --0

a=p_p._ +p+p.+

(b) Conductor

(a) Liquid or gas

Fig. 6-4

A somewhat more complex conduction occurs in semiconductors such as germanium and silicon. In the crystal structure each atom has four covalent bonds with adjacent atoms. However, at room temperature, and upon influx of energy from some external source such as light, electrons can move out of the position called for by the covalent bonding. This creates an electron-hole pair available for conduction. Such materials are called intrinsic semiconductors. Electron-hole pairs have a short lifetime, disappearing by recombination. However, others are constantly being formed and at all times some are available for conduction. As shown in Fig. 6-4(c), the conductivity a consists of two terms, one for the electrons and another for the holes. In practice, impurities, in the form of valence-three or valence-five elements, are added to create p-type and n-type semiconductor materials. The intrinsic behavior just described continues, but is far overshadowed by the presence of extra electrons in n-type, or holes in p-type. materials. Then, in the conductivity a, one of the densities, p, or Ph , will far exceed the other. EXAMPLE 2. Determine the conductivity of intrinsic germanium at room temperature. At 300 K there are 2.5 x 10 1" electron-hole pairs per cubic meter. The electron mobility is I'~ = 0.38 m2 /V · s and the hole mobility is "'" = 0.18 m2 /V · s. Since the material is not doped, the numbers of

electrons and holes are equal. a= N,e(IJ, + p 11 )

= (2.5 X

1019)(1.6 X 10- 19)(0.38 + 0.18) = 2.24 S/m

CHAP. 6)

CURRENT, CURRENT DENSITY, AND CONDUCfORS

6.6 CURRENT I Where current density J crosses a surface S, as in Fig. 6-5, the current I is obtained by integrating the dot product of J and dS. dl=l· dS

Of course, J need not be uniform overS and Sneed not be a plane surface.

Fig. 6-S EXAMPLE 3. Find the current in the circular wire shown in Fig. 6-6 if the current density is J = 15(1- e- 1000')•. (A/m 2 ). The radius of the wire is 2 mm.

ciS ....

---

......

Fig. 6-6

A cross section of the wire is chosen for S.

Then

dl=J·dS

= 15(1- e- 1000')8, • rdrdrpa, and

2xf.002 15(1 - e -•cro..>)r dr drp

L 0

= 1.33 X 10- 4 A= 0.133 rnA

Any surface S which has a perimeter that meets the outer surface of the conductor all the way around will have the same total current, I = 0.133 rnA, crossing it.

CURRENT, CURRENT DENSITY, AND CONDUCTORS

[CHAP. 6

6.7 RESISTANCE R If a conductor of uniform cross-sectional area A and length t, as shown in Fig. 6-7, has a voltage difference V between its ends, then

E=t

aV J=-

and

{

assuming that the current is uniformly distributed over the area A.

The total current is then

= aAV

I=JA

{

Since Ohm's law states that

V = IR,

the resistance is {

R= aA

(Q)

(Note that 1 s-• = 1 Q; the siemens was formerly known as the mho.) This expression for resistance is generally applied to all conductors where the cross section remains constant over the length t. However, if the current density is greater along the surface area of the conductor than in the center, then the expression is not valid. For such nonuniform current distributions the resistance is given by R=

J · dS

aE · dS

If E is known rather than the voltage difference between the two faces, the resistance is given by

f f

E·dl R=---aE·dS

The numerator gives the voltage drop across the sample, while the denominator gives the total current I.

Fig.6-7

EXAMPLE 4. Find the resistance between the inner and outer curved surfaces of the block shown in Fig. 6-8, 7 where the material is silver for which a == 6.17 x 10 S/m. If the same current I crosses both the inner and outer curved surfaces,

k J ==-a, r

and

k E==-a, ar

CHAP. 6)

CURRENT, CURRENT DENSITY, AND CONDUCTORS

J-0.05m

Fig. 6-8

Then (5" = 0.0873 rad}, k

3.0

R= .05 [

-a,Âˇ dra,

o.z or [.0873 k -a,Âˇ r dtj> dza, o

In 15 u(O. 05)(0.0873)

1.01 x

w-s g

tO.t"'Q

6.8 CURRENT SHEET DENSITY K At times current is confined to the surface of a conductor, such as the inside walls of a waveguide. For such a current sheet it is helpful to define the density vector K (in A/m), which gives the rate of charge transport per unit transverse length. (Some books use the notation Is.) Figure 6-9 shows a total current of /, in the form of a cylindrical sheet of radius r, flowing in the positive z direction. In this case, I K=-a 2~rr

at each point of the sheet.

For other sheets, K might vary from point to point.

Fig. 6-9

In general, the current flowing through a contour C within a current sheet is obtained by integrating the normal component of K along the contour.

EXAMPLE 5. A thin conducting sheet lies in the z = 0 plane for 0 < x < 0.05 m. An a>' directed current of 25 A is sinusoidally distributed across the sheet, with linear density zero for x = 0 and x = 0.05 m and maximum at x = 0.025 m (see Fig. 6-10). Obtain an expression forK.

CURRENT, CURRENT DENSITY, AND CONDUCTORS

[CHAP. 6

Fig. 6-10 The data give

K = (k sin 20nx)ay

(A/m). for an unknown constant k.

I = 25 = Ky dx = k or

Then

sin 20nx dx

25 = k/lOn or k = 250n A/m.

6.9 CONTINUITY OF CURRENT Current I crossing a general surface S has been examined where J at the surface was known. Now, if the surface is closed, in order for net current to come out there must be a decrease of positive charge within:

rJ. J · dS = I = - dQdt = - ~at f p dv where the unit normal in dS is the outward-directed normal.

fl·dS ~v

Dividing by

~v,

a pdv =---at

As ~v-+0, the left side by definition approaches V · J, the divergence of the current density, while the right side approaches -apt at. Thus

V·l=--

This is the equation of continuity for current. In it p stands for the net charge density, not just the density of mobile charge. As will be shown below, ap/at can be nonzero within a conductor only transiently. Then the continuity equation, V · J = 0, becomes the field equivalent of Kirchhoff's current law. which states that the net current leaving a junction of several conductors is zero. In the process of conduction, valence electrons are free to move upon the application of an electric field. So, to the extent that these electrons are in motion, static conditions no longer exist. However, these electrons should not be confused with net charge, for each conduction electron is balanced by a proton in the nu"leus such that there is zero net charge in every ~v of the

CHAP. 6)

CURRENT, CURRENT DENSITY, AND CONDUCTORS

material. Suppose, however, that through a temporary imbalance a region within a solid conductor has a net charge density p 0 at time t = 0. Then, since J = aE = (a/ c )0,

v-~D=- ap

c at Now, the divergence operation consists of partial derivatives with respect to the spatial coordinates. If a and c are constants, as they would be in a homogeneous sample, then they may be removed from the partial derivatives.

;<V. D)=-

a ap -p=--

ap a -+-p=O

The solution to this equation is Thus p decays exponentially, with a time constant T = cIa, also known as the relaxation time. At t = T, p has decayed to 36.8% of its initial value. For a conductor T is extremely small, on the order of 10- 19• This confirms that free charge cannot remain within a conductor and instead is distributed evenly over the conductor surface. EXAMPLE 6. Determine the relaxation time for silver, given that o = 6.17 x 107 S/m. Po is placed within a silver block, find p after one. and also after five, time constants. Since E= E 0 , E

=- = 0

10-"J6.1r 6.17 X 107

If charge of density

1.43 X 10 '" S

Therefore at at

t = r: t = 5-r:

P = Poe-•

= 0.368p0

= Pne - = 6.74 x 5

10- -'p"

6.10 CONDUCTOR-DIELECTRIC BOUNDARY CONDITIONS Under static conditions all net charge will be on the outer surfaces of a conductor and both E and D are therefore zero within the conductor. Because the electric field is a conservative field, the line integral of E • dl is zero for any closed path. A rectangular path with corners 1, 2, 3, 4 is shown in Fig. 6-11.

fE·dl+ fE·dl+ fE·dl+ fE·di=O If the path lengths 2 to 3 and 4 to 1 are now permitted to approach zero, keeping the interface between them, then the second and fourth integrals are zero. The path from 3 to 4 is within the 1

~ectric

~1:===r----~~~ctor 4

Fig. 6-11

CURRENT, CURRENT DENSITY, AND CONDUCTORS

conductor where E must be zero.

[CHAP. 6

r r

This leaves

E· dl=

E,dt=O

where £, is the tangential component of E at the surface of the dielectric. can be chosen arbitrarily,

Since the interval 1 to 2

E,= D,=O

at each point of the surface. To discover the conditions on the normal components, a small, closed, right circular cylinder is placed across the interface as shown in Fig. 6-12. Gauss' law applied to this surface gives

fo· or

D • dS +

top

dS=

D • dS +

bottom

Qenc

D • dS

s1de

Ps dS

The third integral is zero since, as just determined, D, = 0 on either side of the interface. The second integral is also zero, since the bottom of the cylinder is within the conductor, where D and E are zero. Then,

D · dS =

top

Dn dS

top

= J.

'J: Ps dS

which can hold only if and

E =Ps n

Fig. 6-U

EXAMPLE 7. The electric field intensity at a point on the surface of a conductor is given by O.Zax- 0.3ay- 0.2a, (V /m). Find the surface charge density at the point. Supposing the conductor to be surrounded by free space,

En= ±lEI= ±0.412 V/m p, =

(36w-") ±0.412) = ±3.64 pC/m .7r (

The ambiguity in sign arises from that in the direction of the outer normal to the surface at the given point.

In short, under static conditions the field just outside a conductor is zero (both tangential and normal components) unless there exists a surface charge distribution. A surface charge does not imply a net charge in the conductor, however. To illustrate this, consider a positive charge at the origin of spherical coordinates. Now if this point charge is enclosed by an uncharged conducting

CURRENT, CURRENT DENSITY, AND CONDUCfORS

CHAP. 6)

spherical shell of finite thickness, as shown in Fig. 6-13(a), then the field is still given by

E= +Q a 4.1rET2

except within the conductor itself, where E must be zero. The coulomb forces caused by +Q attract the conduction electrons to the inner surface, where they create a Ps 1 of negative sign. Then the deficiency of electrons on the outer surface constitutes a positive surface charge density PszÂˇ The electric flux lines '11, leaving the point charge +Q, terminate at the electrons on the inner surface of the conductor, as shown in Fig. 6-13(b ). Then electric flux lines 'II originate once again on the positive charges on the outer surface of the conductor. It should be noted that the flux does not pass through the conductor and the net charge on the conductor remains zero.

Psi

(a)

(b)

Fig. 6-13

Solved Problems 6.1. ....+

An AWG #12 copper conductor has an 80.8mil diameter. A 50-foot length carries a current of 20 A. Find the electric field intensity E, drift velocity U, the voltage drop, and the resistance for the 50 foot length.

Mathcad

Since a mil is 1 ~ inch, the cross-sectional area is 2

A = :rt [(0.0808in)(2.54xl0in m)] = 3.31 x 10 _ 6 m2

Then For copper,

=A= 3.31 x to-"= 6.04 x Int. A/m2

a= 5.8 x 107 S/m. Then J

E=-= 0

6.04 X 10" =1.04x 10 'V/m 5.8 X 107

V = Ef = (1.04 X 10- 1)(50)(12)(0.0254) = 1.59 V

R =I=

1.59 = 7.95 x 10- 2 g 20

The electron mobility in copper is IJ = 0.0032 m2 /V Âˇ s, and since is 5.8 x 107 p =;= 0.0032

1.81 x 1011' C/m 3

a= piJ,

the charge density

CURRENT, CURRENT DENSITY, AND CONDUCfORS

(CHAP. 6

From J = pU the drift velocity is now found as J 6.05 X 10" _ U=-= ow=3.34x 10 4 m/s p 1. 81 X 1

With this drift velocity an electron takes approximately 30 seconds to move a distance of 1 centimeter in the #12 copper conductor.

6.2.

What current density and electric field intensity correspond to a drift velocity of 5.3 x w- 4 m/s in aluminum? For aluminum, the conductivity is

J = pU =; U =

a= 3.82 x 10 S/m

and the mobility is 1-' = 0.0014 m2 /V. s.

3.83 x 107 0.00 (5.3 X 10- 4 ) = 1.45 X 107 A/m 2 14

J E =- =- = 3. 79 X 10-l v /m a 1-'

6.3.

A long copper conductor has a circular cross section of diameter 3.0 mm and carries a current of lOA. Each second, what percent of the conduction electrons must leave (to be replaced by others) a 100 mm length? Avogadro's number is N = 6.02 x lW6 atoms/kmol. The specific gravity of copper is 8.96 and the atomic weight is 63.54. Assuming one conduction electron per atom, the number of electrons per unit volume is N. = ( 6. 02 x 1<Y6 atoms)( 1 kmol )(8 .96 x 103 kg)( 1 electron) atom kmol 63.54 kg m3 = 8.49 x 1Q2" electrons/m3

The number of electrons in a 100 mm length is N=

3 2 )

1t (

3 X 10 2

(0.100)(8.49 X lW") = 6.00 X t(¥2

A 10-A current requires that C) ( 1 ( 10-; 1. 6 x 10 pass a fixed point.

electron) C

= 6.25 x 10 electrons/s

Then the percent leaving the 100 mm length per second is 6.25 X 10 19 (100) = 0.104% per s _ x 6 00 1022

6.4.

What current would result if all the conduction electrons in a !-centimeter cube of aluminum passed a specified point in 2.0 s? Assume one conduction electron per atom. The density of aluminum is 2. 70 x IO-' kg/m 3 and the atomic weight is 26.98 kg/kmol. N. = (6.02 x

and

6.5.

tl.Q !lt

1=-=

1<¥6)(26 ~98 )(2. 70 x

103 ) = 6.02 x

Then

l<Y" electrons/m3

(6.02 X 102" electrons/m')(IO - 2 m)-'(1.6 X 10- 19 C/electron) 2s

4.82kA

What is the density of free electrons m a metal for a mobility of 0.0046 m2 /V · s and a conductivity of 29.1 MS/m?

CHAP. 6}

CURRENT, CURRENT DENSITY, AND CONDUCfORS

Since o = 1-'P·

a 29.1 x 10" p=-= =6.33 x 109 C/m3 ll 0.0046 6.33X 109

N~ = 1.6 X

and

6.6.

w-19

3. 96 X 1<Y" electrons/m

Find the conductivity of n-type germanium (Ge) at 300 K, assuming one donor atom in each lOS atoms. The density of Ge is 5.32 x 103 kg/m3 and the atomic weight is 72.6 kg/kmol. The carriers in an n-type semiconductor material are electrons. Since 1 kmol of a substance contains 6.02 x lOU atoms, the carrier density is given by 1 kmol)( N~= ( 6.02X ln2ll vatoms)( -- - - 5.32 X 1o-' -kg)(electrons) kmol 72.6 kg m3 10" atoms =

4.41

1Cf0 electrons/m3

The intrinsic concentration n1 for Ge at 300 K is 2.5 x 1019 m- 3 • The mass-action law, then gives the density of holes:

n:,

N = ~

Because N~~ Nh, at 300 K is

(2.5 X 1019)2 = 1.42 x 1018 holes/m3 4.41 X 1Cf0

conductivity will be controlled by the donated electrons, whose mobility

o= N.,ep., 6.7.

N,N~

= (4.41 X

Hf0 )(1.6 X 10- 19)(0.38) = 26.8 S/m

A conductor of uniform cross section and 150 m long has a voltage drop of 1.3 V and a current density of 4.65 x lOS A/m2 • What is the conductivity of the material in the conductor? Since E=V/f and J=aE, 4.65 X lW =

6.8.

aG~~)

a= 5.37 x 107 S/m

A table of resistivities gives 10.4 ohm· circular mils per foot for annealed copper. the corresponding conductivity in siemens per meter?

What is

A circular mil is the area of a circle with a diameter of one mil ( 1o- 3 in).

1 cir mil=

.n[ (10;

in)(0.0254~)

= 5.07

10- 10 m2

The conductivity is the reciprocal of the resistivity. in)(0.0254:m)( 1 cir mil ) =<=5.78x107 S/m a= ( - 1- n ft. . )( 1210.4~~ ·ctrmd ft m 507 . x 10 10 m2

6.9.

An A WG #20 aluminum wire has a resistance of 16.7 ohms per 1000 feet. conductivity does this imply for aluminum? From wire tables, a #20 wire has a diameter of 32 mils. A=.n [

2 32 x 10- 3 ] (0.0254) =5.19x10- 7 m2 2

t = (1000 ft)(12 in/ft)(0.0254 m/in) = 3.05 x 1if m

What

CURRENT, CURRENT DENSITY, AND CONDUCfORS

Then from

R = f/aA,

(CHAP. 6

3.05 X l{f a= (16. 7)(5.19 x w-7) = 35.2 MS/m

6.10. In a cylindrical conductor of radius 2 mm, the current density varies with the distance from • the axis according to Find the total current I. I=

l·dS=

JdS=

211[.002

1 0

lWe-400rrdrdtl>

e -400.-

]0.002

= 2.tt(1W) [ ( -40()) 2 ( -400r- 1)

= 7.51 rnA 0

6.11. Find the current crossing the portion of the y = 0 plane defined by 0.1 m and -0.002::5 z ::s; 0.002 m if

= Hf lxl a,.

JJ · r- r-•

-0.1 ::s; x ::s;

(A/m2 )

002

dS =

-0.002

llf lxl a,· dx dza, = 4 rnA

-0 I

6.12. Find the current crossing the portion of the x = 0 plane defined by n/4m and -0.01s,z::s;O.Olm if

J I=

6.13. Given J = 103 sin Ba, cal shell r = 0.02 m.

I • dS =

A/m2

= 100 cos 2yax

[

ln/4

.01

-o_ot

-n

-;r /4 ::5 y

::5

(A/m 2)

100 COS 2yax • dy dzax = 2.0 A

in spherical coordinates, find the current crossing the spheri-

Since I and dS = r 2 sin 8 d(J dtJ>a,

are radial, I=

L2"l" o

IW(0.02fsin 2 8d8dtJ> = 3.95 A

6.14. Show that the resistance of any conductor of constant cross-sectional area A and length f is given by R::::: f/oA, assuming uniform current distribution. A constant cross section along the length t results in constant E, and the voltage drop is

E·di=Ef

If the current is uniformly distributed over the area A, I=

where o is the conductivity.

Then, since

I R

l•dS=JA=aEA

= V/1, f R=-

CURRENT, CURRENT DENSITY, AND CONDUCTORS

CHAP. 6)

6.15. Determine the resistance of the insulation in a length f of coaxial cable, as shown in Fig. 6-14.

(f)

)

Fig. 6-14

Assume a total current I from the inner conductor to the outer conductor. distance r,

Then, at a radial

I 1 J=-=A 2nrt

E=-2nurt

and so

The voltage difference between the conductors is then V.b="

1" b

I I b --dr=--ln2nurf 2nof a

and the resistance is

v 1 b R=-=--ln1

2not

6.16. A current sheet of width 4 m lies in the z = 0 plane and contains a total current of 10 A in a direction from the origin to (1, 3, 0) m. Find an expression for K. At each point of the sheet, the direction of K is the unit vector

a.. + 3a,.

v1o and the magnitude of K is !J1 A/m. Thus

K= IO(a. + 3a,) A/m 4

v1o

6.17. As shown in Fig. 6-15, a current lr follows a filament down the z axis and enters a thin conducting sheet at z = 0. Express K for this sheet.

Fig. 6-15

CURRENT, CURRENT DENSITY, AND CONDUCTORS

(CHAP. 6

Consider a circle in the z = 0 plane. The current lr on the sheet spreads out uniformly over the circumference 2:rrr. The direction of K is a,. Then lr K=-a

2:rtr '

6.18. For the current sheet of Problem 6.17 find the current in a 30" section of the plane (Fig. 6-16).

However, integration is not necessary, since for uniformly distributed current a 30" segment will contain 30"/360° or 1/12 of the total.

6.19. A current I (A) enters a thin right circular cylinder at the top, as shown m Fig. 6-17. Express K if the radius of the cylinder is 2 em.

r Fia· 6-17 On the top, the current is uniformly distributed over any circumference 2:rrr, so that I

znr•'

(A/m)

Down the side, the current is uniformly distributed over the circumference 2:rr(0.02 m), so that I K=--(-a.) 0.04:rt

(A/m)

6.20. A cylindrical conductor of radius 0.05 m with its axis along the z axis has a surface charge density Ps = p 0 /z (C/m 2). Write an expression forE at the surface. Since D"

= p.,

E" = p,/E0 •

At (0.05, t/), z), E=Ea=.!!E...a " ., E z ,

(V/m)

6.11. A conductor occupying the region x

5 has a surface charge density

Write expressions for E and D just outside the conductor.

CHAP. 6)

CURRENT, CURRENT DENSITY. AND CONDUCTORS

The outer normal is -a..

Then, just outside the conductor, Pu D = D,( -a.)= p,( -a,)=,~ (-a,) vy~ + z 2 E=

and

Pn (-a) EnYY 2 + Z 2 x

6.22. Two concentric cylindrical conductors, '" = 0.01 m and rh = 0.08 m, have charge v-f+ densities P.m = 40 pC/m2 and p,h, such that D and E fields exist between the two cylinders but are zero elsewhere. See Fig. 6-18. Find p,h and write expressions for D and E between Mathcad the cylinders.

Fig. 6-18

for

By symmetry, the field between the cylinders must be radial and a function of r only. ra<r<rb,

1d V·D=--(rD,)=O To evaluate the constant c, use the fact that

rD, =c

rdr

D, = D, = p,.

c = (0.01 )(40 x lO

)

=4 x

Then,

r = ru + 0.

10 "C/m

and so

4Xl0" a, r

(C/m 2 )

and

4.52 x

E 11

E=-=

w-z

(V/m)

The density p,b is now found from

Supplementary Problems 6.23.

Find the mobility of the conduction electrons in aluminum. given a conductivity 38.2 MS/m and conduction electron density 1.70 X 102" m 3 . Ans. 1.40 X 10--' m 2 /V · s

6.24.

Repeat Problem 6.23 (a) for copper, where a= 58.0 MS/m silver. where a=61.7MS/m and N,=7.44x 102"m' Ans. (a) 3.21 x 10 'm 2 /V · s; (b) 5.18 x 10 'm 2 /V · s

6.25.

Find the concentration of holes, N, in p-type germanium, where is /l~.=0.18m 2 /V·s. Ans. 3.47XI02 -'m-'

and

N,. = 1.13 x 1029 m-

a= Hf S/m

;

(b) for

and the hole mobility

CURRENT, CURRENT DENSITY, AND CONDUCfORS N~,

(CHAP. 6

6.26.

Using the data of Problem 6.25, find the concentration of electrons, n; = 2.5 x 10 19 m- 3 • Ans. 1.80 x 1015 m- 3

6.27.

Find the electron and hole concentrations in n-type silicon for which o= 10.0S/m, O.l3m 2 /V · s, and n, = 1.5 x lO"'m- 3 • Ans. 4.81 x IefOm- 3 , 4.68 x 1011 m- 3

6.28.

Determine the number of conduction electrons in a 1-meter cube of tungsten, of which the density is 18.8 x 1W kg/m 3 and the atomic weight is 184.0. Assume two conduction electrons per atom. Ans. 1.23 x Ief9

6.19.

Find the number of conduction electrons in a 1-meter cube of copper if o =58 MS/m and 1-' = 3.2 x 10- 3 m2 /V · s. On the average, how many electrons is this per atom? The atomic weight is 63.54 and the density is 8.96 x IW kg/m 3 • Ans. 1.13 x 1029 , 1.33

6.30.

A copper bar of rectangular cross section 0.02 by 0.08 m and length 2.0 m has a voltage drop of 50 mV. Find the resistance, current, current density, electric field intensity, and conduction electron drift velocity. Ans. 2l.61JQ, 2.32kA, 1.45MA/m2 , 25mV/m, 0.08mm/s

6.31.

An aluminum bus bar 0.01 by 0.07 m in cross section and of length 3m carries a current of 300 A. Find the electric field intensity, current density, and conduction electron drift velocity. Ans. 1.12 X 10- 2 v /m, 4.28 X Hf A/m2 • 1.57 X w-s m/s

6.32.

A wire table gives for A WG #20 copper wire at 20 oc the resistance 33.31 Q/km. What conductivity (in S/m) does this imply for copper? The diameter of A WG #20 is 32 mils. Ans. 5.8 x 107 S/m

6.33.

A wire table gives for AWG #18 platinum wire the resistance 1.21 x 10- 3 Q/cm. What conductivity (in S/m) does this imply for platinum? The diameter of AWG #18 is 40 mils. Ans. 1.00 x 107 S/m

6.34.

What is the conductivity of A WG #32 tungsten wire with a resistance of 0.0172 Q/cm? The diameter ofAWG#32is8.0mils. Ans. 17.9MS/m

6.35.

Determine the resistance per meter of a hollow cylindrical aluminum conductor with an outer diameter of 32 mm and wall thickness 6 mm. Ans. 53.4 tJIJ/m

6.36.

Find the resistance of an aluminum foil 1.0 mil thick and 5.0 em square (a) between opposite edges on a square face, (b) between the two square faces. Ans. (a) 1.03mQ; (b) 266pQ

6.37.

Find the resistance of 100ft of A WG #4/0 conductor in both copper and aluminum. has a diameter of 460 mils. Ans. 4.91 mQ, 7.46 mQ

6.38.

Determine the resistance of a copper conductor 2 m long with a circular cross section and a radius of 1 mm at one end increasing linearly to a radius of 5 mm at the other. Ans. 2.20 mQ

6.39.

Determine the resistance of a copper conductor 1 m long with a square cross section and a side I mm at one end increasing linearly to 3 mm at the other. Ans. 5. 75 mQ

6.40.

Develop an expression for the resistance of a conductor of length t if the cross section retains the same shape and the area increases linearly from A to kA over f.

Ans

if the intrinsic concentration is

~-'•

An A WG #4/0

- t (Ink) -oA k -1

6.41.

Find the current density in an A WG #12 conductor when it is carrying its rated current of 30 A. A #12 wire has a diameter of 81 mils. Ans. 9.09 x 10" A/m 2

6.42.

Find the total current in a circular conductor of radius 2 mm if the current density varies with r according to J = lW/r (A/m2 ). Ans. 4.tt A

CHAP. 6)

CURRENT, CURRENT DENSITY, AND CONDUCfORS

6.43.

In cylindrical coordinates, I= lOe- 1oo.-a• (A/m 2 ) for the region 0.01 :S r :S 0.02 m, 0 < z :S 1m. Find the total current crossing the intersection of this region with the plane tP = const. Ans. 2.33 x 10- 2 A

6.44.

Given a current density

in spherical coordinates, find the current crossing the conical strip Ans. 1.38 x 104 A

fJ = n/4,

0.001 ::S r ::S 0.080 m.

6.45.

Find the total current outward directed from a 1-meter cube with one corner at the origin and edges Ans. 3.0 A parallel to the coordinate axes if I= 2x 2ax + 2xy 1ay + 2xya, (A/m2 ).

6.46.

As shown in Fig. 6-19, a 0.03 m, and at fJ = tt/2 the spherical shell and in 265 Ans. -.- a8 (A/m), sm fJ

current of 50 A passes down the z axis, enters a thin spherical shell of radius enters a plane sheet. Write expressions for the current sheet densities K in the plane. 7.96 - a , (A/m) r

Fig. 6-19

6.47.

A filamentary current of I (A) passes down the z axis to z = S x 10- 2 m where it enters the portion 0 :S tP :S tt/4 of a spherical shell of radius S x 10- 2 m. Find K for this current sheet. 80/ Ans. -.-na8 (A/m) 1tSID u

6.48.

A current sheet of density K = 20a. A/m lies in the plane x = 0 and a current density I= lOa. A/m2 also exists throughout space. (a) Find the current crossing the area enclosed by a circle of radius 0.5 m centered at the origin in the z = 0 plane. (b) Find the current crossing the square lxi:S0.25m, IYI::S0.25m, z=O. Ans. (a)27.9A;(b)l2.5A

6.49.

A hollow, thin-walled, rectangular conductor 0.01 by 0.02 m carries a current of 10 A in the positive x direction. Express K. Ans. 167ax A/m

6.50.

A solid conductor has a surface described by x + y =3m and extends toward the origin. surface the electric field intensity is 0.35 V /m. Express E and D at the surface and find p, . Ans. ±0.247(ax+a,) V/m, ±2.19Xl0- 12(ax+ay} C/rn 2 , ±3.10xl0- 12 C/m 2

6.51.

A conductor that extends into the region a surface charge density

z < 0 has one face in the plane z = 0,

p, = 5 X 10- 10e -Hlr sin 2 t/)

in cylindrical coordinates.

At the

over which there is

(C/m 2)

Find the electric field intensity at (0.15 m, tt/3, 0).

Ans.

9.45a. V/m

94 6.52.

CURRENT, CURRENT DENSITY, AND CONDUCfORS

(CHAP. 6

A spherical conductor centered at the origin and of radius 3 has a surface charge density Po cos2 8. Find E at the surface.

Ans.

Po cos 2 fJa, Eo

6.53.

The electric field intensity at a point on a conductor surface is given by E = 0.2a.. - 0.3ay0.2a. V /m. What is the surface charge density at the point? Ans. Âą3.65 pC/m 2

6.54.

A spherical conductor centered at the origin has an electric field intensity at its surface E = 0.53(sin 2 tP)a. V/m in spherical coordinates. What is the charge density where the sphere meets the y axis? Ans. 4.69 pC/m2

Chapter 7 Capacitance and Dielectric Materials 7.1 POLARIZATION P AND RELATIVE PERMITIIVITY

£,

Dielectric materials become polarized in an electric field, with the result that the electric flux density D is greater than it would be under free-space conditions with the same field intensity. A simplified but satisfactory theory of polarization can be obtained by treating an atom of the dielectric as two superimposed positive and negative charge regions, as shown in Fig. 7-l(a). Upon application of an E field the positive charge region moves in the direction of the applied field and the negative charge region moves in the opposite direction. This displacement can be represented by an electric dipole moment, p = Qd, as shown in Fig. 7-l(c).

,F()

---

(a)

: +

(b)

E (c)

F1g. 7-1

For most materials, the charge regions will return to their original superimposed positions when the applied field is removed. As with a spring obeying Hooke's law, the work done in the distortion is recoverable when the system is permitted to go back to its original state. Energy storage takes place in this distortion in the same manner as with the spring. A region liv of a polarized dielectric will contain N dipole moments p. Polarization P is defined as the dipole moment per unit volume:

P = lim Np (C/m 2 ) .av-o liv

This suggests a smooth and continuous distribution of electric dipole moments throughout the volume, which, of course, is not the case. In the macroscopic view, however, polarization P can account for the increase in the electric flux density, the equation being D=£0 E+P This equation permits E and P to have different directions, as they do in certain crystalline dielectrics. In an isotropic, linear material E and P are parallel at each point, which is expressed by P = Xr£oE

(isotropic material)

where the electric susceptibility Xr is a dimensionless constant. D = £o(l

where

£, = 1 + x..

+ X.. )E = £o£,E

is also a pure number.

Then,

(isotropic material)

Since D = £E (Section 3.4), £

£=, £o whence £,is called the relative permittivity.

(Compare Section 2.1.) 95

CAPACITANCE AND DIELECfRIC MATERIALS

(CHAP. 7

EXAMPLE 1. Find the magnitudes of D and P for a dielectric material in which E = 0.15 MV /m =4.25. Since £, = x. + 1 = 5.25,

and

w-9

D = £o£,£ = - - (5.25)(0.15 36.n

Hf') = 6. 961JC/m2

w-9

P = X~£oE = (4.25)(0.15 x Hf') = 5.641JC/m2 36.n

7.2 CAPACITANCE Any two conducting bodies separated by free space or a dielectric material have a capacitance between them. A voltage difference applied results in a charge +Q on one conductor and - Q on the other. The ratio of the absolute value of the charge to the absolute value of the voltage difference is defined as the capacitance of the system: C= ~

(F)

where 1 farad( F) = 1 C/V. The capacitance depends only on the geometry of the system and the properties of the dielectric(s) involved. In Fig. 7-2, charge +Q placed on conductor 1 and -Q on conductor 2 creates a flux field as shown. The D and E fields are therefore also established. To double the charges would simply double D and E, and therefore double the voltage difference. Hence the ratio Q/V would remain fixed.

t F1g. 7-2

EXAMPLE 2. Find the capacitance of the parallel plates in Fig. 7-3, neglecting fringing. Assume a total charge +Q on the upper plate and -Q on the lower plate. This charge would normally be distributed over the plates with a higher density at the edges. By neglecting fringing, the problem is simplified and uniform densities p. = ±Q/A may be assumed on the plates. Between the plates D is uniform, directed from + p. to - p•.

and The potential of the upper plate with respect to the lower plate is obtained as in Section 5.3.

Then C = Q/V = EoE,A/d. Notice that the result does not depend upon the shape of the plates but rather the area, the separation distance, and the dielectric material between the plates.

CHAP. 7]

CAPACITANCE AND DIELECfRIC MATERIALS

Fig. 7-3

7.3 MULTIPLE-DIELECfRIC CAPACITORS When two dielectrics are present in a capacitor with the interface parallel to E and D, as shown in Fig. 7-4(a), the equivalent capacitance can be obtained by treating the arrangement as two capacitors in parallel [Fig. 7-4( b)].

--==-v

(b)

(a)

F1g. 7-4

When two dielectrics are present such that the interface is normal to D and E, as shown in Fig. 7-S(a), the equivalent capacitance can be obtained by treating the arrangement as two capacitors in series [Fig. 7-S(b)).

Ceq

1 Cz

Er2d1

+ E,1d2

-=-+-=-=_..;;..-~...::;

EoErlEr2A

The result can be extended to any number of dielectrics such that the interfaces are all normal to D and E: the reciprocal of the equivalent capacitance is the sum of the reciprocals of the individual capacitances.

CAPACITANCE AND DIELECTRIC MATERIALS

[CHAP. 7

-=-v (b)

(a)

FlJ. 7-5 EXAMPLE 3. A parallel-plate capacitor with area 0.30 m2 and separation 5.5 mm contains three dielectrics with interfaces normal to E and D, as follows: £,1 = 3.0, d 1 = 1.0 mm; £r2 = 4.0, d2 = 2.0 mm; £,3 = 6.0, d3 = 2.5 mm. Find the capacitance. Each dielectric is treated as making up one capacitor in a set of three capacitors in series. C _ £0 £,1A _ £o(3.0)(0.30) _

Similarly,

C2 =5.31nF and

1 -=

ceq

F 7. 96 n

C3 =6.37nF; whence

1 7.96 x 10

1 5.31 x 10

1 6.37 x 10-

9 +---~ 9

Ceq= 2.12 nF

7.4 ENERGY STORED IN A CAPACITOR By Section 5.8, the energy stored in the electric field of a capacitor is given by

WE=~J D·Edv where the integration may be taken over the space between the conductors with fringing neglected. If this space is occupied by a dielectric of relative permittivity E., then D = E0 E,E, giving WE =

E0 E,E2 dv

It is seen that, for the same E field as in free space, the presence of a dielectric results in an increase

in stored energy by the factor energy is given by

E, >

In terms of the capacitance C and the voltage V this stored WE=!CV 2

and the energy increase relative to free space is reflected in C, which is directly proportional to

E,.

7.5 FIXED-VOLTAGE D AND E

A parallel-plate capacitor with free space between the plates and a constant applied voltage V, as shown in Fig. 7-6, has a constant electric field intensity E. With fringing neglected,

CAPACITANCE AND DIELECTRIC MATERIALS

CHAP. 7]

L t Fig. 7-6

Now, when a dielectric with relative permittivity

fills the space between the plates,

E=E0

because the voltage remains fixed, whereas the permittivity increases by the factor

t=r.

EXAMPLE 4.

of voltage.

A parallel-plate capacitor with free space between the plates is connected to a constant source Determine how WE, C, Q, and p_, change as a dielectric of E, = 2 is inserted between the plates. Relationship

Explanation

WE=2WEn C=2Cr, p, = 2p,o Q=2Qo

Section 7.4 C=2WE/V 2 P.v=D.. Q=psA

Insertion of the dielectric causes additional charge in the amount Q0 to be pulled from the constant-voltage source.

7.6 FIXED-CHARGE D AND E The parallel-plate capacitor in Fig. 7-7 has a charge +Q on the upper plate and -Q on the lower plate. This charge could have resulted from the connection of a voltage source V which was subsequently removed. With free space between the plates and fringing neglected,

In this arrangement there is no way for the charge to increase or decrease, since there is no conducting path to the plates. Thus, when a dielectric material is inserted between the plates,

D=D0

1 E=-E0 Er

Fig. 7-7 EXAMPLE 5.

A charged parallel-plate capacitor in free space is kept electrically insulated as a dielectric of relative permittivity 2 is inserted between the plates. Determine the changes in wE. c. and v. Relationship WE=4WEo V= 4Vo C=2C0

(See Problem 7.20.)

Explanation

DÂˇE= !DuÂˇEo V=Ed C=Q/V

CAPACITANCE AND DIELECTRIC MATERIALS

100

[CHAP. 7

7.7 BOUNDARY CONDITIONS AT THE INTERFACE OF TWO DIELECTRICS If the conductor in Figs. 6-11 and 6-12 is replaced by a second, different, dielectric, then the same argument as was made in Section 6.10 establishes the following two boundary conditions:

(1) The tangential component of E is continuous across a dielectric interface. Dtl

and

In symbols,

D,z

(2) The normal component of D has a discontinuity of magnitude IPsl across a dielectric interface. If the unit normal vector is chosen to point into dielectric 2, then this condition can be written D, 1 - D, 2 = -p,

and

Generally the interface will have no free charges

D, 1 =D, 2

,...I+

sa.

Math cad

EXAMPLE 6. Given that E, = 2a, - 3ay + 5az 02 and the angles o, and 02.

(Ps = 0),

so that

and V /m

at the charge-free dielectric interface of Fig. 7-8 find

Fig. 7-8

The interface is a z =canst. plane. The x and y components are tangential and the z components are normal. By continuity of the tangential component of E and the normal component of 0:

2a, -

3ay +

2a. -

3ay + Ez2az

5az

0 1 = E11 E,1E 1 = 4Eu&, - 6E 0 ay + 10Eu&, 02 =

Dx2ax + Dy2ay + 10Eua,

The unknown components are now found from the relation

D2 =

E 11 E,2 E 2 •

from which

10 E,2

£,2=-=2

The angles made with the plane of the interface are easiest found from E, ·a,= lEd cos (90°- 0,) 5 = V38 sino,

o, =

54.2°

E2 · a, = IE2I cos (90°- 0 2) 2 = Vl7 sin fJ 2

02 = 29.0°

101

CAPACITANCE AND DIELECfRIC MATERIALS

CHAP. 7]

A useful relation can be obtained from D •• /£o£.~

v'E~,

+ E: 1

D.z/£o£r2

v'E~z+ E;z

In view of the continuity relations, division of these two equations gives tan 6, = E:r2 tan 6 2 €,1

Solved Problems 7.1.

Find the polarization pin a dielectric material wtih

2.8 if D = 3.0 X

w- 7 8 C/m2 •

Assuming the material to be homogeneous and isotropic,

P=x.EoE Since D = £ 0 £,E

and

x. = E,- 1, P=

7.2..

(£·~ 1)D =

1.93 x 10- 7aC/m2

Determine the value of E in a material for which the electric susceptibility is 3.5 and P=2.3x 10-7aC/m 2 • Assuming that P and E are in the same direction, 1 E=-P=7.42x HfaV/m XeEo

7.3.

Two point charges in a dielectric medium where E, = 5.2 interact with a force of 8.6 x 10-3 N. What force could be expected if the charges were in free space? Coulomb's law, F = Q,Q 2 /(4.1l£0 E,d2 ), shows that the force is inversely proportional to free space the force will have its maximum value.

5.2

-J

Fm••=T(8.6x 10 )=4.47X 10

7.4.

,!!!

-2

is a dielectric material

+ 6az C/m2

find E 2 and the angles 0 1 and 0 2 • The x components are normal to the interface: D .. and £, are continuous.

3 4 6 E 1 =-a. --ay +-a. Eo

Region 1, defined by x < 0, is free space, while region 2, x > 0, for which £,2 = 2.4. See Fig. 7-9. Given 0 1 = 3a,. - 4ay

E,.

102

CAPACITANCE AND DIELECTRIC MATERIALS

(CHAP. 7

Fig. 7-9

whence

1.25

En€,.2

€o

£.2=--=-

Dy2

= -4E,2 = -9.6

To find the angles: D, ·ax= IDd cos (90°- e,) 3 = \161 sine, e, = 22.6°

7.5.

In the free-space region x < 0 the electric field intensity is E 1 = 3ax + 5ay- 3az V /m. The region x > 0 is a dielectric for which e-,2 = 3. 6. Find the angle 6 2 that the field in the dielectric makes with the x = 0 plane. The angle made by E, is found from E, ·a, =lEd cos (90°- 8,) 3= v'43sin e, e, =27.2° Then, by the formula developed in Example 6, tan

7.6.

e2 =-tan e, = 0.1428 E,2

A dielectric free-space interface has the equation 3x + 2y + z = 12m. The origin side of the interface has e-,1 = 3.0 and E 1 = 2ax + 5az V /m. Find E 2 • The interface is indicated in Fig. 7-10 by its intersections with the axes. the free-space side is

= n

3ax +2ay +a, Vl4

The projection of E, on an is the normal component of E at the interface.

The unit normal vector on

CAPACITANCE AND DIELECTRIC MATERIALS

CHAP. 7]

103

Fig. 7-10

Then 11 E, 1 =-=a, =2.36ax + 1.57ay +0.79a. v'l4 Ett = E 1 - E, 1 = -0.36ax- 1.57ay + 4.2la. = E 12 0, 1 =

£ 0 E, 1 E, 1

(7.08ax + 4.71ay + 2.37a.) = 0, 2

1 E, 2 =- 0, 2 = 7.08ax + 4. 7laY + 2.37a, Eo

and finally E 2 = E, 2 + E 12 = 6. 72ax + 3.14ay + 6.58a,

7.7.

V /m

Figure 7-11 shows a planar dielectric slab with free space on either side. constant field E 2 within the slab, show that E 3 = E 1 •

;.. Fig. 7-11

By continuity of E, across the two interfaces,

By continuity of D, across the two interfaces (no surface charges), and so

£,3= £,,

Consequently, E 3 = E 1 •

7.8 •

(a)

Show that the capacitor of Fig. 7-4(a) has capacitance

..-J+

Mat head

(b)

_ E"ot:,lAl eq d

E"oE"r2A2 _ C

+ Cz

Show that the capacitor of Fig. 7-5(a) has reciprocal capacitance

Ceq

t:oEnA/dt

1 l 1 =-+E"ot:,zA/dz C1 C2

Assuming a

CAPACITANCE AND DIELECTRIC MATERIALS

104 (a)

[CHAP. 7

Because the voltage difference Vis common to the two dielectrics, D2

--=--=-a EoErl EoEr2 d n

and

where an is the downward normal to the upper plate. the two sections of the upper plate are

Since

Dn = p,,

the charge densities on

and the total charge is Q -- Psi A

(b)

+ p,2A 2 --

v(EoE,,AI

Thus, the capacitance of the system, Ceq= Q/V, Let +Q be the charge on the upper plate. Then

+ EuEr2A2) d

has the asserted form.

Q D=-a A n

everywhere between the plates, so that

The voltage differences across the two dielectrics are then

and From this it is seen that

7.9. ,.-J+

1/Ceq = V /Q

has the asserted form.

Find the capacitance of a coaxial capacitor of length L, where the inner conductor has radius a and the outer has radius b. See Fig. 7-12.

Mathcad

Fig. 7·12

With fringing neglected, Gauss' law requires that D ac 1/r between the conductors (see Problem 6.22). At r =a, D = Pso where p, is the (assumed positive) surface charge density on the inner conductor. Therefore,

a D = p,-a,

and the voltage difference between the conductors is V.h = a

a (

p,a b -p,a - a ) · d ra =--In€u€,r ,

Eo€,

105

CAPACITANCE AND DIELECTRIC MATERIALS

CHAP. 7]

Q = p,(2:rraL),

The total charge on the inner conductor is

and so

g = 2:rrE

0 E,L In (b/a)

7.10. In the capacitor shown in Fig. 7-13, the region between the plates is filled with a dielectric having ÂŁ, = 4.5. Find the capacitance. z

Fig. 7-13 With fringing neglected, the D field between the plates should, in cylindrical coordinates, be of the form D = D 41 a41 , where D 41 depends only on r. Then, if the voltage of the plate cp =a with respect to the plate cp = 0 is v;h

Thus,

D 41 = -E0 E,\t;ira,

and the charge density on the plate p, = D, = -D4> =

cp =a

E 0:~Vo

The total charge on the plate is then given by

hf.'2EEl/,

I p,dS= L o

E 0 E,V0 h

,.,

~drdz

'â&#x20AC;˘

rex

=---In-

Hence

C= Q = E 0 E,hln~

v;,

When the numerical values are substituted (with a converted to radians), one obtains

C = 7. 76 pF.

7.11. Referring to Problem 7.10, find the separation d which results in the same capacitance when the plates are brought into parallel arrangement, with the same dielectric in between. With the plates parallel C=

E0 E,A

106

CAPACITANCE AND DIELECfRIC MATERIALS

[CHAP. 7

so that a(rz- r1) In (rz/r1) Notice that the numerator on the right is the difference of the arc lengths at the two ends of the capacitor, while the denominator is the logarithm of the ratio of these arc lengths. For the data of Problem 7.10, ar1 = 0.087 mm, ar2 = 2.62 mm, and d = 0. 74 mm.

7.12. Find the capacitance of an isolated spherical shell of radius a. The potential of such a conductor with a zero reference at infinity is (see Problem 2.34)

Q C=-=4JrE0 a

Then

7.13. Find the capacitance between two spherical shells of radius a separated by a distance d ~ a. As an approximation, the result of Problem 7.12 for the capacitance of a single spherical shell, 4Jr£0 a, may be used. From Fig. 7-14 two such identical capacitors appear to be in series.

-=-+c cl c2 c1cz C1+Cz

C = - - - = 2HE 0 a

®)) . ((® Fig. 7-14

7.14. Find the capacitance of a parallel-plate capacitor containing two dielectrics, £,1 = ,... 1+ 1.5 and £ 12 = 3.5, each comprising one-half the volume, as shown in Fig. 7-15. Here, A =2m 2 and d = 10-3 m.

Mathcad

Similarly,

C2 = 31.0 nF.

Then C = C 1 + C 2 = 44.3 nF

Fig. 7-15

CAPACITANCE AND DIELECTRIC MATERIALS

CHAP. 7]

107

7.15. Repeat Problem 7.14 if the two dielectrics each occupy one-half of the space between the .,.. 1+

sa.

plates but the interface is parallel to the plates.

Mathe ad

= EoE,A = E 0 E,A =

(8.854

d/2

10 )(1.5) = . nF 53 1 10 3 /2 X

Similarly, C 2 = 124 nF. Then

c,c, c,+c

C = ---- = 37.2 nF

7.16. In the cylindrical capacitor shown in Fig. 7-16 each dielectric occupies one-half the volume.

Find the capacitance.

L Fig. 7-16

The dielectric interface is parallel to D and E, so the configuration may be treated as two capacitors in parallel. Since each capacitor carries half as much charge as a full cylinder would carry, the result of Problem 7.9 gives C=C,+C2=

where Eravg = ~( E,1 + E,2 ). permittivity.

HEnE,, L

In (bfa)

.1rEoEr2L +~~-,-

2HE 0 €, avgL

In (b/a)

The two dielectrics act like a single dielectric having the average relative

7.17. Find the voltage across each dielectric in the capacitor shown in Fig. 7-17 when the applied

sa. ,.../+

voltage is 200 V. £

Mat head

5(1)

0 cl = - = 5000£,, w-3

1000£ Cz=---11 3 and

-//_..--

...-"'~3 mm

.C./eo-~~/,.~· Fig. 7-17

lmm

108

CAPACITANCE AND DIELECfRIC MATERIALS

[CHAP. 7

The D field within the capacitor is now found from 9

D = p = Q = CV = (2.77 X 10- )(200) " s A A 1

5_54 X 10 _7 C/m2

Then,

£1 = - - = 1.25 X 10 V/m

£2

l:oEr1

=- = Eo

6.25 x 104 V/m

from which

7.18. Find the voltage drop across each dielectric in Fig. 7-18. where t:r 1 = 2.0 and ~:r2 = 5.0. The inner conductor is at r 1 = 2 em and the outer at r2 = 2.5 em, with the dielectric interface halfway between.

Fig. 7-18

The voltage division is the same as it would be for full right circular cylinders. The segment shown, with angle 0', will have a capacitance C¥/2H times that of the complete coaxial capacitor. From Problem 7.9, C 1 = (~) 2HEoE, 1 L = C¥L(l. 5 X 10 2n In (2.25/2.0) C 2 = aL(4.2

Since Q =

cl VI= c2 v2

and

10- 111 )

v, + v2 = v. ~

=----

(F)

it follows that

c. + c2 c V =-- V = 2

10 )

+ C2

4.2 (100) = 74 v 1.5 + 4.2 1.5 (100)=26V l.S + 4.2

7.19. A free-space parallel-plate capacitor is charged by momentary connection to a voltage source V, which is then removed. Determine how WF., D, E. C, and V change as the plates are moved apart to a separation distance d 2 = 2d 1 without disturbing the charge. Relationship

Explanation

D2=D, £2=£.

D=Q/A E= D/E 0

WE2=2WEI C2= ~c~ l/2 = 2Vt

C= EA/d V=Q/C

WE= ~

f Ene dv. and the volume is doubled

CHAP. 7)

109

CAPACITANCE AND DIELECTRIC MATERIALS

7.20. Explain physically the energy changes found in (a) Problem 7.19, (b) Example 5. External work (in the amount WE 1) is done on the system in forcing apart the oppositely charged plates. This work shows up as an increase in internal energy (stored in theE field). (b) The charged plates draw the dielectric slab into the gap. Thus the system performs work (in the amount ~WEo) on the surroundings-specifically, on whatever is guiding the slab into position. The internal energy suffers a corresponding decrease. (a)

7.21. A parallel-plate capacitor with a separation d = 1.0 em has 29 kV applied when free space v-J+

;sa.

Mathe ad

is the only dielectric. Assume that air has a dielectric strength of 30 kV/cm. Show why the air breaks down when a thin piece of glass ( E"r = 6.5) with a dielectric strength of 290 kV /em and thicknesses d2 =0.20cm is inserted as shown in Fig. 7-19.

Atr, E

I.Ocm

Glus,e,

Fig. 7-19

The problem becomes one of two capacitors in series, E0 A C I = 8Xl0

C, =

E 0 E,A

= 125EIV'~

, =

2x10-

3250E 0 A

Then, as in Problem 7.18, VI=

3250 (29) = 27.93 kV 125 + 3250

so that ÂŁ,=

27.93 kV =34.9kV/cm 0 .80 em

which exceeds the dielectric strength of air.

7.22. Find the capacitance per unit length between a cylindrical conductor of radius a= 2.5 em .--1+

sa.

Math cad

and a ground plane parallel to the conductor axis and a distance

= 6.0 m

from it.

A useful technique in problems of this kind is the method of images. Take the mirror image of the conductor in the ground plane, and let this image conductor carry the negative of the charge distribution on the actual conductor. Now suppose the ground ~lane is removed. It is clear that the electric field of the two conductors obeys the right boundary condition at the actual conductor, and, by symmetry, has an equipotential surface (Section 5.6) where the ground plane was. Thus, this field is the field in the region between the actual conductor and the ground plane. Approximating the actual and image charge distributions by line charges +pt and -pe. respectively, at the conductor centers, one has (see Fig. 7-20): Potential at radius a due +p 1 = - ( +Pr) In a 2.n:E 0 Potential at point P due to

-p = - ( 2.n:Eo -pr) In (2h- a) 1

110

CAPACITANCE AND DIELECTRIC MATERIALS

(CHAP. 7

1 I

I I

\ \\\ \:l/ I II hI \ ' \ -..'r<./ It ' .... - --~~ ./ - .-.. . . . -~-p, I

Fig. 7-20

The potential due to -p, is not constant over r =a, the surface of the actual conductor. But it is very nearly so if a~ h. To this approximation, then, the total potential of the actual conductor is p,

2.Jr£o

2HEo

2.Jr£n

2.Jr£o

V. =--lna+--ln(2h-a)"" ---lna+--ln a

2.Jr£o

=--In-

Similarly, the potential of the image conductor is - V.,. Thus, the potential difference between the conductors is 2V.,, so that the potential difference between the actual conductor and the ground plane is ~(2V..) = V.,. The desired capacitance per unit length is then C L

Q/L V.,

- = - - =-

V..

2.JrE =---"-0

In (2h/a)

For the given values of a and h, CIL =9.0pFim. The above expression for C I L is not exact, but provides a good approximation when a ~ h practical case). An exact solution gives

(the

In (dla)

Supplementary Problems 7 .23.

Find the magnitudes of D, P, and £, for a dielectric material in which 4.25. Ans. 6.97~JCim , 5.641JCim2 , 5.25

E = 0.15 MVI m and

x, =

7.Z4.

In a dielectric material with £, =3.6, Ans. 8.94kVIm, 206nCim2 , 2.6

7.15.

Given E=-3a.+4ay-2az

VIm in

region z > 0, for which

= 6.5.

£,

= 285 nCim 2 • the

Ans.

Find the magnitudes of E, P, and

region z < 0, where £, = 2. 0, 4 -3•r +4•)' -6.5 - •z VIm

find

x,. in

the

CHAP. 7]

111

CAPACITANCE AND DIELECTRIC MATERIALS

7.26.

Given that D = 2ax- 4a} + l.Sa, C/m 2 in the region x > 0, which is free space, find P in the Aru. 1.6a,- 16ay + 6a, C/m 2 region x < 0, which is a dielectric with E, = S.O.

7.27.

Region 1, z < 0 m, 2.S. And region 3,

Ans.

is free space where D =Sa, + 7a, C/m 2. Region 2, E, = 3.0. Find E 2 , P 2 , and 6 3 •

0 < z ~ 1 m,

has

z >1m, has

7 ) -1 ( Say +-a, E0 2.S

(V/m), 7.Say +4.2a,

, 2S.02o C!m-,

7.28.

The plane interface between two dielectrics is given by 3x + z = S. On the side including the origin, D1 = (4.Sax + 3.2&,)10- 7 and E,1 = 4.3, while on the other side, E,2 = 1.80. Find £ 1 , £2, D2' and 62. Ans. 1.4S X 10\ 3.37 X 104 , S.37 X 10- 7 , 83.06°

7.29.

A dielectric interface is described by 4y + 3z =12m. The side including the ongm is free space where 0 1 = Bx + 3ay + 2a, p,C/m 2 • On the other side, E,2 = 3.6. Find D 2 and 6 2 • Ans. S.14 p,C/m 2 , 44.4°

7.30.

Find the capacitance of a parallel-plate capacitor with a dielectric of Ans. S.43 nF separation 4.S mm.

7.31.

A parallel-plate capacitor of 8.0 nF has an area 1.51 m2 and separation 10 mm. What separation would Ans. 1.67 mm be required to obtain the same capacitance with free space between the plates?

7.32.

Find the capacitance between the inner and outer curved conductor surfaces shown in Fig. 7-21. Neglect fringing. Ans. 6.86 pF

= 3.0,

area 0.92 m2, and

Fig. 7-21 7.33.

Find the capacitance per unit length between a cylindrical conductor 2. 7S inches in diameter and a Ans. 8.99pF/m (note units) parallel plane 28ft from the conductor axis.

7.34.

Double the conductor diameter in Problem 7.33 and find the capacitance per unit length. Ans. 10.1 pF/m

v-f+

Mathe ad

7.35.

Find the capacitance per unit length between two parallel cylindrical conductors in air, of radius l.S em Ans. 6.92 pF/m and with a center-to-center separation of 8S em.

7.36.

A parallel-plate capacitor with area 0.30 m2 and separation S.S mm contains three dielectrics with interfaces normal to E and D, as follows: E,1 = 3.0, d 1 = 1.0 mm; E,2 = 4.0, d 2 = 2.0 mm; E,3 = 6.0, d 3 = 2.S mm. Find the capacitance. Ans. 2.12 nF

7.37.

With a potential of 1000 V applied to the capacitor of Problem 7.36, find the potential difference and potential gradient (electric field intensity) in each dielectric. Ans. 267V, 267kV/m; 400V, 200kV/m; 333V, I33kV/m

7.38.

Find the capacitance per unit length of a coaxial conductor with outer radius 4 mm and inner radius O.S mm if the dielectric has €, = S.2. Ans. 139 pF/m

112

CAPACITANCE AND DIELECTRIC MATERIALS

[CHAP. 7

7.39.

Find the capacitance per unit length of a cable with an inside conductor of radius 0. 75 em and a cylindrical shield of radius 2.25 em if the dielectric has E, = 2. 70. Ans. 137 pF/m

7.40.

The coaxial cable in Fig. 7-22 has an inner conductor radius of 0.5 mm and an outer conductor radius of 5 mm. Find the capacitance per unit length with spacers as shown. Ans. 45.9 pF/m € ss I ~-f) )1----;:-T-2--Q

_____., f----1 0mm

r= ~

~SOmm Fig. 7-22

7.41.

A parallel-plate capacitor with free space between the plates is charged by momentarily connecting it to a constant 200-V source. After removal from the source a dielectric of E, = 2.0 is inserted. completely filling the space. Compare the values of WE, D, E, P.• , V, and C after insertion of the dielectric to the values before. Partial Ans. V2 = ~ V1

7.42.

A parallel-plate capacitor has its dielectric changed from E, 1 = 2.0 to E,2 = 6.0. It is noted that the stored energy remain~ fixed: W2 = W1 . Examine the changes, if any, in V, C, D, E, Q, and p•. Partial Ans. Ps2 = Y3 Pst

7.43.

A parallel-plate capacitor with free space between the plates remains connected to a constant voltage source while the plates are moved closer together, from separation d to ~d. Examine the changes in Q, P.• , C, D, E, and WE. Partial Ans. D2 = 2D 1

7.44.

A parallel-plate capacitor with free space between the plates remains connected to a constant voltage source while the plates are moved farther apart, from separation d to 2d. Express the changes in D, E, Q, p. , C, and WE . Partial Ans. D2 = ~D 1

7.45.

A parallel-plate capacitor has free space as the dielectric and a separation d. Without disturbing the charge Q, the plates are moved closer together, to d/2, with a dielectric of E, = 3 completely filling Partial Ans. ll; = kV1 the space between the plates. Express the changes in D, E, V, C, and WE.

7.46.

A parallel-plate capacitor has free space between the plates. Compare the voltage gradient in this free space to that in the free space when a sheet of mica, E, = 5.4, fills 20% of the distance between the Ans. 0.84 plates. Assume the same applied voltage in each case.

7.47.

A shielded power cable operates at a voltage of 12.5 kV on the inner conductor with respect to the cylindrical shield. There are two insulations; the first has E,1 = 6.0 and is from the inner conductor at r = 0.8 em to r = 1.0 em, while the second has E,2 = 3.0 and is from r = 1.0 em to r = 3.0 em, the inside surface of the shield. Find the maximum voltage gradient in each insulation. Ans. 0.645 MV/m. 1.03 MV/m

Fig. 7-23

CHAP. 7] 7.48.

CAPACITANCE AND DIELECfRIC MATERIALS

113

A shielded power cable has a polyethylene insulation for which £, = 2.26 and the dielectric strength is 18.1 MV/m. What is the upper limit of voltage on the inner conductor with respect to the shield when the inner conductor has a radius 1 em and the inner side of the concentric shield is at radius of 8 em? Am. 0.376 MV

7.49.

For the coaxial capacitor of Fig. 7-16, a= 3 em, b = 12 em, £,1 = 2.50, El'2 = 4.0. Find E 1 , D,, and 0 2 if the voltage difference is SOV. Partial Ans. ~= ±(36.1/r)a, (V/m)

7.50.

In Fig 7-23, the center conductor, r1 = 1 mm, is at 100 V with respect to the outer conductor at r3 = 100 mm. The region 1 < r <50 mm is free space, while 50< r < 100 mm is a dielectric Ans. 91.8 V, 8.2 V with E, = 2.0. Find the voltage across each region.

7.51.

Fi.nd the stored energy per unit length in the two regions of Problem 7.50. Ans. 59.9nJ/m, 5.30nJ/m

Chapter 8 Laplace's Equation 8.1 INfRODUCfiON

Electric field intensity E was determined in Chapter 2 by summation or integration of point charges, line charges, and other charge configurations. In Chapter 3, Gauss' law was used to obtain D, which then gave E. While these two approaches are of value to an understanding of electromagnetic field theory, they both tend to be impractical because charge distributions are not usually known. The method of Chapter 5, where E was found to be the negative of the gradient of V, requires that the potential function throughout the region be known. But it is generally not known. Instead, conducting materials in the form of planes, curved surfaces, or lines are usually specified and the voltage on one is known with respect to some reference, often one of the other conductors. Laplace's equation then provides a method whereby the potential function V can be obtained subject to the conditions on the bounding conductors.

8.2 POISSON'S EQUATION AND LAPLACE'S EQUATION D

In Section 4.3 one of Maxwell's equations, V • D = p, and -VV=E,

was developed. substituting £E =

V· £(-VV)= p If throughout the region of interest the medium is homogeneous, then

may be removed from the

partial derivatives involved in the divergence, giving p V·VV=--

which is Poisson's equation. When the region of interest contains charges in a known distribution p, Poisson's equation can be used to determine the potential function. Very often the region is charge-free (as well as being of uniform permittivity). Poisson's equation then becomes V 2 V=O

which is Laplace's equation.

8.3 EXPLICIT FORMS OF LAPLACE'S EQUATION

Since the left side of Laplace's equation is the divergence of the gradient of V, these two operations can be used to arrive at the form of the equation in a particular coordinate system. Cartesian Coordinates.

VV=-a ax +-a ay y +-a az z X

and, for a general vector field A,

114

CHAP. 8)

115

LAPLACE'S EQUATION

Hence, Laplace's equation is

CyHndriall Coordinates.

av or

1 av r a,p

av az

VV=-a +--&,p +-a 1

and

1 aA.p

oAz a,p + az

V·A=;a/rA,)+;

so that Laplace's equation is

Spherical Coordinates.

1av

VV=-a +--a6 + - - - a ar r r ao r sin () a,p and

aA<~>

2 V·A=--(rA ) - - - - ( A 6 sin8)+---r2 or r , sin () ao r sin () of/J

so that Laplace's equation is

aj + r 2sin1 () aoa (sinOa~ + 1 ao r 2 sin

2 1 a (r V2 V=-r 2 or or

()

~v = 0 1

ol/J

8.4 UNIQUENESS mEOREM Any solution to Laplace's equation or Poisson's equation which also satisfies the boundary conditions must be the only solution that exists; it is unique. At times there is some confusion on this point due to incomplete boundaries. As an example, consider the conducting plane at z = 0, as shown in Fig. 8-1, with a voltage of 100 V. It is clear that both

ltJ = and

+ 100

V2 = 100

satisfy Laplace's equations and the requirement that V = 100 when Z = 0. The answer is that a single conducting surface with a voltage specified and no reference given does not form the complete boundary of a properly defined region. Even two finite parallel conducting planes do not form a complete boundary, since the fringing of the field around the edges cannot be

.. ~

.. : ~.: ~}:- ~-~~-:

·.

V•IOOV

Fig. 8-1

LAPLACE'S EQUATION

116

(CHAP. 8

determined. However, when parallel planes are specified and it is also stated to neglect fringing, then the region between the planes has proper boundaries.

8.5 MEAN VALUE AND MAXIMUM VALUE mEOREMS Two important properties of the potential in a charge-free region can be obtained from Laplace's equation: At the center of an included circle or sphere, the potential Vis equal to the average of the values it assumes on the circle or sphere. (See Problems 8.1 and 8.2.) (2) The potential V cannot have a maximum (or a minimum) within the region. (See Problem 8.3.)

(1)

It follows from (2) that any maximum of V must occur on the boundary of the region. since V obeys Laplace's equation,

Now,

~v + cfv + cfv = 0 2 2 2

so do av I ax. av I ay, and av I az. Thus, the cartesian components of the electric field intensity take their maximum values on the boundary.

8.6 CARTESIAN SOLUTION IN ONE VARIABLE Consider the parallel conductors of Fig. 8-2, where V = 0 at z =d. Assuming the region between the plates is charge-free,

= 0

and

= 100 V

v2 v = cfv2 + cfv2 + cfv2 =O ax

V= IOOV

Fig. 8-2

With fringing neglected, the potential can vary only with z.

Then

d2V -= 0 2

Integrating,

V=Az +B The boundary condition V = 0 d gives A = 100ld. Thus

z = 0 requires that B = 0.

v = too(~)

(V)

And

V = 100

117

LAPLACE'S EQUATION

CHAP. 8)

The electric field intensity E can now be obtained from 0 100 a 0 E=-VV=-(ava + OYvay + OZva)=-!...(too:.)a =OX z 0Z d z d z

(V/m)

Then

dOO

-daz

(C/m2 )

At the conductors,

where the plus sign applies at z = d and the minus at z = 0.

8.7 CARTESIAN PRODUCT SOLUTION When the potential in cartesian coordinates varies in more that one direction, Laplace's equation will contain more than one term. Suppose that Vis a function of both x andy, and has the special form V = X(x)Y(y). This will make possible the separation of the variables. ~(XY) + ~(XY) =

ox 2

oy 2

becomes or Since the first term is independent of y, and the second of x, each may be set equal to a constant. However the constant for one must be the negative of that for the other. Let the constant be a 2 â&#x20AC;˘ 1 d 2X X dx

- - =2 a 2

1 d2Y Y dy 2

--=-a2

The general solution for X (for a given a) is X= A 1 eax +A 2 e-ax or, equivalently, X= A 3 cosh ax+ A 4 sinh ax and the general solution for Y (for a given a) is Y = B 1ei"Y + B 2 e-jay

or, equivalently,

Y= B 3 cosay + B 4 sinay Therefore, the potential function in the variables x andy can be written

or V = (A3 cosh ax

+ A 4 sinh ax )(B 3 cos ay + B 4 sin ay)

Because Laplace's equation is a linear, homogeneous equation, a sum of products of the above form-each product corresponding to a different value of a-is also a solution. The most general solution can be generated in this fashion.

118

LAPLACE'S EQUATION

Three-dimensional solutions, V = X(x)Y(y)Z(z), there are two separation constants.

(CHAP. 8

of similar form can be obtained, but now

8.8 CYLINDRICAL PRODUCf SOLUTION If a solution of the form

R(r)~(

!!..(r dR) + RZ d ~ + Rtt> d Z dr dr r d,P dz

Dividing by

R~Z

q, )Z(z)

V =

is assumed, Laplace's equation becomes 2

= O

and expanding the r-derivative, 1 d 2R 1 dR 1 d2~ 1 d 2Z - - 2+ - - +2- - = ---= -b 2 2 R dr Rr dr r ~ d,P Z dz 2

The r and q, terms contain no z and the z term contains neither r nor constant, -b 2 , as above. Then

This equation was encountered in the cartesian product solution.

z Now the equation in r and

q, may be r 2 d 2R

The resulting equation in

They may be set equal to a

The solution is

cl cosh bz + c2 sinh bz

further separated as follows:

r dR R dr

--+ --+ b2r2 = 2 R dr

q,.

1 d 2 tt> - - -2 = a2

tt> d,P

q,,

has solution

The equation in r,

is a form of Bessel's differential equation. functions.

Its solutions are in the form of power series called Bessel

R = Csla(br) + C6Na(br) where

la(br) =

L~ -'(-1)m(br/2t+2m ----1 -'---'--~Â

m=O

and

m. r(a + ffl + 1)

Na(br) =(cos an)Ja_(br)- J _a(br) sman

The series la(br) is known as a Bessel function of the first kind, order a; if a = n, an integer, the gamma function in the power series may be replaced by (n + m)!. Na(br) is a Bessel function of the second kind, order a; if a= n, an integer, N.,(br) is defined as the limit of the above quotient as a-n. The function Na(br) behaves like In r near r = 0 (see Fig. 8-3). Therefore, it is not involved in the solution ( C 6 = 0) whenever the potential is known to be finite at r = 0.

CHAP. 8)

LAPLACE'S EQUATION

119

(a)

(b)

Fig. 8-3

For integral order nand large argument x, the Bessel functions behave like damped sine waves: ln(x)= /£cos

(x -~- ~Jt)

N"(x)= /£sin

(x -~- n;)

See Fig. 8-3.

8.9 SPHERICAL PRODUCT SOLUllON Of particular interest in spherical coordinates are those problems in which V may vary with rand 8 but not with l/J. For product solution V = R(r)e(8), Laplace's equation becomes 2

(1d r d r 2r dR) ( R dr 2 + R dr + d8 2 +

de)

e tan 8 d8

The separation constant is chosen as n(n + 1), where n is an integer, for reasons which will become apparent. The two separated equations are 2 dR 2 d R r dr +2r dr -n(n + 1)R =0

and

d 29

d 82 +tan

de d +n(n+1)e=O 8 8

This equation in r has the solution R

= Ctr" + c2,-<n+t)

The equation in 8 possesses (unlike Bessel's equation) a polynomial solution of degree n in the variable ; =cos 8, given by 1 d" P"(;) = 2"n! d;" (;2 -1)"

n =0, 1, 2, ....

The polynomial Pn(;) is the Legendre polynomial of order n. There is a second, independent solution, Q" ( ;), which is logarithmically infinite at ; = ± 1 (i.e., 8 = 0,

Jt).

Solved Problems 8.1.

As shown in Fig. 8-4(a), the potential has the value ltl on 1/n of the circle, and the value 0 on the rest of the circle. Find the potential at the center of the circle. The entire region is charge-free.

LAPLACE'S EQUATION

120

[CHAP. 8 V= V1

V=O

V= V1 (a)

(b)

Fig. 8-4

Call the potential at the center ¥.-. Laplace's equation allows superposition of solutions. If n problems of the type of Fig. 8-4(a) are superposed, the result is the problem shown in Fig. 8-4(b). Because of the rotational symmetry, each subproblem in Fig. 8-4(b) gives the same potential, V... at the center of the circle. The total potential at the center is therefore nYc. But, clearly, the unique solution for Fig. 8-4(b) is V = V1 everywhere inside the circle, in particular at the center. Hence, or

8.2.

V:=v. c

Show how the mean value theorem follows from the result of Problem 8.1. Consider first the special case shown in Fig. 8-5, where the potential assumes n different values on n equal segments of a circle. A superposition of the solutions found in Problem 8.1 gives for the potential at the center

•

• •

F1g.8-S

121

LAPLACE'S EQUATION

CHAP. 8]

which is the mean value theorem in this special case. With

~q:, =

2n/n,

Now, letting n-+"',

12"

1 V,=V(q:,)dq:, 2n " which is the general mean value theorem for a circle. Exactly the same reasoning, but with solid angles in place of plane angles, establishes the mean value theorem for a sphere.

8.3.

Prove that within a charge-free region the potential cannot attain a maximum value Suppose that a maximum were attained at an interior point P. Then a very small sphere could be centered on P, such that the potential V,. at P exceeded the potential at each point on the sphere. Then V,. would also exceed the average value of the potential over the sphere. But that would contradict the mean value theorem.

8.4.

Find the potential function for the region between the parallel circular disks of Fig. 8-6. Neglect fringing Since Vis not a function of r or q:,, Laplace's equation reduces to

and the solution is V = Az + R. The parallel circular disks have a potential function identical to that for any pair of parallel planes. For another choice of axes, the linear potential function might be Ay + B or Ax + B. z

Fig. 8-6

8.5. .,...,+

Two parallel conducting planes in free space are at y = 0 and y = 0.02 m, and the zero voltage reference is at y = 0.01 m. If D = 253ay nC/m2 between the conductors, determine the conductor voltages.

Math cad

From Problem 8.4,

V = Ay + B.

Then D E=-= Eo

-vv = -Aa

253 X 10

----=-=-a =-Aa 8.854 x w-u y y whence A = -2.86 X 104 V /m.

Then,

0 = (-2.86 X 104 )(0.01) + B V = -2.86

and Then, for y = 0,

V = 286 V

and for

= 2.86 X

l0 y + 2.86 X 102 4

y = 0.02,

V = -286 V.

(V)

102 V

122

8.6. ,rJ+

LAPLACE'S EQUATION

[CHAP. 8

The parallel conducting disks in Fig. 8-7 are separated by 5 mm and contain a dielectric for which E, = 2.2. Determine the charge densities on the disks.

Mal head

V:o :!SOV

- - E, "'

2 2 ---, -- l

v-IOOV

Fig. 8-7 Since

V=Az+B. llV 250-100 A =- = llz 5 x 10 ·3

and

-vv =

€oE,E

Since Dis constant between the disks, and

3 X 104 V /m

-3 x lifa, V/m

= -5.84 X

p, = ±5.84

w·-

a, C/m 2

at a conductor surface,

w-? C/m

+ on the upper plate, and - on the lower plate. 8.7.

Find the potential function and the electric field intensity for the region between two concentric right circular cylinders, where V = 0 at r = 1 mm and V = 150 V at r = 20 mm. Neglect fringing. See Fig. 8-8.

ISO V

.... y

Fig. 8-8 The potential is constant with cJ> and z.

Then Laplace's equation reduces to

~!!_ (r dV\ = rdr . d-; J

Integrating once, dV r-==A

and again,

V ==A In r

+ B. Applying the boundary conditions, O=A lnO.OOl + B

150 =A In 0.020 + B

which give A= 50.1,

B = 345.9.

thus

= 50. lin r + 345.9

and

8.8.

123

LAPLACE'S EQUATION

CHAP. 8)

(V)

50 1 Âˇ (-a,) (V/m) r

In cylindrical coordinates two tP = const. planes are insulated along the z axis, as shown in Fig. 8-9. Neglect fringing and find the expression for E between the planes, assuming a potential of 100 V for tP = a and a zero reference at tP = 0.

V=IOOV

V=O

~ 4>=0

4>=a Fig. 8-9

This problem has already been solved in Problem 7.10; here Laplace's equation will be used to obtain the same result. Since the potential is constant with rand z, Laplace's equation is 1d2 V rd4l

--=0

Integrating, V =AlP +B.

Applying the boundary conditions, O=A(O) + B

whence

A=100

100=A(a) + B

B=O

Thus 100 E=-VV=-!~(too.!)a.=a rdtP a ra â&#x20AC;˘

and

8.9.

(V/m)

In spherical coordinates, V = 0 for r = 0.10 m and V = 100 V for r = 2.0 m. Assuming free space between these concentric spherical shells, find E and D. Since V is not a function of 0 or tP. Laplace's equation reduces to

Integrating gives

LAPLACE'S EQUATION

124

[CHAP. 8

and a second integration gives

-A

V=-+B

The boundary conditions give

-A

0=-+B 0.10

whence A= 10.53 V · m,

-A

and

100=-+B 2.00

B = 105.3 V. Then V = -10. r

+ 105.3

(V)

dV 10.53 E=-VV=--a = - - - a

D = eoE =

-9.32 x to-n a, r2

(V/m)

(C/m 2 )

8.10. In spherical coordinates, V = -25 V on a conductor at r = 2 em and V = 150 V ~ at r = 35 em. The space between the conductors is a dielectric for which Er = 3.12. I.!! Find the surface charge densities on the conductors. From Problem 8.9,

-A

V=-+B

The constants are determined from the boundary conditions

-A

-25=-+B 0.02

giving

150=-+B 0.35

-3.71

v = - - + 160.61 (V) r d (-3.71 ) -3.71 E= -VV= - - --+160.61 a,=--a, 2

D= On a conductor surface,

E"oE",E

-0.103

=- r--a, 2

(V/m)

(nC/m 2 )

Dn = p • . -0.103

at r=0.02m:

p. = (0.02Y

at r=0.35m:

p. = (0.

= -256nC/m

+0.103 I 2 ) = +0.837 nC m 35 2

8.11. Solve Laplace's equation for the region between coaxial cones, as shown in Fig. 8-10. A potential V1 is assumed at 8 1 , and V = 0 at 8 2 • The cone vertices are insulated at r=O. •

'*'

The potential is constant with r and 4>.

Integrating

Laplace's equation reduces to

125

LAPLACE'S EQUATION

CHAP. 8)

...

Fig. 8-10

V = A In (tan ~)

and

The constants are found from 0 = A In (tan

~2 ) + B

Hence

8.12. InProblem8.1l,let 6 1 =10°, 6 2 =30°, At what angle 0 is the voltage 50 V?

and

V1 =100V.

Findthevoltageat

6=20°.

Substituting the values in the general potential expression gives

Mathcad

V = -89.34[ln

Then, at For

e = 200,

(0_ 2~e) tan-

100 -89.34ln (tan ) = 37.40 V 0.268 2 50= -89.341n (tan e/ ) 0.268

V =50V,

Solving gives

(tan~) -In 0.268] = -89.34ln

= 17.41°.

8.13. With reference to Problems 8.11 and 8.12 and Fig. 8-11, find the charge distribution on the conducting plane at 62 = 90°. Mathcad

The potential is obtained by Substituting sion of Problem 8.1 1. Thus

e2 = 90°, e, = 10°, and V, = 100 V in the expres-

In (tan~) V=100--ln (tan 5°) Then

1 dV -100 41.05 E=---all= all=--a11 r de (r sin e) In (tan 5°) r sine D-

E- 3.63 x 10-to 8 r sin e "

(C/ z) m

126

LAPLACE'S EQUATION

[CHAP. 8

Fig. 8-11 On the plane 8 = 90째, sin 8 = 1 the direction of D requires that the surface charge on the plane be negative in sign. Hence,

p_, = -

3.63 x

w-w

(C/m2 )

8.14. Find the capacitance between the two cones of Fig. 8-12.

Assume free space .

a .-f+

Math cad

Fig. 8-12

with

If fringing is neglected, the potential function is given by the expression of Problem 8.11 8 1 = 75째, 8 2 = 105째. Thus

(tan~) -In (tan 52.SO)

v = vl =--=-----=---=------,-----. In (tan 37.SO) -In (tan 52.SO) = ( -1.89V1) In

(tan~) + const.

from which D -_

(

1 dV rde

Eo - - - 8 6

) _ 1.89E0 V1 . a, rsme

The charge density on the upper plate is then _ D _ 1.89E0 Vt Ps - n - r sin 75째

127

LAPLACE'S EQUATION

CHAP. 8]

so that the total charge on the upper plate is

and the capacitance is C = Q /V1 = 12.28£o .

8.15. The region between two concentric right circular cylinders contains a unifonn charge density p. Use Poisson's equation to find V. Neglecting fringing, Poisson's equation reduces to

!~(rd!'\ = =e rdr

d-;)

~(rdY\=_pr

d-;)

dV pr r-=--+A dr 2£

Integrating,

dV dr

pr A 2£ r pr2

-=--+V=--+Ainr+B

4£

Note that static problems involving charge distributions in space are theoretical exercises, since no means exist to hold the charges in position against the coulomb forces.

8.16. The region 1l z 1l --<-<2 Zo 2

has a charge density p = w-s cos (z/Zo) (C/m3 ). Elsewhere the charge density is zero. Find VandE from Poisson's equation, and compare with the results given by Gauss' law. Since Vis not a function of x or y, Poisson's equation is

(z I Zo) = 10- z~ cos +Az+ 8 £ 8

Integrating twice, and

E= -VV =

(10-sZos~n (z/Zo)- A

(V)

)•z

(V/m)

But, by the symmetry of the charge distribution, the field must vanish on the plane z = 0. Therefore A = 0 and E

= 10- Zosin (z/Zo) a £

(V/m)

A special gaussian surface centered about z = 0 is shown in Fig. 8-13. D cuts only the top and bottom surfaces, each of area A. Furthermore, since the charge distribution is symmetrical about z = 0, D must be antisymmetrical about z = 0, so that D,op = Daz , Dbouom = D( -az)·

o[ dS+Dl dS=fz JJto- cos(Z/Zo)dxdydz 8

top

bottom

-z

2DA = 2ZoA10- 8 sin (Z/Zo)

128

LAPLACE'S EQUATION

---

[CHAP. 8

ffzo/2 '• ..

.'· . .:::TL~ ·o~J ... -rrz0 /2

Fig. 8-13

or Then, for

= Zo10- 8 sin (z/:zo)

for

0 < z < :JrZo/2

-n:Zo/2 < z < n:Zo/2,

and E = D/£ agrees with the result from Poisson's equation.

8.17. A potential in cylindrical coordinates is a function of r and 4J but not z. Obtain the separated differentia) equations for Rand~. where V = R(r)~(4J), and solve them. The region is charge-free. Laplace's equation becomes d 2R cl> dR R d 2 c!> cl> - + - - +2 - - - 0 dr2 r dr r dqi r 2 d 2R

dR 1 d 2 cl> dr +'R c1r = -idqi r

The left side is a function of r only, while the right side is a function of 4> only; therefore, both sides are equal to a constant, a 2 • 2 2 r d R r dR --+--=a2 Rdr Rdr

d 2R ldR a 2R -+ ----=0 dr2 rdr r

or with solution R = C 1r" + C2 r-".

with solution cl> =c) cos at/J +

•

Also,

c.. sin a.p.

8.18. Given the potential function V = V0 (sinh ax)(sin az) (see Section 8.7), determine the shape 1 and location of the surfaces on which V = 0 and V = V0 • Assume that a > 0. 81

Since the potential is not a function of y, the equipotential surfaces extend to ±oo in the y direction. Because sin az = 0 for z = n:rr/a, where n = 0, 1, 2, ... , the planes z = 0 and z = :rr/a are at zero potential. Because sinh ax= 0 for x = 0, the plane x = 0 is also at zero potential. The V = 0 equipotential is shown as a heavy broken line in Fig. 8-14. The V = Yo equipotential has the equation

Yo = Yo(sinh ax)(sin az)

1 sinaz

Sinax=--

When values of z between zero and :rr/a are substituted, the corresponding x coordinates are readily

129

LAPLACE'S EQUATION

CHAP. 8)

:r---~----------~::::::::::::::::~ V=O

---o~--L_------~-----===~2~::::::::~3:_~x a

Fig. 8-14

obtained.

For example:

1.57 1.57

1.02 2.12

0.67 2.47

0.49 2.65

0.28 2.86

0.10 3.04

0.88

1.0

1.25

1.50

2.00

3.00

The equipotential, which is symmetrical about z = n/'la, is shown as a heavy curve in Fig. 8-14. Because vis periodic in z, and because V( -x, -z) = V(x, z), the whole xz plane can be filled with replicas of the strip shown in Fig. 8-14.

8.19. Find the potentia) function for the region inside the rectangular trough shown in Fig. 8-15.

Fig. 8-15 The potential is a function of x and z, of the form (see Section 8.7)

V =(C. cosh az + C 2 sinh az)(C3 cos ax+ C 4 sin ax) The conditions V = 0 at x = 0 and z = 0 require the constants C 1 and C3 to be zero. Then since V = 0 at x = c, a= nn/c, where n is an integer. Replacing C2 C4 by C, the expression becomes

V=Csinh~sin~ c

or more generally, by superposition, ~ . nnz . nn:x V= £.. Cnsmh-sm"-• c c The final boundary condition requires that

~ (cnsmh. nn;~ sm. nn:x v.o= £.., .. -•

(O<x<c)

LAPLACE'S EQUATION

130

[CHAP. 8

Thus the constants b, =- C, sinh (mrd/c) are determined as the coefficients in the Fourier sine series for f(x) .. V0 in the range 0 < x <c. The well-known formula for the Fourier coefficients, 2 [ f(x) sin-dx nm b, =c 0 c

2Vo1c sm-dx= . nm b,=c o c 4Vo/mr

gives The potential function is then

=L , odd

for O<x<c,

, = 1, 2, 3, ... n even , odd

4V0 sinh (n.nz/c) sin~ n.n sinh (n.nd/c) c

O<z<d.

8.10. Identify the spherical product solution V

(Section 8.9, with C 1 = 0,

= 1)

_ C2

n (

-~rl COS

>_ C2 cos 8 -

r'l

with a point dipole at the origin.

Figure 8-16 shows a finite dipole along the z axis, consisting of a point charge +Q at z = +d/2 and a point charge -Qat z = -d/2. The quantity p = Qd is the dipole moment (Section 7.1). The potential at point Pis

A point dipole at the origin is obtained in the limit as d--+ 0.

For small d,

Therefore, in the limit,

which is the spherical product solution with c2 =p/4nÂŁo. Similarly, the higher-order Legendre polynomials correspond to point quadrupoles, octupoles, etc.

z p

Fig. 8-16

Supplementary Problems 8.11.

In cartesian coordinates a potential is a function of x only. At x = -2.0 em, V = 25.0 V and E = 1.5 x let'( -a.. ) V/m throughout the region. Find Vat x = 3.0 em. Aru. 100 V

CHAP. 8)

131

LAPLACE'S EQUATION

8.22.

In cartesian coordinates a plane at z = 3. 0 em is the voltage reference. Find the voltage and the charge density on the conductor z = 0 if E = 6.67 x 103 a, VIm for z > 0 and the region contains a dielectric for which E, = 4.5. Ans. 200 V, 266 nC/m 2

8.23.

In cylindrical coordinates. V = 75.0 V at r = 5 mm and V =0 at r=60mm. voltage at r = 130 mm if the potential depends only on r. Ans. -23.34 v

8.24.

Concentric, right circular, conducting cylinders in free space at r = 5 mm and r = 25 mm have voltages of zero and V0 , respectively. If E = -8.28 x 103 a, V /m at r = 15 mm, find V0 and the charge density on the outer conductor. Ans. 200 V, +44 nC/m 2

8.25.

For concentric conducting cylinders, V = 75 V at r = 1 mm and V in the region between the cylinders, where t:, = 3.6. Ans. (798/r)a,

= 0 at

Find

the

= 20 mm. Find D

(pC/m 2 )

8.26.

Conducting planes at cp = 10" and cp = 0" in cylindrical coordinates have voltages of 75 V and zero, respectively. Obtain D in the region between the planes, which contains a material for which E, = 1.65. Ans. ( -6.28/r)a, (nC/m2 )

8.27.

Two square conducting planes 50 em on a side are separated by 2.0 em along one side and 2.5 em along the other (Fig. 8-17). Assume a voltage difference and compare the charge density at the center of one plane to that on an identical pair with a uniform separation of 2.0 em. Ans. 0.89

l_ 2 Oc;m

50 em E

T Fig. 8-17 8.28.

The voltage reference is at r = 15 mm in spherical coordinates and the voltage is V0 at r = 200mm. Given E=-334.7a,V/m at r=llOmm, find V0 . The potential is a function of r only. Ans. 250 V

8.29.

In spherical coordinates, V = 865 V at r =50 em and E = 748.2a, V /m mine the location of the voltage reference if the potential depends only on r.

8.30.

With a zero reference at infinity and V = 45.0 V at r = 0.22 m in spherical coordinates, a dielectric of E, = 1. 72 occupies the region 0.22 < r < 1.00 m and free space occupies r > 1.00 m. Determine D at r = 1.00 Âą 0 m. Ans. 8.55V/m, 14.7V/m

8.31.

In Fig. 8-18 the cone at

e = 45Â°

at r = 85 em. DeterAns. r = 250 em

has a voltage V with respect to the reference at

Fig. 8-18

e = 30Â°.

LAPLACE'S EQUATION

132 0.25 m and 6 = 30°, Ans. 125.5V 8.32.

E = -2.30 x Hfa8 V /m.

Determine the voltage difference V.

In Problem 8.31 determine the surface charge densities on the conducting cones at 30o and 45°, if -12.5 2 8.84 2 e, = 2.45 between the cones. Ans. - - (nC/m ), (nC/m ) r

8.33.

(CHAP. 8

Find E in the region between the two cones shown in Fig. 8-19.

Ans.

0.288¥1 rsin 6

(V/m)

l'lg. 8-19

8.34.

In cylindrical coordinates, p = 111/r (pC/m 3). Given that V = 0 at r 50 V at r = 3.0 m due to this charge configuration, find the expression for E.

Ans.

( 12.5- ~

)a,

= 1.0 m

and

V ==

(V/m)

8.35.

Determine E in spherical coordinates from Poisson's equation, assuming a uniform charge density p. pr Ans. ( £ -~ a, 3

8.36.

Specialize the solution found in Problem 8.35 to the case of a uniformly charged sphere. Ans. See Problem 2.54.

8.37.

Assume that a potential in cylindrical coordinates is a function of r and z but not t/J, V = R(r)Z(z). Write Laplace's equation and obtain the separated differential equations in rand z. Show that the solutions to the equation in r are Bessel functions and that the solutions in z are exponentials or hyperbolic functions.

8.38.

Verify that the first five Legendre polynomials are

Po(cos 6) = 1

P1 (cos 6) =cos 6 P2 (cos 6) = !{3 cos2 6 -1) P3 (cos 6) = !{5 cos3 6- 3 cos 6) P.(cos 6) = A(35 cos4 6- 30 cos 2 6 + 3) and graph them against

C= cos 6.

Ans.

See Fig. 8-20.

LAPLACE'S EQUATION

CHAP. 8)

133

Fig. 8-ZO

8.39.

Obtain E for Problem 8.18 and plot several values on Fig. 8-14. Note the orthogonality of E and the AILS. E = - Yoa[(coshax)(sin az)a.. +(sinh ax)(cos az)az) equipotential surfaces.

8.40.

Given V= Vu(coshax)(sinay), where a>O, determine the shape and location of the surfaces on which V = 0 and V =Yo. Make a sketch similar to Fig. 8-14. AILS. See Fig. 8-21.

3 a

V=O

2 a

V=O Fig. 8-ll

t~J=O

Fig. 8-U

t/J=w

3 a

134

LAPLACE'S EQUATION

(CHAP. 8

8.41.

From the potential function of Problem 8.40, obtain E and plot several values on the sketch of the equipotential surfaces, Fig. 8-21. AILS. E =- V0 a((sinhax)(sin ay)a.., +(cosh ax)(cos ay)ay)

8.41.

Use a superposition of the product solutions found in Problem 8.17 to obtain the potential function for 4V0 r"- (a2 /r)" . thesemicircularstripshowninFig.8-22. AILS. V= L b" ( 2 /b)"smmp nocld

n:n

Chapter 9 Ampere's Law and the Magnetic Field 9.1 INTRODUCfiON

A static magnetic field can originate from either a constant current or a permanent magnet. This chapter will treat the magnetic fields of constant currents. Time-variable magnetic fields, which coexist with time-variable electric fields, will be examined in Chapters 12 and 13.

9.2 BIOT-SAVART LAW A differential magnetic field strength, dH, results from a differential current element I dl. The field varies inversely with the distance squared, is independent of the surrounding medium, and has a direction given by the cross product of I dl and aR. This relationship is known as the Biot-Savart law:

dH=ldiXaR (A/m) 4~rR 2

The direction of R must be from the current element to the point at which dH is to be determined, as shown in Fig. 9-1.

,,,, ,, ,,

Fig. 9-1

Current elements have no separate existence. All elements making up the complete current filament contribute to H and must be included. The summation leads to the integral form of the Biot-Savart law:

A closed line integral is required to ensure that all current elements are included (the contour may close at oo). EXAMPLE 1. An infinitely long, straight, filamentary current I along the z axis in cylindrical coordinates is

135

136 shown in Fig. 9-2.

AMPERE'S LAW AND THE MAGNETIC FIELD A point in the

[CHAP. 9

z = 0 plane is selected with no loss in generality. dH

In differential form,

I dz az X (ra,- zaz) 4n(r2 + z2f12 I dz ra., 4n(r2 + z 2)3'2

The variable of integration is z. integrating.

Since

a<~>

does not change with z, it may be removed from the integrand before

Fig. 9-2 This important result shows that H is inversely proportional to the radial distance. The direction is seen to be in agreement with the "right-hand rule" whereby the fingers of the right hand point in the direction of the field when the conductor is grasped such that the right thumb points in the direction of the current.

EXAMPLE 2. An infinite current sheet lies in the z

0 plane with

K = Ka.,

as shown in Fig. 9-3. Find H.

////// K

Fig. 9-3 The Biot-Savart law and considerations of symmetry show that H has only an x component, and is not a function of x or y. Applying Ampere's law to the square contour 12341, and using the fact that H must be

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9]

137

antisymmetric in z,

H · dl

Thus, for all z > 0,

= (H)(2a) + 0 + (H)(2a) + 0 = (K)(2a)

= (K /2)a" .

H=!5_ 2

More generally, for an arbitrary orientation of the current sheet, H=!Kxa"

Observe that H is independent of the distance from the sheet. Further, the directions of H above and below the sheet can be found by applying the right-hond rule to a few of the current elements in the sheet.

9.3 AMPERE'S LAW The line integral of the tangential component of the magnetic field strength around a closed path is equal to the current enclosed by the path:

fu ·

dl=lenc

At first glance one would think that the law is used to determine the current I by an integration. Instead, the current is usually known and the law provides a method of finding H. This is quite similar to the use of Gauss' law to find D given the charge distribution. In order to utilize Ampere's law to determine H there must be a considerable degree of symmetry in the problem. Two conditions must be met: 1. At each point of the closed path His either tangential or normal to the path. 2. H has the same value at all points of the path where His tangential. The Biot-Savart law can be used to aid in selecting a path which meets the above conditions. In most cases a proper path will be evident. EXAMPLE 3. Use Am¢re's law to obtain H due to an infinitely long, straight filament of current I. The Biot-Savart law shows that at each point of the circle in Fig. 9-2 H is tangential and of the same magnitude. Then

fn·dl=H(2nr)=l so that

I H=-a 2nr "'

9.4 CURL

The curl of a vector field A is another vector field. Point Pin Fig. 9-4 lies in a plane area AS bounded by a closed curve C. In the integration that defines the curl, C is traversed such that the

Fig.9-4

138

AMPERE'S LAW AND THE MAGNETIC FIELD

(CHAP. 9

enclosed area is on the left. The unit normal a,, determined by a right-hand rule, is as shown in the figure. Then the component of the curl of A in the direction a, is defined as

fA·dl

(curiA)•a, = lim--t.s-.o AS In the coordinate systems, curl A is completely specified by its components along the three unit vectors. For example, the x component in cartesian coordinates is defined by taking as the contour C a square in the x = const. plane through P, as shown in Fig. 9-5.

fA·dl (curl A)· a..,=

lim

t.y

t.z-.o

Ay Az

Fig. 9-S

If A= A..,a.., +A yay + A .. a..

at the comer of AS closest to the origin (point 1), then

aA .. aA ay az

(curiA) ·a..,=---Y

and

The y and z components can be determined in a similar fashion.

Combining the three components,

aA.. aAy) aAa .. ) + (aAy aA..,) curiA= (--a . . , + (aA.., ----a.. ay

ax y

(cartesian)

A third-order determinant can be written, the expansion of which gives the cartesian curl of A.

a.., curl A=

...

a a a ax ay az Ax Ay A..

the elements of the second row are the components of the del operator. This suggests (see Section 1.3) that V X A can be written for curl A. As with other expressions from vector analysis, this

CHAP. 9)

AMPERE'S LAW AND THE MAGNETIC FIELD

139

convenient notation is used for curl A in other coordinate systems, even though V is defined only in cartesian coordinates. Expressions for curl A in cylindrical and spherical coordinates can be derived in the same manner as above, though with more difficulty. curiA=(! aA .. _ aA•)a,+ (aA, _ aA .. )a +![a(rA) _ r aq, az az ar r ar

aA,]a.. (cylindrical) aq, curl A= _1_ [a( At sin 8) _ aA 8 ]a +! [-1- aA, _ a(rA•)]a +! [ a(rA 8 ) _ aA,] rsin 8 a8 aq, r sin 8 aq, ar r ar a8 .. (spherical) 8

Frequently useful are two properties of the curl operator: (1)

The divergence of a curl is the zero scalar; that is, V· (Vx A)=O

for any vector field A. (2)

The curl of a gradient is the zero vector; that is, VX(Vf)=O

for any scalar function of position f (see Problem 9.20). Under static conditions, E = - VV,

and so, from (2), VXE=O

9.5 RELATIONSffiP OF J AND H In view of Am¢re's law, the defining equation for (curl H)x (see Section 9.4) may be rewritten as (curlH)·ax=

lim

~=lx

ll.yll.z--+Oliy liz

where lx = dlxldS is the area density of x-directed current. Thus the x components of curl Hand the current denisty J are equal at any point. Similarly for the y and z components, so that VXH=J

This is one of Maxwell's equations for static fields. will produce J for that region.

If H is known throughout a region, then V X H

A long, straight conductor cross section with radius a has a magnetic field strength H (/r/2na 2 )a. within the conductor (r <a) and H = (//2nr)a. for r >a. Find J in both regions. Within the conductor,

EXAMPLE 4.

which corresponds to a current of magnitude I in the cross-sectional area tra 2• Outside the conductor, J= VXH=

+ z direction which is distributed uniformly over the

-~(_!_)a, +!~(.!...)a. =0 az 2trr

r ar 2n

140

AMPERE'S LAW AND THE MAGNETIC FIELD

(CHAP. 9

9.6 MAGNETIC FLUX DENSITY B LikeD, the magnetic field strength H depends only on (moving) charges and is independent of the medium. The force field associated with H is the magnetic flux density B, which is given by B=pH

where

= p 0 p,

is the permeability of the medium.

The unit of B is the tesla,

N 1T=1-A·m

The free-space permeability p 0 has a numerical value of 4Jt x 10-7 and has the units henries per meter, Him; p, , the relative permeability of the medium, is a pure number very near to unity, except for a small group of ferromagnetic materials which will be treated in Chapter 11. Magnetic flux, 4>, through a surface is defined as 41>=

B·dS

The sign on Cl> may be positive or negative depending upon the choice of the surface normal in dS. The unit of magnetic flux is the weber, Wb. The various magnetic units are related by 1 T= 1 Wb/m2

1H=1 Wb/A

EXAMPLE 5. Find the flux crossing the portion of the plane 4> = n/4 defined by 0.01 < r < 0.05 m and 0 < z <2m (see Fig. 9-6). A current filament of 2.50 A along the z axis is in the a. direction. #Jol B=#JoH=-a41 :JrT

dS=drdza 41 ~=

12f.os ° o

0.01

#A 1 - a41 ·drdza41 2:JrT

21J 0 l 0.05 =--ln2n 0.01 = 1.61 x 10- 6 Wb

or 1.61 IJWb

It should be observed that the lines of magnetic flux Cl> are closed curves, with no starting point or termination point. This is in contrast with electric flux 'I'. which originates on positive charge

8 X

2.50 A

GIAP. 9]

AMPERE'S LAW AND THE MAGNETIC FIELD

141

and terminates on negative charge. In Fig. 9-7 all of the magnetic flux ~ that enters the closed surface must leave the surface. Thus B fields have no sources or sinks, which is mathematically expressed by V·B=O (see Section 4.1). dS

Fig. 9-7

9.7 VECTOR MAGNETIC POTENTIAL A Electric field intensity E was first obtained from known charge configurations. Later, electric potential V was developed and it was found that E could be obtained as the negative gradient of V, i.e., E = - VV. Laplace's equation provided a method of obtaining V from known potentials on the boundary conductors. Similarly, a vector magnetic potential, A, defined such that VXA=B serves as an intermediate quantity, from which B, and hence H, can be calculated. Note that the definition of A is consistent with the requirement that V • B = 0. The units of A are Wb/m or T·m. If the additional condition V·A=O is imposed, then vector magnetic potential A can be determined from the known currents in the region of interest. For the three standard current configurations the expressions are as follows. current filament: sheet current: volume current:

A=J,Illdl T 4nR

A-J

llKdS s 4nR

A= JIll dv v 4nR

Here, R is the distance from the current element to the point at which the vector magnetic potential is being calculated. Like the analogous integral for the electric potential (see Section 5.5), the above expressions for A presuppose a zero level at infinity; they cannot be applied if the current distribution itself extends to infinity. EXAMPLE 6. Investigate the vector magnetic potential for the infinite, straight, current filament I in free space. In Fig. 9-8 the current filament is along the z axis and the observation point is (x, y, z). The particular

142

AMPERE'S LAW AND THE MAGNETIC FIELD

[CHAP. 9

current element I dl ==I dlaz

is shown, where tis the running variable along the z axis.

It is clear that the integral

- [ 1J0 ldl

A-

4 R az 1[

does not exist, since, when t is large, R = t. This is a case of a current distribution that extends to infinity. It is possible, however, to consider the differential vector potential

and to obtain from it the differential B. Thus, for the particular current element at dA= and

t = 0,

1J0 ldl a 4n(xz + yz + z2)112 z

1J0 ldl[ -y X dB == V X dA ==-a+ a ] (x2 + y2 + zJ)J/2 x (x2 + y2 + z2)312 Y 4.7r

This result agrees with that for dH = (1/l'o) dB given by the Biot-Savart law. For a way of defining A for the infinite current filament, see Problem 9.17.

9.8 STOKES' mEOREM Consider an open surfaceS whose boundary is a closed curve C. Stokes' theorem states that the integral of the tangential component of a vector field F around C is equal to the integral of the normal component of curl F over S:

fF路dl= L(VXF)路dS IfF is chosen to be the vector magnetic potential A, Stokes' theorem gives

fA路dl=

B路dS=fll

Solved Problems 9.1.

Find H at the center of a square current loop of side L. Choose a cartesian coordinate system such that the loop is located as shown in Fig. 9-9.

143

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9)

L/2

t, -L/2

-L/2

L/2

Fig. 9-9 symmetry, each half-side contributes the same amount to Hat the center. L/2, y = - L/2, the Biot-Savart law gives for the field at the origin

For the half-side O::sx ::s

dH =(I dx a.-) x [-xa.- + (L/2)ay) 4n[x 2 + (L/2) 2)312 I dx(L/2)a.

Therefore, the total field at the origin is Lf2

H= 8 o

I dx(L/2)a. 4n[x2 + (L/2)3]312

2V21

2V21 = nL a. = nL a" where a" is the unit normal to the plane of the loop as given by the usual right-hand rule.

9.2.

A current filament of 5.0 A in the ay direction is parallel to the y axis at x =2m, z = (Fig. 9-10). Find Hat the :rigin .

~cad -2m

. -· Fig. 9-10 The expression for H due to a straight current filament applies, I H=-a

2nr •

where

r=2V2 and (use the right-hand rule)

a.. +a.

a=--

•

v'2

Thus

9.3.

[CHAP. 9

AMPERE'S LAW AND THE MAGNETIC FIELD

144

5.0 2n:(2VZ)

(ax+ az) = (0. 2Sl)(ax +az) A/m V2

A current sheet, K = lOaz A/m, lies in the x = 5 m -lOaz A/m, is at x = -5 m. Find Hat all points.

plane and a second sheet,

In Fig. 9-11 it is apparent that at any point between the sheets, sheet. Then, for -5 <x < 5, H = 10( -ay) A/m. Elsewhere H = 0.

t K

K X a,= -KaY

for each

lOa,

a,. y--

lOa,

-a,

Fig. 9-11

9.4.

A thin cylindrical conductor of radius a, infinite in length, carries a current /. points using Ampere's law.

Find H at all

The Biot-Savart law shows that H has only a tP component. Furthermore. Hq, is a function of r only. Proper paths for Ampere's law are concentric circles. For path 1 shown in Fig. 9-12.

H â&#x20AC;˘ dl = 2n:rH4> =

/cnc

and for path 2,

H Âˇ dl = 2n:rH<~> = I

Thus, for points within the cylindrical conducting shell, H = 0, and for external points, (I/2n:r)aq,, the same field as that of a current filament I along the axis.

Fig. 9-U

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9]

9.5.

145

Determine H for a solid cylindrical conductor of radius a, where the current I is uniformly distributed over the cross section. Applying

Am~re's

law to contour 1 in Fig. 9-13,

fn·dl=lcnc H(2nr) = {::) lr H=2~·~ na

For external points,

H= (l/2nr)a~.

Fig. 9-13

9.6.

In the region 0 < r < 0.5 m, in cylindrical coordinates, the current density is

J = 4.5e-2raz (A/m 2 ) and J = 0 elsewhere.

Use Ampere's law to find H.

Because the current density is symmetrical about the origin, a circular path may be used in Thus, for r < 0.5 m,

Am~re's law, with the enclosed current given by ~ J • dS.

H~(2nr)= L:zn H

For any

•

4.5e-2rrdrdt/J

1.125 =-,(1- e- 2r- 2re- ')a~ 2

(A/m)

r :2::0.5 m the enclosed current is the same, 0.594n A. Then H~(2nr) =

9.7.

0.594n

Find H on the axis of a circular current loop of radius a. the loop .

0.297 a

(A/m)

Specialize the result to the center of

For the point shown in Fig. 9-14,

R= -aa,+ha. dH = (Ia dt/J •~) X ( -aa, + ha.) = (Ia dt/J)(aa. + ha,) 4n(a 2 + h~ 312 4n(a 2 + h 2 ) 312 Inspection shows that diametrically opposite current elements produce r components which

146

AMPERE'S LAW AND THE MAGNETIC FIELD

[CHAP. 9

z h

...... 9-14

cancel.

Then,

At h =0,

9.8.

H = (I/2a)a,.

A current sheet, located at y = 0, at (0, 0, 1.5) m.

K = 6.0ax A/m, lies in the z = 0 plane and a current filament is z = 4 m, as shown in Fig. 9-15. Determine I and its direction if H = 0 z

·"'

K"' 6.0ax

Fig. 9-15 Due to the current sheet, 6.0 1 H=-KXa =-(-a 2 n 2 )' )A/m For the field to vanish at (0, 0, 1.5) m, IHI due to the filament must be 3.0 A/m. I

IHI= 2:rcr I

J.O =2:rc(2.5) I= 47.1 A

To cancel the H from the sheet, this current must be in the a.. direction, as shown in Fig. 9-15.

9.9.

Given A= (y cos ax )ax + (y + ex)az ,

find V X A at the origin.

a..

lly

a ax

a ay

y cos ax

a az y+e-

VXA=

At (0,0,0), VXA=a.. -ay -a•.

= a.. - e•a,. -cos axa,

147

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9]

9.10. Calculate the curl of H in cartesian coordinates due to a current filament along the z axis with current I in the az direction. From Example 1, H-~a _.!._(-ya.., +xay) - 2nr • - 2n x2 + y 2

and so

VXH=

a..

a ox

-Oy

a oz

-y

x2+y2 x2+y2

x-) - -y-)] az VXH= [- a ~ ax 2 + Y2 aya ~2 + Y2

=0 except at x = y

= 0.

This is consistent with V X H =J.

9.11. Given the general vector field A = 5r sin <J>az (2, :Tr, 0).

in cylindrical coordinates, find curl A at

Since A has only a z component, only two partials in the curl expression are nonzero. V X A=.; a~ (5r sin 4' )a, - :, (5r sin sin 4' )a. = 5 cos t/Ja, - 5 sin t~Ja.

VxAI

Then

-5a,

(2.n.O)

A= 5e-r cos <Par- 5 cos <J>az

9.12. Given the general vector field curl A at (2, 3:~r/2, 0).

VX A= !~(-5cos t/J)a, + r~

in cylindrical coordinates, find

[~(5e-'cos t/')-~(-5cos t/')]a _.!~(5e-• cos t/J)az ~

•

=(;sin 4' )a,+ (;e-'sin 4' )az Then

9.13. Given the general vector field (2, :Tr/2, 0). V

1 [ A= ---:-- r~ 8

= -2.50a,- 0.34az (2.3JII2.0)

A = 10 sin 6a9

in spherical coordinates, find V X A at

a (10 SID. 8) ]a,+-1a . 10 sin 8 a.1. (lOr SID 8)a. =--a., ~ r~ r

VxAI

Then

=5a• (2.,12.0)

9.14. A circular conductor of radius r 0 = 1 em has an internal field 1<t ( 2sinar--cosar 1 r ) H=a.,.

where a = rc /2r0 •

(A/m)

Find the total current in the conductor.

AMPERE'S LAW AND THE MAGNETIC FIELD

148

law.

There are two methods: (1) to calculate J = V X H The second is simpler here.

J;_'ll

H • dl =

Am~re's

( '~ sin~- 2'~ cos ~)ro dt~J

8 X lltr~

and then integrate; (2) to use

104 4

/cnc = ,(

(CHAP. 9

1'&

~ A

1'&

9.15. A radial field H=

2.39

106 COS

4>ar A/m

exists in free space. Find the magnetic flux q, crossing the surface defined by rt/4, o~ z ~1m. See Fig. 9-16. 3.00 B = #loll= -cos t/Ja,

(T)

l fn/4

-rc /4 ~ 4> ~

(3-·-cost/' ()() )

-n/4

a,·rdt~Jdza.

=4.24Wb Since B is inversely proportional to r (as required by V • B = 0), radial distance is chosen, the total flux will be the same.

it makes no difference what

( I - - - -'--'=':·: I

I I

I I I I

I I

Fag. 9-16

9.16. In cylindrical coordinates, B = (2.0/r)~ (T). pJanesurfacedefinedby 0.5~r~2.5m and = =

Determine the magnetic flux crossing the SeeFig.9-17.

O~z~2.0m.

B · dS

12.o L2.s 2.0 ~. drdz &.p o

.s r

( 2.5)

=4.0 In- =6.44Wb 0.5

9.17. Obtain the vector magnetic potentia) A in the region surrounding an infinitely Jong, straight, filamentary current I.

149

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9}

l.O

Fag. 9-17

As shown in Example 6. the direct expression for A as an integral cannot be used. relation

However, the

Pol VXA=B=-a

21TT â&#x20AC;˘

may be treated as a vector differential equation for A. 4' component of the cylindrical curl is needed.

Since B possesses only a

4' component, only the

aA, aA, Pol ---==-

2rcr

It is evident that A cannot be a function of z, since the filament is uniform with z. dA, dr

Pol

--=-

Pol A =--lnr+C â&#x20AC;˘ 2rc

2rcr

The constant of integration permits the location of a zero reference. expression becomes

With A.

'ro)

Pol A= - ( In- a.

2rc

9.18. Obtain the vector magnetic potential A for the current sheet of Example 2. For z >0, PoK VXA=B=-a 2 X

aA.

aAy

PoK

---==--

whence As A must be independent of x and y, dAy PoK --=-dz 2

Thus, for

z > 0,

For z < 0,

Then

change the sign of the above expression.

r = r0

the

AMPERE'S LAW AND THE MAGNETIC FIELD

150

[CHAP. 9

9.19. Using the vector magnetic potential found in Problem 9.18, find the magnetic flux crossing the rectangular area shown in Fig. 9-18. z

K X

Fag. 9-18 Let the zero reference be at

Zo = 2,

so that Po )K A= --(z-2

In the line integral

A is perpendicular to the contour on two sides and vanishes on the third (z = 2). Thus, cp =

r- A· dl = - #Jo2 (1 - 2) 10 K dy = #JoK Jy-0 2

Note how the choice of zero reference simplified the computation. By Stokes' theorem it is V x A, and not A itself, that determines Cl»; hence the zero reference may be chosen at pleasure.

9.20. Show that the curl of a gradient is zero. From the definition of curl A given in Section 9.4, it is seen that curl A is zero in a region if

for every closed path in the region.

But if A= V/,

where f is a single-valued function,

fA· dl= fvt· dt=fdt =O (see Section 5.6).

Supplementary Problems 9.11.

Show that the magnetic field due to the finite current element shown in Fig. 9-19 is given by H

9.11.

= _..!_ (sin a 1 4 rcr

sin a 2 )a...

Obtain dH at a general point (r, 6, tj>) in spherical coordinates, due to a differential current element I dl /dlsin 6 at the origin in the positive z direction. Ans. a., rcr 4 2

CHAP. 9]

151

AMPERE'S LAW AND THE MAGNETIC FIELD

l'lg. 9-19 9.23.

Currents in the inner and outer conductors of Fig. 9-20 are uniformly distributed. show that for b :s r :s c,

Use

Am~eÂˇs

law to

Fig. 9-20

9.24.

Two identical circular current loops of radius r = 3 m and 1 = 20 A are in parallel planes, separated on their common axis by 10m. Find H at a point midway between the two loops. Ans. 0. 908an A/m

9.15.

A current filament of 10 A in the +y direction lies along the y axis, and a current sheet, 2.0... A/m, is located at z = 4 m. Determine H at the point (2, 2, 2) m. Ans. 0.398a.. + 1.0.,- 0.398a. A/m

9.16.

Show that the curl of (xa.. +yay + za.)/(x 2

9.27.

Given the general vector A= (-cosx)(cosy)a.,

9.28.

Given the general vector A= (cosx)(siny)a.. + (sinx)(cosy)ay, Ans. 0

9.29.

Given the general vector A= (sin 2(/.l)aâ&#x20AC;˘ Ans. 0.5a.

9.30.

Given the general vector A= e-2z(sin ~(jl)a4> (0.800, tr/3, 0.500). Ans. 0.368a, + 0.230a.

+ y 2 + z 2 ) 312 is zero. (Hint: V X E = 0.) find the curl of A at the origin.

Ans.

find the curl of A everywhere.

in cylindrical coordinates, find the curl of A at (2, tr/4, 0).

in cylindrical coordinates, find the curl of A at

152

AMPERE'S LAW AND THE MAGNETIC FIELD

9.31.

Given the general vector A = (sin 4' )a, + (sin 8)a.., point (2, :rc/2, 0). Ans. 0

9.32.

Given the general vector A= 2.50a8 + 5.008.; (2.0, :rc/6, 0). Ans. 4.33a,- 2.50a8 + l.25a.,

9.33.

Given the general vector A

in spherical coordinates, find the curl of A at the

in spherical coordinates, find the curl of A at

_2cos8 -

[CHAP. 9

sinO

- - J - a,+-]-~

show that the curl of A is everywhere zero.

9.34.

A cylindrical conductor of radius 10- 2 m has an internal magnetic field H

= (4.77 x tlt>(i- 3 Ans.

What is the total current in the conductor?

9.35.

In cylindrical coordinates,

J = llf(cos2 2r)a,

x':

)a...

5.0A

in a certain region.

Ans.

and then take the curl of H and compare with J.

9.36.

(A/m)

Obtain H from this current density r sin4r cos4r 1 ) H=llf ( - + - - + - - - - a 4 8 32r 32r â&#x20AC;˘

In cartesian coordinates a constant current density, J = lc}Ay , exists in the region -a :s z :s a. See Fig. 9-21. Use Am~re's law to find H in all regions. Obtain the curl of Hand compare with J.

Ans.

{

10 aa~ l 0 za, -10 aa~

z >a -a :s z :sa

z<-a

curlH=J

-a

Fig. 9-21

9.37.

Compute the total magnetic flux 5 X 10- 2 m if

02 B = Âˇ (sin2 r

Ans. 9.38.

plane in cylindrical coordinates for r :s

crossing the z = 0

4' )a.

(T)

3.14 X 10- 2 Wb

Given that B = 2.so(sin ~)e- 2Ya. find the total magnetic flux crossing the strip z = 0,

9.39.

y 2:0,

(T) 0 :s x :s 2m.

Ans.

1.59Wb

A coaxial conductor with an inner conductor of radius a and an outer conductor of inner and outer radii

153

AMPERE'S LAW AND THE MAGNETIC FIELD

CHAP. 9)

b and c, respectively, carries current I in the inner conductor.

crossing a plane

4> = const. between the conductors.

Find the magnetic flux per unit length 1-lol b Ans. -In2tr n

9.40.

One uniform current sheet, K =Kelly, is at z = b > 2 and another, K = K0 ( -a,.), is at z = -b. Find the magnetic flux crossing the area defined by x = const., -2 :s x :s 2, 0 :s y :s L. Assume free space. Ans. 41-loKoL

9.41.

Use the vector magnetic potential from Problem 9.17 to obtain the flux crossing the rectangle

const., r1 :s r :s r0 ,

9.41.

0 :s z :s L,

due to a current filament I on the z axis.

Ans.

4> =

PalL In~ 2tr

Given that the vector magnetic potential within a cylindrical conductor of radius a is p 0 /r2

A=---a 4tra 2 • find the corresponding H.

Ans.

poll = curl A

Kaa,. ,

9.43.

One uniform current sheet, K = K0 ( -ay ), is located at x = 0 and another, K = at x = a. Find the vector magnetic potential between the sheets. Ans. (p0 KoX + C)ay

9.44.

Between the current sheets of Problem 9.43 a portion of a z =const. plane is defined by O:sx :sb and 0 :s y :sa. Find the flux ell crossing this portion, both from J B · dS and from ~A · dl. Ans. abpoKo

Chapter 10 Forces and Torques in Magnetic Fields 10.1 MAGNETIC FORCE ON PARTICLES A charged particle in motion in a magnetic field experiences a force at right angles to its velocity, with a magnitude proportional to the charge, the velocity, and the magnetic flux density. The complete expression is given by the cross product

F=QUXB Therefore, the direction of a particle in motion can be changed by a magnetic field. The magnitude of the velocity, U, and consequently the kinetic energy, will remain the same. This is in contrast to an electric field, where the force F = QE does work on the particle and therefore changes its kinetic energy. If the field B is uniform throughout a region and the particle has an initial velocity normal to the field, the path of the particle is a circle of a certain radius r. The force of the field is of magnitude F = IQI UB and is directed toward the center of the circle. The centripetal acceleration is of magnitude ru 2 r = U2 /r. Then, by Newton's second law,

IQIUB=mr

r=--

IQIB

Observe that r is a measure of the particle's linear momentum, mU. EXAMPLE 1. Find the force on a particle of mass 1. 70 x 10- 27 kg and charge 1.60 x 10- 19 C if it enters a field B = 5 mT with an initial speed of 83.5 km/s. Unless directions are known for B and U0 , the particle's initial velocity, the force cannot be calculated. Assuming that U 0 and Bare perpendicular, as shown in Fig. 10-1,

F=IQIUB = (1.60 X 10- 1 ~(83.5 X 1W)(5 X 10- 3 ) = 6.68 x

w- 17 N

L Uo

X X

(8 into page}

Fig. 10.1 EXAMPLE 2. revolution.

For the particle of Example 1, find the radius of the circular path and the time required for one mU (1. 70 x 10- 27)(83.5 x lW) r = IQI B = (1.60 X w-19)(5 X w-3) =0.177 m 2~rr

T=-=13 u .31'S

154

CHAP. 10)

155

FORCES AND TORQUES IN MAGNETIC FIELDS

10.2 ELECfRIC AND MAGNETIC ftELDS COMBINED

When both fields are present in a region at the same time, the force on a particle is given by F=Q(E+UXB)

This Lorentz force, together with the initial conditions, determines the path of the particle. EXAMPLE 3. In a certain region surrounding the origin of coordinates, B = 5.0 x 10-"a, T

and E =

5.0.. V/m. A proton (Qp = 1.602 x 10- 19 C, mp = 1.673 x 10- 27 kg) enters the fields at the origin with an

initial velocity U 0 =2.5 x 1lfa.. m/s. Describe the proton's motion and give its position after three complete revolutions. The initial force on the particle is F0 = Q(E + U0 X B)= Qp(Ea. - U0 Bay)

The z component (electric component) of the force is constant, and produces a constant acceleration in the z direction. Thus the equation of motion in the z direction is

z = ~atz=! (QPE\,z 2

m;J

The other (magnetic) component, which changes into -QPUBa,, produces circular motion perpendicular to the

z axis, with period T

= 2nr = 2nmP U

QPB

The resultant motion is helical, as shown in Fig. 10-2. z

Fig. 10.2

After three revolutions, x

=y =0

and

z =! (QPE\(3T)z = 18nzE~P =37.0 m 2

m;J

QPB

10.3 MAGNETIC FORCE ON A CURRENf ELEMENf

A frequently encountered situation is that of a current-carrying conductor in an external magnetic field. Since I= dQ/dt, the differential force equation may be written dF = dQ(U X B)=(! dt)(U X B)= !(dl X B)

where dl = U dt is the elementary length in the direction of the conventional current /. If the conductor is straight and the field is constant along it, the differential force may be integrated to give F=ILB sin 8

The magnetic force is actually exerted on the electrons that make up the current I. However, since the electrons are confined to the conductor, the force is effectively transferred to the heavy

156

FORCES AND TORQUES IN MAGNETIC FIELDS

[CHAP. 10

lattice; this transferred force can do work on the conductor as a whole. While this fact provides a reasonable introduction to the behavior of current-carrying conductors in electric machines, certain essential considerations have been omitted. No mention was made, nor will be made in Section 10.4, of the current source and the energy that would be required to maintain a constant current I. Faraday's law of induction (Section 12.3) was not applied. In electric machine theory the result will be modified by these considerations. Conductors in motion in magnetic fields are treated again in Chapter 12; see particularly Problems 12.10 and 12.13. EXAMPLE 4. Find the force on a straight conductor of length 0.30 m carrying a current of 5.0 A in the direction, where the field is B = 3.50 x 10- 3 (a.. - a,) T.

-a.

F=!(LXB) = (5.0)((0.30)( -a.) X 3.50 X 10- 3 (a_. -a,))

=7.42

xw-l(-a.vi"a") N

The force, of magnitude 7.42 mN, is at right angles to both the field B and the current direction, as shown in Fig. 10-3.

10.4 WORK AND POWER

The magnetic forces on the charged particles and current-carrying conductors examined above result from the field. To counter these forces and establish equilibrium, equal and opposite forces, Fa. would have to be applied. If motion occurs, the work done on the system by the outside agent applying the force is given by the integral W=

finall

Fa ·dl

•nitia11

A positive result from the integration indicates that work was done by the agent on the system to move the particles or conductor from the initial location to the final, against the field. Because the magnetic force, and hence Fa, is generally nonconservative, the entire path of integration joining the initial and final locations of the conductor must be specified. EXAMPLE 5. Find the work and power required to move the conductor shown in Fig. 10-4 one full revolution in the direction shown in 0.02 s, if B = 2.50 x 10- 3ar T and the current is 45.0 A. F =!(I X B)= 1.13 x 10- 2 a., N

and so F.. = -1.13 X 10- 2a., N. W= =

F.. ·dl

f" (

-1.13 X 10- 2 )a., • r d,Pa.,

=-2.13 X 10and P= W/t= -0.107 W.

CHAP. lO)

FORCES AND TORQUES IN MAGNETIC FIELDS

157

t 0 10m

.. ... ...-X

r-003m

Fig. 1{1-4

The negative sign means that work is done by the magnetic field in moving the conductor in the direction shown. For motion in the opposite direction, the reversed limits will provide the change of sign, and no attempt to place a sign on r dcpaq, should be made.

10.5 TORQUE The moment of a force or torque about a specified point is the cross product of the lever arm about that point and the force. The lever arm, r, is directed from the point about which the torque is to be obtained to the point of application of the force. In Fig. 10-5 the force at P has a torque about 0 given by

T=rXF where T has the units N · m. torque from energy.)

(The units N · m/rad have been suggested, in order to distinguish

F I

p y ~ ~

Fig. 1()..5

In Fig. 10-5, T lies along an axis (in the xy plane) through 0. If P were joined to 0 by a rigid rod freely pivoted at 0, then the applied force would tend to rotate P about that axis. The torque T would then be said to be about the axis, rather than about point 0. EXAMPLE 6. A conductor located at x = 0.4 m, y = 0 and 0 < z < 2.0 m carries a current of 5.0 A in the a, direction. Along the length of the conductor B = 2.5a, T. Find the torque about the z axis.

F =/(LX B)= 5.0(2.0a,

2.5ax) = 25.0ay N

T=r X F =0.4ax X 25.0ay = lO.Oa, N · m

10.6 MAGNETIC MOMENT OF A PLANAR COIL Consider the single-turn coil in the z = 0 plane shown in Fig. 10-6, of width w in the x direction and length C along y. The field B is uniform and in the +x direction. Only the

158

FORCES AND TORQUES IN MAGNETIC FIELDS

[CHAP. 10

Fig. 10-6

Âąy-directed currents give rise to forces.

For the side on the left,

F= /(fa,. X Ba... ) = -Bifaz and for the side on the right, The torque about the y axis from the left current element requires a lever arm r = -(w/2)a... ; the sign will change for the lever arm to the right current element. The torque from both elements is

where A is the area of the coil. It can be shown that this expression for the torque holds for a flat coil of arbitrary shape (and for any axis parallel to they axis). The magnetic moment m of a planar current loop is defined as /Aa, , where the unit normal a, is determined by the right-hand rule. (The right thumb gives the direction of a, when the fingers point in the direction of the current.) It is seen that the torque on a planar coil is related to the applied field by

T=mX8 This concept of magnetic moment is essential to an understanding of the behavior of orbiting charged particJes. For example, a positive charge Q moving in a circular orbit at a velocity U, or an angular velocity co, is equivalent to a current I= (co/2rc)Q, and so gives rise to a magnetic moment

m=-QAa 2rc "

as shown in Fig. 10-7. More important to the present discussion is the fact that in the presence of a magnetic field B there will be a torque T = m X 8 which tends to tum the current loop until m and 8 are in the same direction, in which orientation the torque will be zero.

Fig. 10-7

FORCES AND TORQUES IN MAGNEI1C FIELDS

CHAP. 10)

159

Solved Problems 10.1.

A conductor 4 m long lies along the y axis with a current of 10.0 A in the a,. direction. the force on the conductor if the field in the regions is 8 = 0.05ax T.

Find

F = IL X B = 10.0(4a,. X 0.05a.. ) = -20a, N

10.2.

A conductor of length 2.5 m located at

z = 0,

x = 4m

carries a current of 12.0 A in the

-a,. direction. Find the uniform 8 in the region if the force on the conductor is 1.20 X w- 2 N in the direction (-ax + a. )/VZ. From F=ILXB, 2

(1.20 x 10- )(-a.vi

8 ')

a.. a,. a, 0 -(12.0)(2.5) 0 B.. B,. B,

whence

B,=B"=

= -30B,a.. + 30B.. a,

w-" 'lfi T

the y component of B may have any value.

10.3.

A current strip 2 em wide carries a current of 15.0 A in the ax direction, as shown in Fig. 10-8. Find the force on the strip per unit length if the uniform field is 8 = 0.20., T.

Fig. 10-8

In the expression for dF, I dl may be replaced by K dS. dF=(KdS)xB 15.0) = ( 0.02 dx dy(0.20)a. F= [

LL 150.0dxdya.

.01 0.01

r.=3.0a,N/m

10.4.

Find the forces per unit length on two long, straight, parallel conductors if each carries a current of 10.0 A in the same direction and the separation distance is 0.20 m. Consider the arrangement in cartesian coordinates shown in Fig. 10-9. The conductor on the left creates a field whose magnitude at the right-hand conductor is B

= 1-'ol = (4.7t X 10-7){10.0) 2.1fr

2.7t(O. 20)

_5 T

160

FORCES AND TORQUES IN MAGNETIC FIELDS

(CHAP. 10

fla. 10.9 and whose direction is -aâ&#x20AC;˘ . Then the force on the right conductor is F =/La, X B( -a,)= ILB( -a.. )

L = 10-4 ( -a.. ) N/m

and

An equal but opposite force acts on the left-hand conductor. The force is seen to be one of attraction. Two parallel conductors carrying current in the same direction will have forces tending to pull them together.

10.5.

A conductor carries current I parallel to a current strip of density K0 and width w, as shown in Fig. 10-10. Find an expression for the force per unit length on the conductor. What is the result when the width w approaches infinity?

From Problem 10.4, the filament Ko dx shown in Fig. 10-10 exerts an attractive force

dF =IB =/llo(K0 dx) L a, 21tr a, on the conductor. Adding to this the force due to the similar filament at -x, the components in the x direction cancel, giving a resultant

Integrating over the half-width of the strip, F P.olKoh -=--(-a.) L 1t

Lwn -dx- = (llolKo W) (-a.) --arctano

hz+xz

The force is one of attraction, as expected. As the strip width approaches infinity, F I L-+ (llol K0 /2)(- a,).

10.6.

Find the torque about the y axis for the two conductors of length t, separated by a fixed distance w, in the uniform field 8 shown in Fig. 10-11.

CHAP. 10]

FORCES AND TORQUES IN MAGNETIC FIELDS

161

Fig. 10-11

The conductor on the left experiences the force F 1 =Ita,. X Ba... =Bit( -az) the torque of which is

The force on the conductor on the right results in the same torque. The sum is therefore T = Bllw( -a,.)

10.7.

A D' Arsonval meter movement has a uniform radial field of B = 0. 10 T and a restoring spring with a torque T = 5.87 x 10- 5 8 (N · m), where the angle of rotation is in radians. The coil contains 35 turns and measures 23 mm by 17 mm. What angle of rotation results from a coil current of 15 rnA? The shaped pole pieces shown in Fig. 10-12 result in a uniform radial field over a limited range of deflection. Assuming that the entire coil length is in the field, the torque produced is

T = nBllw = 35(0.10)(15 x w- 3 )(23 x w- 3 )(17 x w- 3 ) = 2.05 X 10- 5 N · m

This coil turns until this torque equals the spring torque. 2.05

10- 5 = 5.87 X

w-se

e = 0. 349 rad

or 20°

Fig. 10-12

10.8.

The rectangular coil in Fig. 10-13 is in a field

8 = 0.05 ax:n,ayT Find the torque about the z axis when the coil is in the position shown and carries a current of 5.0A. m = IAa,

= 1.60 x 10- 2a.. a.. +a,.

T=mXB= 1.60x 10-2a.. X0.05 y'2 = 5.66 X 10-"az N · m

If the coil turns through 45°, the direction of m will be (a... + a,.)/v'i and the torque will be zero.

162

FORCES AND TORQUES IN MAGNETIC FIELDS

(CHAP. 10

lifi :~m ·= ; ; ;: : ~; ; ;=: : ,.= ·=:; ; ;: : :~x:i ot: - - - - Y

Fig. 10-13

10.9.

Find the maximum torque on an 85-turn, rectangular coil, 0.2 m by 0.3 m, carrying a current of 2.0 A in a field B = 6.5 T. T....,. = nBllw = 85(6.5)(2.0)(0.2)(0.3) = 66.3 N · m

10.10. Find the maximum torque on an orbiting charged particle if the charge is 1.602 x 10- 19 C, the circular path has a radius of 0.5 x 10- 10 m, the angular velocity is 4.0 x 1016 rad/s, and B =0.4 X 10-3 T. The orbiting charge has a magnetic moment W 4X 1016 m= 1t'QAa, =~(1.602 X 10- 19)1t'(0.5 2

X 10- ~ a"

=8.01

10-24a,

A· m2

Then the maximum torque results when a" is normal to B. T_,. = mB

= 3.20 X 10-27 N. m

10.11. A conductor of length 4 m, with current held at 10 A in the a,. direction, lies along the y axis between y =±2m. If the field is 8 = 0.05ax T, find the work done in moving the conductor parallel to itself at constant speed to x = z =2m. For the entire motion, F=ILXB = -2.0ar The applied force is equal and opposite, F., =2.0ar Because this force is constant, and therefore conservative, the conductor may be moved first along z, then in the x direction, as shown in Fig. 10-14. Since F.. is completely in the z direction, no work is

Fig. 10-14

CHAP. 10]

FORCES AND TORQUES IN MAGNETIC FIELDS

done in moving along x.

163

Then, W=

(20a,) • dza, = 4.0 J

10.12. A conductor lies along the z axis at -1.5 s z s 1.5 m and carries a fixed current of 10.0 A in the -az direction. See Fig. 10-15. For a field

= 3.0 X l0- 4 e-o.2xay (T)

1.5 ~

~ ~ ~ ~

~~------'1 ....

-1.5

Fig. 10-15

find the work and power required to move the conductor at constant speed to x = 2.0 m, y = 0 in 5 x w- 3 s. Assume parallel motion along the x axis. F = IL X B = 9.0 X w-le-0.2x.%

Then F.,= -9.0 x l0- 3e-o.z...ax

and W=

(-9.0 X l0- 3e-o.z...a.. ) · dxax

= -1.48

x w- 2 1

The field moves the conductor, and therefore the work is negative.

P=-= t

-1.48 x

SxlO

The power is given by

= -2.fT/W

10.13. Find the work and power required to move the conductor shown in Fig. 10-16 one full turn • in the positive direction at a rotational frequency of N revolutions per minute, if B = Boar (B 0 a positive constant).

Fig. 10-16

164

FORCES AND TORQUES IN MAGNETIC FIELDS

[CHAP. 10

1be force on the conductor is

F= IL X B =/La. X Boa,= BolLa• so that the applied force is

Fa= B 0 1L( -a.) The conductor is to be turned in the •• direction. revolution is W=

Therefore, the work required for one full

BoiL(-a.) · rdf/>a•

= -2:~rrB 01L

Since N revolutions per minute is N f(IJ per second, the power is

= _ 2:~rrB0 ILN (IJ

The negative signs on work and power indicate that the field does the work. The fact that work is done around a closed path shows that the force is nonconservative in this case.

10.14. In the configuration shown in Fig. 10-16 the conductor is 100 mm long and carries a constant 5.0 A in the az direction. If the field is B = -3.5 sin ¢a,

and r=25mm, find the work done in moving the conductor at constant speed from tjJ =0 to tjJ = rc, in the direction shown. If the current direction is reversed for rc < ¢ < 2rc, what is the total work required for one full revolution? F=ILXB=-1.75xl0- 3 sinf/>a•

Fa= 1. 75 Then

10- sin

4>•.

1.75 X 10- 3 sin f/>a. • rdf/>a. = 87.5 1Jl

If the current direction changes when the conductor is between same. The total work is 175 1Jl.

and 2n:, the work will be the

10.15. Compute the centripetal force necessary to hold an electron (m~ = 9.107 x 10- 31 kg) in a circular orbit of radius 0.35 x 10- 10 m with an angular velocity of 2 x 1016 rad/s.

10.16. A uniform magnetic field 8 = 85.3az pT exists in the region x ~ 0. If an electron enters this field at the origin with a velocity U0 = 450a_.. km/s, find the position where it exits. Where would a proton with the same initial velocity exit? r.

meUo =--=3.00x

• IQIB

w- z m

The electron experiences an initial force in the a,. direction and it exits the field at x = z =0, y = 6cm. A proton would turn the other way, and part of the circular path is shown at P in Fig. 10-17. With mP = l840me,

and the proton exits at x

=z = 0,

y = -110m.

165

FORCES AND TORQUES IN MAGNETIC FIELDS

CHAP. 10]

Fig. 10·-17

10.17. If a proton is fixed in position and an electron revolves about it in a circular path of radius 0.35 x 10-to m, what is the magnetic field at the proton? The proton and electron are attracted by the coulomb force, Q2

F=--

4Jr£or2

which furnishes the centripetal force for the circular motion.

Thus

or Now, the electron is equivalent to a current Joop I= (ru/2Jr)Q. loop is, from Problem 9.7,

The field at the center of such a

Substituting the value of w found above, B=

(~-to/4Jr)Q2 2

r V4Jr£ 0 m,r

(10-7)(1.6 X 10-19)2 (0.35 X 10

10 2 )

10 9 )(9.1

10 31 )(0.35

10 1 ~

35T

Supplementary Problems 10.18. A current element 2m in length lies along the y axis centered at the origin. The current is 5.0 A in the a,. direction. If it experiences a force 1.50(a_. + a.)/V2 N due to a uniform field 8, determine B. Ans. 0.106(-a_.+a.) T 10.19. A magnetic field, 8 = 3.5 x 10- 2a.. T, exerts a force on a 0.30-m conductor along the x axis. If the conductor current is 5.0 A in the -a.. direction, what force must be applied to hold the conductor in position? Ans. -5.25 x l0- 211y N 10.20. A current sheet, K = 30.01ly A/m, lies in the plane z = -5 m and a filamentary conductor is on the y axis with a current of 5.0 A in the a,. direction. Find the force per unit length. Ans. 94.2~-tN/m (attraction) 10.21. A conductor with current I pierces a plane current sheet K orthogona1ly, as shown in Fig. 10-18. the force per unit length on the conductor above and below the sheet. Ans. ±~-toKI/2 10.22. Find the force on a 2-m conductor on the z axis with a current of S.O A in the a. direction, if 8

Ans.

-60a_. + 20ay N

= 2.0._. + 6.0.)'

Find

166

FORCES AND TORQUES IN MAGNETIC FIELDS

[CHAP. 10

Fig. 10-18

10.23. Two infinite current sheets, each of constant density Ko , are parallel and have their currents oppositely directed. Find the force per unit area on the sheets. Is the force one of repulsion or attraction? Ans. po/(f,/2 (repulsion) 10.14. The circular current loop shown in Fig. 10-19 is in the plane z = h, parallel to a uniform current sheet, K = K~, , at z 0. Express the force on a differential length of the loop. Integrate and Ans. dF= Vap0 K0 cos t/>dtP(-az) show that the total force is zero.

K Fig. 10-19 10.15. Two conductors of length t normal to 8 are shown in Fig. 10-20; they have a fixed separation w. that the torque about any axis parallel to the conductors is given by Bltw cos 6.

Show

Fig. 10-20

10.16. A circular current loop of radius r and current I lies in the z = 0 plane. Find the torque which results if the current is in the a. direction and there is a uniform field B = B0 (a,. + az)/v2. Ans. (nr2B 0 ItJ2)ay 10.17. A current loop of radius r = 0. 35 m is centered about the x axis in the plane x = 0 and at (0, 0, 0.35) m the current is in the -a,. direction at a magnitude of 5.0 A. Find the torque if the uniform Ans. 1.70 X 10- 4 ( -a,.) N · m field is B = 88.4(a., + az) pT. 10.28. A current of 2.5 A is directed generally in the a., direction about a square conducting loop centered at the origin in the z = 0 plane with 0.60 m sides parallel to the x and y axes. Find the forces and the torque on the loop if B = 15ay mT. Would the torque be different if the loop were rotated through 45° in the z = 0 plane? Ans. 1.35 x 10- 2( -a,.) N · m; T = m X B 10.19. A 200-turn, rectangular coil, 0.30 m by 0.15 m with a current of 5.0 A, is in a uniform field 0.2 T. Find the magnetic moment m and the maximum torque. Ans. 45.0 A · m2 , 9.0 N · m

CHAP. 10)

FORCES AND TORQUES IN MAGNETIC FIELDS

167

10.30. Two conductors of length 4.0 m are on a cylindrical shell of radius 2.0 m centered on the z axis, as shown in Fig. 10-21. Currents of 10.0 A are directed as shown and there is an external field 8 = O.S11x T at ¢ = 0 and 8 = -O.Sax T at ¢ =Jr. Find the sum of the forces and the torque about the axis. Ans. -40a} N, 0

Fig. 10-21

10.31. A right circular cylinder contains 550 conductors on the curved surface and each has a current of constant magnitude 7.5 A. The magnetic field is 8 = 38 sin ¢a, mT. The current direction is a, for 0 < ¢ < 1t and -a, for 1t < ¢ < 27r (Fig. 10-22). Find the mechanical power required if the Ans. 60.2 W cylinder turns at 1600 revolutions per minute in the -a<~> direction.

.......

Fig. 10-22

10.32. Obtain an expression for the power required to tum a cylindrical set of n conductors (see Fig. 10-22) against the field at N revolutions per minute, if B = B 0 sin 2¢a, and the currents change direction in B 0 nltrN each quadrant where the sign of 8 changes. Ans. (W) 60 10.33. A conductor of length t lies along the x axis with current I in the ax direction. Find the work done in turning it at constant speed, as shown in Fig. 10-23, if the uniform field is B =Boa,. Ans. nBo/ 21/4

Fig. 10-23

168

FORCES AND TORQUES IN MAGNETIC FIELDS

[CHAP. 10

10.34. A rectangular current loop, of length t along the y axis, is in a uniform field 8 = Boa. , as shown in Fig. 10-24. Show that the work done in moving the loop along the x axis at constant speed is zero. z

10.35. For the configuration shown in Fig. 10-24, the magnetic field is 8 =

Bo(sin ';;)•.

Find the work done in moving the coil a distance w along the x axis at constant speed, starting from the Ans. -4B0 Ifw/n location shown.

10.36. A conductor of length 0.25 m lies along the y axis and carries a current of 25.0 A in the ay direction. Find the power needed for parallel translation of the conductor to x = S.Om at constant speed in 3.0 s if the uniform field is 8 = 0.06a. T. Ans. -0.625 W 10.37. Find the tangential velocity of a proton in a field 1 em. Ans. 14.4 m/s

B = 30 #lT if the circular path has a diameter of

10.38. An alpha particle and a proton (Q... = 2QP) enter a magnetic field B =lilT with an initial speed Uo =8.5 m/s. Given the masses 6.68 X 10- 27 kg and 1.673 X 10- 27 kg for the alpha particle and Ans. 177 mm,_ 88.8 mm the proton, respectively, find the radii of the circular paths. 10.39. If a proton in a magnetic field completes one circular orbit in 2.35 ~-ts, what is the magnitude of 8? Ans. 2.79 X w-zr 10.40. An electron in a field B = 4.0 X w-z T has a circular path 0.35 X w-•o m in radius and a maximum Ans. 2.0 x 1016 rad/s torque of 7.85 x 10- 26 N · m. Determine the angular velocity 10.41. A region contains uniform 8 and E fields in the same direction, with B = 650 ~-tT. An electron follows a helical path, where the circle has a radius of 35 mm. If the electron had zero initial velocity in the axial direction and advanced 431 mm along the axis in the time required for one full circle, find the Ans. 1.62 kV /m magnitude of E.

Chapter 11 Inductance and Magnetic Circuits 11.1 INDUCTANCE The inductance L of a conductor system may be defined as the ratio of the linking magnetic flux to the current producing the flux. For static (or, at most, low-frequency) current 1 and a coil containing N turns, as shown in Fig. 11-1,

N4>

L=1

The units on L are henries, where 1 H = 1 Wb/ A. Inductance is also given by L = M l, where )., the flux linkage, is N4> for coils with N turns or simply 4.> for other conductor arrangements.

Fig. ll-1

.--J+

Mathcad

It should be noted that L will always be the product of the permeability f.l of the medium (units on f.l are H/m) and a geometrical factor having the units of length. Compare the expressions for resistance R (Chapter 6) and capacitance C (Chapter 7) . EXAMPLE 1.

Find the inductance per unit length of a coaxial conductor such as that shown in Fig.

con t

' t

Fig. ll-2

169

170

INDUCfANCE AND MAGNETIC CIRCUITS

[CHAP. 11

11-2. Between the conductors, I H=-•

2:trr •

B= IJola. 2Irr The currents in the two conductors are linked by the ftux across the surface

A= ~ 0

4> = const. For a length t,

1" -drdz IJol 1-'olf b =--ln"

2nr

2:tr

L #lo b -=-In- (H/m) t 2:tr a

and

EXAMPLE 2. Find the inductance of an ideal solenoid with 300 turns, t = 0.50 m, and a circular cross section of radius 0.02 m. The turns per unit length is n = 300/0.50 = 600, so that the axial field is B = #loH = llo6001 (Wb/m2) Then

!:_ = N4.> = N(!!.)A = 300(600#-'0 }:tr(4 X 10-4 ) t

=5681-'H/m

or L

= 284~JH.

In Section 5.8 an imagined bringing-in of point charges from infinity was used to derive the energy content of an electric field:

WE=!L

D•Edv

2 "0'

There is no equivalent in a magnetic field to the point charge, and consequently no parallel development for its stored energy. However, a more sophisticated approach yields the completely analogous expression

WH=!L 2

Comparing this with the formula

WH = !LP L=

B·Hdv

"ol

from circuit analysis yields

B·H

L --dv p "ol

~ 1.11!!,

EXAMPLE 3. Checking Example 1,

val

(

#-'of

• H dv = 1-'oi~i:z,.l" ~ P ) b -B 12 12 0 0 " 41f r rdrdt/>dz =-In2:tr a

11.2 STANDARD CONDUCI'OR CONFIGURATIONS Figures 11-3 through 11-7 give exact or approximate inductances of some common noncoaxial arrangements.

171

INDUCf ANCE AND MAGNETIC CIRCUITS

CHAP. 11)

\ turns

Fig. ll-3. Toroid, square cross section

J1o.'V2S L ""---

(H)

brr (assuming average flux density at average radius r)

Fig. ll-4. Toroid, general cross sectionS

.!:.. = {

P.o cosh- 1 .!L

(H/m}

Ford>>a,

radius

.!.:._ ""

d-----

f.J.o

(H/m)

_!__

(H/m)

Fig. 11-5. ParaUel conductors of radius a

.!:.. = t

f.J.o cosh- 1

2rr

,.,~In 4 2rr

(H/m}

Fig. 11-6. Cylindrical conductor parallel to a ground plane

.....

Fig. 11-7. Long solenoid of small cross-sectional areaS

11.3 FARADAY'S LAW AND SELF-INDUCTANCE Consider an open surface S bounded by a closed contour C. If the magnetic flux cp linking S varies with time, then a voltage v around C exists; by Faraday's law, dcp dt

v=--

As was shown in Chapter 5, the electrostatic potential or voltage, V, is well-defined in space and is

172

INDUCTANCE AND MAGNETIC CIRCUITS

(CHAP. 11

associated with a conservative electric field. By contrast, the induced voltage v given by Faraday's law is a multivalued function of position and is associated with a nonconservative field (electromotive force). More about this in Chapter 12. Faraday's law holds in particular when the flux through a circuit element is changing because the current in that same element is changing: dcpdi di v= - - - = - L di dt dt

In circuit theory, L is called the self -inductance of the element and v ts called the voltage of self-inductance or the back-voltage in the inductor.

11.4 INTERNAL INDUCTANCE Magnetic ftux occurs within a conductor cross section as well as external to the conductor. This internal ftux gives rise to an internal inductance, which is often small compared to the external inductance and frequently ignored. In Fig. ll-8(a) a conductor of circular cross section is shown, with a current I assumed to be uniformly distributed over the area. (This assumption is valid only at low frequencies, since skin effect at higher frequencies forces the current to be concentrated at the outer surface.) Within the conductor of radius a, Ampere's law gives lr 2:rra

H=--a 2

and

f.lolr

B=--a 2 2:rra

t t

(a)

{b)

Fig. 11-8

The straight piece of conductor shown in Fig. ll-8(a) must be imagined as a short section of an infinite torus, as suggested in Fig. ll-8(b ). The current filaments become circles of infinite radius. The lines of ftux dcf> through the strip e dr encircle only those filaments whose distance from the conductor axis is smaller than r. Thus, an open surface bounded by one of those filaments is cut once (or an odd number of times) by the lines of dcf>; whereas, for a filament such as 1 or 2, the surface is cut zero times (or an even number of times). It follows that dcf> links only with the fraction :rrr2 /:rra 2 of the total current, so that the total ftux linkage is given by the weighted "sum" A=

and

I (Jrr2) :rra 2

dcf>

ia (Jrr2) 0

f.lolr :rra 2 2:rra 2

edr = f.lolf 8:rr

CHAP. 11)

INDUCfANCE AND MAGNETIC CIRCUITS

173

This result is independent of the conductor radius. The total inductance is the sum of the external and internal inductances. If the external inductance is of the order of~ x 10- 7 H/m, the internal inductance should not be ignored.

11.5 MUTUAL INDUCfANCE In Fig. 11-9 a part t:Pt 2 of the magnetic flux produced by the current it through coil 1 links the N2 turns of coil 2. The voltage of mutUill induction in coil 2 is given by

v 2 = N2 dcpt 2 dt

(negative sign omitted)

'P12

Fig. ll-9

In terms of the mutUill inductance Mt 2= N2C/J12/ It , dcp12 dit dit v2=N2-d. -d =M.2-d lt t t

This mutual inductance will be a product of the permeability p of the region between the coils and a geometrical length, just like inductance L. If the roles of coils 1 and 2 are reversed, di2 v, =M21 dt

The following reciprocity relation can be established:

M12 = M21 .

EXAMPLE 4. A solenoid with N1 = 1000, r1 = l.Ocm, and t. =50 em is concentric within a second coil of N2 = 2000, r2 = 2.0 em, and {2 =50 em. Find the mutual inductance assuming free-space conditions. For long coils of small cross section, H may be assumed constant inside the coil and zero for points just outside the coil. With the first coil carrying a current 11 , H

= (~C::)I.

B = prJ,000/1 ~=

(A/m)

(in the axial direction)

(Wb/m2 )

BA = (prJ,OOOi1)(n x 10-4 )

(Wb)

Since H and B are zero outside the coils, this is the only flux linking the second coil.

11.6 MAGNETIC CIRCUITS In Chapter 9, magnetic field intensity H, flux Cl>, and magnetic flux density B were examined and various problems were solved where the medium was free space. For example, when Ampere's law

174

INDUCTANCE AND MAGNETIC CIRCUITS

[CHAP. 11

is applied to "the closed path C through the long, air-core coil shown in Fig. 11-10, the result is

fH·di=NI But since the flux lines are widely spread outside of the coil, B is small there. restricted to the inside of the coil, where

The flux is effectively

Nl H=t

Ferromagnetic materials have relative permeabilities IJr in the order of thousands. Consequently, the flux density B = p 0 p,H is, for a given H, much greater than would result in free space. In Fig. 11-11, the coil is not distributed over the iron core. Even so, the Nl of the coil causes a flux Cl> which follows the core. It might be said that the flux prefers the core to the surrounding space by a ratio of several thousand to one. This is so different from the free-space magnetics of Chapter 9 that an entire subject area, known as iron-core magnetics or magnetic circuits, has developed. This brief introduction to the subject assumes that all of the flux is within the core. It is further assumed that the flux is uniformly distributed over the cross section of the core. Core lengths required for calculation of NI drops are mean lengths.

, . . ----- __c '

Fig. 11-11

Fig. ll-10

11.7 THE B-H CURVE A sample of ferromagnetic material could be tested by applying increasing values of H and measuring the corresponding values of flux density B. Magnetization curves, or simply B-H curves, for some common ferromagnetic materials are given in Figs. 11-12 and 11-13. The relative permeability can be computed from the B-H curve by use of p, = B I IJoH. Figure 11-14 shows the extreme nonlinearity of p, versus H for silicon steel. This nonlinearity requires that problems be solved graphically.

11.8 AMPERE'S LAW FOR MAGNETIC CIRCUITS A coil of N turns and current I around a ferromagnetic core produces a magnetomotive force (mmf) given by Nl. The symbol F is sometimes used for this mmf; the units are amperes or ampere turns. Ampere's law, applied around the path in the center of the core shown in Fig. ll-15(a), gives

F=NI=fH·dl =

H·dl+

= H 1 t 1 + H2 t 2 + H 3 f 3

H·dl

HCAim)

Fil· 11-U.

B·H curves, H < 400 A/m

f I·

ft·

1.60

1!-

~41~-~· ~:!

1.40

. ' .• l . ,.

I+-

It •

H1+

1.00

ill!' It"' I. :jill::

·.:1, II tt

: !tU!:t ::;: tr;-;-l;t ~d "1 iJ_ ~L

;

•

::I+

t:-,_:.i l

•

'---'T -n~

"!"'"'

c-rt"• -n-i--T-

....

:j J 'T"

·:t ·:r-:~

.... ::·:::'"i .:rr .~~Hltltit.::-usl·:::..::ltg;_~ !t ~~t ~~H t· ~l ~ · ···· "~-H-~

r. 1.20

it- -. •

\-i 't!: ~:.j:: ;:-:-:-7!--;-; " ... ,. " .. ---- h=j-:: F-4 ~ .::ft:_; ;i-1; :::__. ". :;_:_: •• :: !.:;-:-

••••

....

~£ ~-+

: -:-:

:tr.

"-+~.

::t: &-~

•I

i!!:.:J .• ~···

0.40 ••

...

·t--J:

; --:t .. H 1000

~ l ;j

IH 2000

]000

H(A/m)

FiR· 11-13.

B·H curves, H > 400 A/m

4000

5000

6000

176

INDUCTANCE AND MAGNETIC CIRCUITS

100

200

300

[CHAP. 11

400

II (A/m)

Fig. 11-14 II, (I

â&#x20AC;˘I>

R { )

(b)

(a)

Fig. 11-15

Comparison with Kirchhoff's law around a single closed loop with three resistors and an emf V,

V=V1 +Vz+V:, suggests that F can be viewed as an NI rise and the Ht terms considered NI drops, in analogy to the voltage rise V and voltage drops V1 , V2 and V,. The analogy is developed in Fig. 11-15(b) and (c). Flux in Fig. 11-15(b) is analogous to current/, and reluctance PJl is analogous to resistance R. An expression for reluctance can be developed as follows.

NI drop= Ht = hence

BAC~) =

<1>9R

9R=_!__ (H-1) flA

If the reluctances are known, then the equation

F = N I = ( 9/l 1 + 9/lz + 9RJ) can be written for the magnetic circuit of Fig. 11-15( b).

However, llr must be known for each

CHAP. 11)

INDUCI'ANCE AND MAGNETIC CIRCUITS

177

material before its reluctance can be calculated. And only after B or H is known will the value of This is in contrast to the relation

#l, be known.

R=_!_ oA

(Section 6.7), in which the conductivity o is independent of the current.

11.9 CORES WITH AIR GAPS Magnetic circuits with small air gaps are very common. The gaps are generally kept as small as possible, since the NI drop of the air gap is often much greater than the drop in the core. The ftux fringes outward at the gap, so that the area at the gap exceeds the area of the adjacent core. Provided that the gap length la is less than -fo the smaller dimension of the core, an apparent area, Sa, of the air gap can be calculated. For a rectangular core of dimensions a and b, Sa= (a

+ la)(b + la)

If the total ftux in the air gap is known, Ha and Halo can be computed directly.

For a uniform iron core of length l; with a single air gap, Ampere's law reads NI = H;l; + Halo

laff)

= H;l; + S

llo

fl) is known, it is not difficult to compute the NI drop across the air gap, obtain B; , take H; from the appropriate B-H curve and compute the NI drop in the core, H;l;. The sum is the NI required to establish the ftux fl). However, with NI given, it is a matter of trial and error to obtain B; and fl), as will be seen in the problems. Graphical methods of solution are also available.

If the ftux

11.10 MULTIPLE COILS Two or more coils on a core could be wound such that their mmfs either aid one another or oppose. Consequently, a method of indicating polarity is given in Fig. 11-16. An assumed direction for the resulting ftux fl) could be incorrect, just as an assumed current in a de circuit with two or more voltage sources may be incorrect. A negative result simply means that the ftux is in the opposite direction.

{b)

(o)

..,.. ll-16

178

INDUCTANCE AND MAGNEflC CIRCUITS

(CHAP. 11

11.11 PARALLEL MAGNETIC CIRCUITS The method of solving a parallel magnetic circuit is suggested by the two-loop equivalent circuit shown in Fig. 11-17(b). The leg on the left contains an NI rise and an Nl drop. The Nl drop between the junctions a and b can be written for each leg as follows: F - H 1 f 1 = H2 l 2 == H3 l 3

and the ftuxes satisfy ~. =~2+~3

Different materials for the core parts will necessitate working with several B-H curves. An air gap in one of the legs would lead to ~l; + Hala for the mmf between the junctions for that leg.

a --cJ>J

c2>2t IH2/2

HJ/3

b {b)

la)

..,.. 11-17

The equivalent magnetic circuit should be drawn for parallel magnetic circuit problems. It is good practice to mark the material types, cross-sectional areas, and mean lengths directly on the diagram. In more complex problems a scheme like Table 11-1 can be helpful. The data are inserted directly into the table, and the remaining quantities are then calculated or taken from the appropriate B-H curve.

Table 11-1

Part

Material

Area

ÂŤlÂť

1 2

Solved Problems 11.1.

Find the inductance per unit length of the coaxial cable in Fig. 11-2 if a= 1 mm and b = 3 mm. Assume #l, = 1 and omit internal inductance. L 1-l b 4Jr X 10- 7 -=-In-= ln3=0.22!-lH/m f 2n a 2n

CHAP. 11]

11.2.

INDUCTANCE AND MAGNETIC CIRCUITS

179

Find the inductance per unit length of the parallel cylindrical conductors shown in Fig. 11-5, where d =25ft, a = 0.803 in.

~ = #lo cosh-â&#x20AC;˘ !!_ = (4 x f

10-7 ) cosh-â&#x20AC;˘

25 12 ( ) = 2.37 #lH/m 2(0.803)

The approximate formula gives L #lo d -=-In-= 2.37 #lH/m f

When d/a 2: 10, the approximate formula may be used with an error of less than 0.5%.

11.3.

A circular conductor with the same radius as in Problem 11.2 is 12.5 ft from an infinite conducting plane. Find the inductance.

~ = #lo In~= (2 x 10-7 ) In 25( 12) = 1.18 #lH/m 2n

0.803

This result is ! that of Problem 11.2. A conducting plane may be inserted midway between the two conductors of Fig. 11-5. The inductance between each conductor and the plane is 1.18 #lH/m. Since they are in series, the total inductance is the sum, 2.37 #lH/m.

11.4.

8!!. u.s.

Assume that the air-core toroid shown in Fig. 11-4 has a circular cross section of radius 4 mm. Find the inductance if there are 2500 turns and the mean radius is r = 20 mm. 2 7 2 2 L = #lN S = (4n x 10- )(2500) n(0.004) = _ mH 3 14 2nr 2n(0.020) Assume that the air-core toroid in Fig. 11-3 has 700 turns, an inner radius of 1 em, an outer radius of 2 em, and height a= 1.5 em. Find L using (a) the formula for square cross-section toroids; (b) the approximate formula for a general toroid, which assumes a uniform H at a mean radius. L = #lo~a In~= (4n X 10-7)(700)2(0.015) In 2 = 1.02 mH 2n ~ 2n 2 7 L = #lo~S = (4n x 10- )(700) (0.01)(0.015) =0. mH 98 2nr 2n(0.015)

(a) (b)

With a radius that is larger compared to the cross section, the two formulas yield the same result. See Problem 11.26.

11.6.

Use the energy integral to find the internal inductance per unit length of a cylindrical conductor of radius a. At a distance

r::; a

from the conductor axis, lr H=2ml2._

whence The inductance corresponding to energy storage within a length f of the conductor is then L

(B . H) dv

_E2.._ [ rlnrf dr =#lof 4n2a 4 0 81r

or L/f= #lo!Bn. This agrees with the result of Section 11.4.

180

11.7.

[CHAP. 11

INDUCTANCE AND MAGNETIC CIRCUITS

The cast iron core shown in Fig. 11-18 has an inner radius of 7 em and an outer radius of 9 em. Find the flux if the coil mmf is 500 A. f = 2.1r(0.08) = 0.503 m

F 500 H=e= o.5o3 =995 A/m

From the B-H curve for cast iron in Fig. 11-13,

B = 0.40 T. 2

$ = BS = (0.40)(0.02) = 0.16 mWb

2cm

Fig- 11-18

11.8.

a v-f+

The magnetic circuit shown in Fig. 11-19 has a C-shaped cast steel part, 1, and a cast iron part, 2. Find the current required in the 150-turn coil if the flux density in the cast iron is B 2 =0A5T.

Mathcad

12cm

1 Scm

Fig. 11-19

The calculated areas are S, = 4 x 10-- 4 m 2

and

~ = 3.6 x 10 -• m

•

f 1 = 0.11 + 0.11 + 0.12 = 0.34m f 2 = 0.12 + 0.009 + 0.009 = 0.138 m

From the B-H curve for cast iron in Fig. 11-13,

H 2 = 1270 A/m.

$ = B 2 S2 = (0.45)(3.6 X 10

)

= 1.62 X 10- 4 Wb

B, =-=0.41 T S, Then, from the cast steel curve in Fig. 11-12, H, = 233 A/m. The equivalent circuit, Fig. 11-20, suggests the equation F = Nl= H 1 f 1 + H2 f 2

150/ = 233(0.34) + 1270(0.138) I= 1.70 A

:VI

Fig. 11-20

The mean lengths are

11.9.

181

INDUCTANCE AND MAGNETIC CIRCUITS

CHAP. 11)

The magnetic circuit shown in Fig. 11-21 is cast iron with a mean length f 1 = 0.44 m and square cross section 0.02 x 0.02 m. The air-gap length is fa = 2 mm and the coil contains 400 turns. Find the current I required to establish an air-gap flux of 0.141 mWb.

Fig. 11-21 The flux 4» in the air gap is also the flux in the core. «~»

B; :s.= From Fig. 11-13, II;= 850 A/m.

Then H;t,

For the air gap,

= 850(0.44) = 374 A

Sa= (0.02 + 0.002)2 = 4.84 X 10- 4 m2 ,

and so

o.t41 x w- 3

«~»

H,t, = lloSa fa= (4.n

Therefore, F

0.141 x to-] X =0.3ST 4 10 4

= H;f; + H,f, = 838 A

3 10 7 )(4.84 X 10- 4 ) ( 2 X 10 ) = 464 A

and F

838

I=N= 400 =2.09A 11.10. Determine the reluctance of an air gap in a de machine where the apparent area is Sa= 4.26 X 10- 2 m2 and the gap length fa= 5.6 mm. 9li = ~ /loS,

5.6 x 10-3 (4.n X 10 7 )(4.26 X 10

= 1.05 X lW H-t 2 )

11.11. The cast iron magnetic core shown in Fig. 11-22 has an area S; = 4 cm 2 and a mean length 0.438 m. The 2-mm air gap has an apparent area Sa= 4.84 cm2 • Determine the air-gap flux~-

...i. 2mm T ..,.. 11-22 The core is quite long compared to the length of the air gap. and cast iron is not a particularly good magnetic material. As a first estimate, therefore, assume that 600 of the total ampere turns are dropped at the air gap, i.e., H,f, = 600 A.

4»

n"t,=-st, llo "

«~»

_ 600(4.n x w- )(4.84 x w-•> _ 7

1.82 X 10

-•

182

INDUCTANCE AND MAGNETIC CIRCUITS Then B; = «<J/S, = 0.46 T,

and from Fig. 11-13, II;== 1340 A/m.

[CHAP. 11

The core drop is then

H;f; = 1340(0.438) = 587 A

so that H;f;

+ H,f, =

1187 A

This sum exceeds the 1000 A mmf of the coil. Consequently, values of B, lower than 0.46 T should be tried until the sum of H,f; and H,f, is 1000 A. The values B; = 0.41 T and «lJ = 1.64 x to-• Wb will result in a sum very close to 1000 A.

ll.U. Solve Problem 11.11 using reluctances and the equivalent magnetic circuit, Fig. 11-23.

Fig. 11-23 From the values of B, and H1 obtained in Problem 11.11,

B-

#loJJ, == ~ == 3.83 x

w-• H/m

Then, for the core, a~~

0.438

""~ = - - (3.83 X 10 4 )(4 X 10 #lo#l,S;

2 86 .

nl>

X lu

H-1

and for the air gap,

!J1t - - " "- #loSa

w-]

(4Jr X 10- )(4.84 X 10- 4 )

=3.29X tlfH- 1

The circuit equation, F=«l.J(~+

gives

«J.J

!Jit,.)

1000 =

2.86 X llf + 3.29 X llf

1.63 x

w-• Wb

The corresponding flux density in the iron is 0.41 T, in agreement with the results of Problem 11.11. While the air-gap reluctance can be calculated from the dimensions and #lo, the same is not true for the reluctance of the iron. The reason is that #l, for the iron depends on the values of B, and H,.

11.13. Solve Problem 11.11 graphically with a plot of

versus F.

Values of H, from 700 through 1100 A/m are listed in the first column of Table 11-2; the corresponding values of 8 1 are found from the cast iron curve, Fig. 11-13. The values of «lJ and H;f; are computed, and H,.t.. is obtained from «l.Jf,llloS... Then F is given as the sum of H;f; and H.t, . Since the air gap is linear, only two points are required. The flux «lJ for F = 1000 A is seen from Fig. 11-24 to be approximately 1.65 x to-•Wb. This method is simply a plot of the trial and error data used in Problem 11.11. However, it is helpful if several different coils or coil currents are to be examined.

183

INDUCTANCE AND MAGNETIC CIRCUITS

CHAP. 11}

Table 11-2 H;f; (A)

H,.f,. (A)

F(A)

w-•

307

388

695

0.335

1.34 x to-•

350

441

79t

900

0.365

t.46 x to-•

395

480

874

1000

0.400

1.60 x

438

526

964

1100

0.420

w-• 1.68 x w-•

482

552

1034

41> (Wb)

H;(A/m)

B;(T)

700

0.295

1.18 x

800

.-.

•I

F(A)

..,.. ll-Z4

11.14. Determine the fluxes fl) in the core of Problem 11.11 for coil mmfs of 800 and 1200 A. a graphical approach and the negative air-gap line.

Use

The 41> versus H;f; data for the cast iron core, developed in Problem 11.13, are plotted in Fig. 11-25. The air-gap 41> versus F is linear. One end of the negative air-gap line for the coil mmf of 800 A is at 41> = 0, F = 800 A. The other end assumes H,.f,. = 800 A, from which 41>

= f.loS,.~H,.f,.)

2.43 x to-• Wb

which locates this end at 41> = 2.43 x to·• Wb, F =0. The intersection of the F = 800 A negative air-gap line with the nonlinear «1» versus F curve for the cast iron core gives 41> = 1. 34 x to-• Wb. Other negative air-gap lines have the same negative slope. For a coil mmf of tOOO A, 41> = 1.63 x to-• Wb and for 1200 A, 41> = 1.85 x Wb.

w-•

11.15. Solve Problem 11.11 for a coil mmf of 1000 A using the B-H curve for cast iron. This method avoids the construction of an additional curve such as the «1» versus F curves of Problems 11.13 and 11.14. Now, in order to plot the air-gap line on the B-H curve of iron, adjustments must be made for the different areas and the different lengths. Table 11-3 suggests the necessary calculations.

F tOOO -=--=2283A/m t, 0.438

INDUCfANCE AND MAGNEfiC CIRCUITS

184

[CHAP. 11

F{A)

Fig. 11-25 Table 11-3

Ba (T)

Ha (A/m)

Ba(~)

(T)

Ha(i)

(A/m)

!__H (~) (A/m) l;

0.10

0.80 X lOS

0.12

363

1920

0.30

2.39 X lOS

0.36

1091

1192

O.SO

3.98 X lOS

0.61

1817

466

The data from the third and fifth columns may be plotted directly on the cast iron B-H curve, as shown in Fig. 11-26. The air gap is linear and only two points are needed. The answer is seen to be B; = 0.41 T. The method can be used with two nonlinear core parts, as well (see Problem 11.16).

H{A/m)

Fig. 11-26

CHAP. 11)

185

INDUCfANCE AND MAGNETIC CIRCUITS

11.16. The magnetic circuit shown in Fig. 11-27 consists of nickel-iron alloy in part 1, where ~~ = lOcm and sl =2.25cm2 • and cast steel for part 2. where fz= 8 em and s2 = 3 cm 2 • Find the flux densities B I and Bz. Math cad

I 2

Fig. 11-27

The data for part 2 of cast steel will be converted and plotted on the 8-H curve for part 1 of nickel-iron alloy ( F /l'1 = 400 A/m). Table 11-4 suggests the necessary calculations.

Table 11-4

82(~)

H2(A/m)

0.33

200

0.44

160

240

0.44

250

0.59

200

0.55

300

0.73

240

160

0.65

350

0.87

2RO

120

0.73

400

0.97

320

0.7R

450

1.04

360

0.83

500

1.11

400

From the graph, Fig. 11-28,

(T)

(A/m)

£_Hz(~)

8 2 (T)

H2(t)

8 1 = 1.01 T.

Then, since

2. 25 x wX 10 4 8 2 = 1.01 ( 3

l'l

(A/m)

8 1 S 1 = 8 2S2 ,

4 )

=0.76T

These values can be checked by obtaining the corresponding H 1 and H 2 from the appropriate 8-H curves and substituting in

11.17. The cast steel parallel magnetic circuit in Fig. 11-29(a) has a coil with 500 turns. The mean lengths are f 2 = ( 1 = 10 em, f 1 = 4 em. Find the coil current if ct>3 = 0.173 mWb.

1 =

Math cad

<l>2

+ >

Since the cross-sectional area of the center leg is twice that of the two side legs, the flux density is the same throughout the core, i.e., 81=8,=8,=

Corresponding to

8 = 1.15 T,

0.173 X 10 3 1.5X10 4

Fig. 11-13 gives

l.15T

H = 1030 A/m.

The N/ drop between points a

186

[CHAP. 11

INDUCTANCE AND MAGNETIC CIRCUITS

!00

H 1 (A/m)

Fig. 11-28 a

I I

J-+-2~m~

'2------ --- --Âˇ---')b

b (b)

(a)

Fig. ll-29 and b is now used to write the following equation [see Fig. ll-29(b)): F -Hf, = Hf2 = Hf3

F=H(f1 + f 2 )= 1030(0.14) = 144.2A

or F

I=N=

Then

144.2 =0.29A 500

11.18. The same cast steel core as in Problem 11.17 has identical 500-turn coils on the outer legs, â&#x20AC;˘ with the winding sense as shown in Fig. ll-30(a). If again cll 3 = 0.173 mWb, find the coil currents. The ftux densities are the same throughout the core and consequently H is the same. The equivalent circuit in Fig. 11-30(b) suggests that the problem can be solved on a per pole basis.

413

B = ~ = 1.15T

f; = H( f,

and

H= l030A/m

+ f 3 ) = 1030(0. 14) = 144.2 A

Each coil must have a current of 0.29 A.

(from Fig. 11-13) /=0.29A

187

INDUCfANCE AND MAGNETIC CIRCUITS

CHAP. 11)

Ht3

I-I

(j)

Htj_OIF,

<fll

(b)

(a)

Fig. 11-30

11.19. The parallel magnetic circuit shown in Fig. ll-31(a) is silicon steel with the same cross-sectional area throughout, S = 1.30 cm2 • The mean lengths are t 1 = t 3 = 25 em, f 2 = 5 em. The coils have 50 turns each. Given that cl> 1 = 90 ~JWb and cl> 3 = 120 ~JWb, find the coil currents. !1>2 = 4>3 - 4>, = 0.30 X 10- 4 Wb 90x 10- 6 B, = 1.30 X 10-4 0.69T

r--4>1 .__. -.- 4>3-- -,

' fl

-----<~>3

<~>.-

HI tiT

H2{2I

F,~

j_H

3 t3

tFJ

<fl2

b (b)

(a)

Fig. 11-31 From Fig. 11-12, H 1 =87A/m. Then, H 1 f 1 =21.8A. Similarly, B 2 =0.23T, H2 =49A/m, H 2 f 2 = 2.5 A; and 8 3 = 0.92 T, H 3 = 140 A/m, H 3 f 3 = 35.0 A. The equivalent circuit in Fig. ll-31(b) suggests the following equations for the Nl drop between points a and b: H,f,- F,. = Hzfz = F;- H3f3 21.8- Fi = 2.5 = F;- 35.0 from which Fi = 19.3 A

and

F; = 37.5 A.

The currents are 11 = 0.39 A

and 13 = 0. 75 A.

11.20. Obtain the equivalent magnetic circuit for Problem 11.19 using reluctances for three legs, and calculate the flux in the core using fi = 19.3 A and F3 = 37.5 A.

92=-~IJofJ,S

From the values of Band H found in Problem 11.19, fJofJ, 1 = 7. 93 X 10- 3 H/m

1-'of.',z

= 4.69 x 10-

H/m

Now the reluctances are calculated: 92, = ___!!._s =2.43 x tOS H-• #Jof.'rt

fJofJr3 = 6.57

10- 3 H/m

188

INDUCfANCE AND MAGNETIC CIRCUITS ~=

8.20 x l(f H-\

~=2.93

[CHAP. 11

x lOS H- 1 â&#x20AC;˘ From Fig. 11-32, I) = 4>3~ + 4>2~

(1)

F.= 4>,9/tt- 4>2~

(2)

4>,

+ 4>2 = 413

-<~>,

(3)

-<1>3

Fig. 11-31 Substituting 4>2 from (3) into (1) and (2) results in the following set of simultaneous equations in 4> 1 and 413: F.=

41,(9/t, + ~)- 4>3~

I)= -4>,~

+ 4>3(~ + ~)

19.3 = 4>,(3.25 X lOS)- 4>3(0.82 X lOS) 37.5 = -4>,(0.82 X lOS)+ 4>3(3. 75

tOS)

Solving, 4> 1 = 89.7 IJWb, 4> 2= 30.3 IJWb, !1>1 = 120 IJWb. Although the simultaneous equations above and the similarity to a two-mesh circuit problem may be interesting, it should be noted that the ftux densities Br. Bv and 8 3 had to be known before the relative permeabilities and reluctances could be computed. But if B is known, why not find the ftux directly from 41 = BS? Reluctance is simply not of much help in solving problems of this type.

Supplementary Problems U.n.

Find the inductance per unit length of a coaxial conductor with an inner radius a = 2 mm outer conductor at b = 9 mm. Assume IJ, = 1. Ans. 0.301 IJH/m

U.n.

Find the inductance per unit length of two parallel cylindrical conductors, where the conductor radius is 1 mm and the center-to-center separation is 12 mm. Ans. 0. 992 ~JH/m

and an

11.23. Two parallel cylindrical conductors separated by 1 m have an inductance per unit length of 2.12 ~JH/m. What is the conductor radius? Ans. 5 mm 11.14. An air-core solenoid with 2500 evenly spaced turns has a length of 1.5 m and radius 2 x 10- 2 m. Ans. 6.58 mH the inductance L.

Find

11.15. A square-cross-section, air-core toroid such as that in Fig. 11-3 has inner radius 5 em, outer radius 7 em and height 1.5 em. If the inductance is 495 IJH, how many turns are there in the toroid? Examine the approximate formula and compare the result. Ans. 700, 704 11.16. A square-cross-section toroid such as that in Fig. 11-3 has r 1 = 80 em, r2 = 82 em, a= 1.5 em, and 700 turns. Find L using both formulas and compare the results. (See Problem 11.5.) Ans. 36.3 IJH (both formulas)

U.n.

A coil with 5000 turns, r 1 = 1.25 em, and f, = l.Om has a core with IJ, =50. A second coil of 500 turns, r2 = 2.0cm, and f 2 = lO.Oem is concentric with the first coil, and in the space between the coils IJ = IJo . Find the mutual inductance. Ans. 7. 71 mH

11.28. Determine the relative permeabilities of cast iron, cast steel, silicon steel, and nickel-iron alloy at a ftux density of 0.4 T. Use Figs. 11-12 and 11-13. Ans. 318, 1384, 5305, 42,440

CHAP. ll)

189

INDUCfANCE AND MAGNETIC CIRCUITS

11.Z9. An air gap of length fa= 2 mm has a flux density of 0.4 T. Determine the length of a magnetic core with the same Nl drop if the core is of (a) cast iron, (b) cast steel, (c) silicon steel. Ans. (a) 0.64cm; (b) 2.77m; (c) 10.6m 11.30. A magnetic circuit consists of two parts of the same ferromagnetic material (IJ, = 4000). Part 1 has f 1 = SO mm, S1 = 104 mm2 ; part 2 has f 2 = 30 mm, ~ = 120 mm2 â&#x20AC;˘ The material is at a part of the curve where the relative permeability is proportional to the flux density. Find the flux 4> if the mmf is 4.0A. Ans. 26.3 IJWb 11.31. A toroid with a circular cross section of radius 20 mm has a mean length 280 mm and a flux 1.50 mWb. Find the required mmf if the core is silicon steel. Ans. 83.2 A

4> =

11.32. Both parts of the magnetic circuit in Fig. 11-33 are cast steel. Part 1 has f 1 = 34 em and S, = 6cm2 ; part2has f 2 =16cm and ~=4cm 2 â&#x20AC;˘ Determinethecoilcurrent/,,if / 2 =0.SA, N 1 = 200 turns, N2 = 100 turns, and 4> = 120 IJWb. Ans. 0.65 A

Fig. 11-33 11.33. The silicon steel core shown in Fig. 11-34 has a rectangular cross section 10 mm by 8 mm and a mean length 150mm. The air-gap length is 0.8mm and the air-gap flux is 801JWb. Find the Ans. S61.2A mmf.

Fig. 11-34 11.34. Solve Problem 11.33 in reverse: the coil mmf is known to be 561.2 A and the air-gap flux is to be determined. Use the trial and error method, starting with the assumption that 90% of the Nl drop is across the air gap. 11.35. The silicon steel magnetic circuit of Problem 11.33 has an mmf of 600 A. flux. Ans. 85.2 IJWb

Determine the air-gap

11.36. For the silicon steel magnetic circuit of Problem 11.33, calculate the reluctance of the iron, !12; , and the reluctance of the air gap, !l'la. Assume the flux 4> = 80 IJWb and solve for F. See Fig. 11Ans. !12, = 0.313 IJH- 1 , !l'la = 6. 70 IJH- 1 , F = 561 A 35.

Fig. 11-35

190

INDUCTANCE AND MAGNETIC CIRCUITS

[CHAP. 11

11.37. A silicon steel core such as shown in fig. 11-34 has a rectangular cross section of area S; = 80 mm2 and an air gap of length f,. = 0.8 mm with area S,. = 95 mm2 • The mean length of the core is 150 mm and the mmf is 600 A. Solve graphically for the flux by plotting 4> versus F in the manner of Problem 11.13. Ans. 851lWb 11.38. Solve Problem 11.37 graphically using the negative air-gap line for an mmf of 600 A.

Ans.

851JWb

11.39. Solve Problem 11.37 graphically in the manner of Problem 11.15, obtaining the flux density in the core. Ans. 1.06T 11.40. A rectangular ferromagnetic core 40 x 60 mm has a flux 4> = 1.44 mWb. An air gap in the core is of length f,. = 2.5 mm. Find the Nl drop across the air gap. Ans. 1079 A 11.41. A toroid with cross section of radius 2 em has a silicon steel core of mean length 28 em and an air gap of length l mm. Assume the air-gap area, S,. , is 10% greater than the adjacent core and find the mmf required to establish an air-gap flux of 1.5 mWb. Ans. 952 A 11.42. The magnetic circuit shown in Fig. 11-36 has an mmf of 500 A. Part 1 is cast steel with f 1 = 340 mm and S1 = 400 mm2 ; part 2 is cast iron with f 2 = 138 mm and ~ = 360 mm 2 • Detennine Ans. 229 IJWb the flux 4>.

Fig. 11-36 11.43. Solve Problem 11.42 graphically in the manner of Problem 11.16.

Ans.

2291-'Wb

11.44. A toroid of square cross section, with r 1 = 2 em, r2 = 3 em, and height a = 1 em, has a two-part core. Part 1 is silicon steel of mean length 7.9em; part 2 is nickel-iron alloy of mean length Ans. w-• Wb 7.9 em. Find the flux that results from an mmf of 17.38 A. 11.45. Solve Problem 11.44 by the graphical method of Problem 11.15. Why is it that the plotting of the second reverse B-H curve on the first is not as difficult as might be expected? Ans. 10- 4 Wb. The mean lengths and cross-sectional areas are the same 11.46. The cast steel parallel magnetic circuit in Fig. 11-37 has a 500-tum coil in the center leg, where the cross-sectional area is twice that of the remainder of the core. The dimensions are f,. = 1 mm, ~ = ~= 150mm 2 , S1 =300mm2 , f 1 =40mm, f 2 = llOmm, and f 3 = 109mm. Find the coil current required to produce an air-gap flux of 125 IJWb. Assume that S., exceeds~ by 17%. Ans. 1.34 A

I I

__ --- -~----- -· I

CHAP. 11)

INDUCfANCE AND MAGNETIC CIRCUITS

191

11.47. The cast iron parallel circuit core in Fig. 11-38 has a 500-turn coil and a uniform cross section of 1.5 cm2 throughout. The mean lengths are f 1 = f 3 = 10 em and f 2 = 4 em. Determine the coil current Ans. 1.05 A necessary to result in a flux density of 0.25 T in leg 3.

Fig. 11-38 11.48. Two identical 500-turn coils have equal currents and are wound as indicated in Fig. 11-39. The cast steel core has a flux in leg 3 of 120#'Wb. Determine the coil currents and the flux in leg 1. Ans. 0.41 A, OWb

0.06

fig. 11-39 11.49. Two identical coils are wound as indicated in Fig. 11-40. The silicon steel core has a cross section of 6 cm2 throughout. The mean lengths are f 1 = f 3 = 14 em and ( 2 = 4 em. Find the coil mmfs if the flux in leg 1 is 0.7 mWb. Ans. 38.5 A

Chapter 12 Displacement Current and Induced EMF 12.1 DISPLACEMENf CURRENf In static fields the curl of H was found to be pointwise equal to the current density Jc. This is conduction current density; the subscript c has been added to emphasize that moving chargeselectrons, photons, or ions--compose the current. If V X H = Jc were valid where the fields and charges are variable with time, then the continuity equation would be V • Jc = V • (V X H)= 0, instead of the correct

V·J = - c at

Hence, James Clerk Maxwell postulated that where

VX H=J.. +lv

lv=a,

With the inclusion of the displacement current density Jv. the continuity equation is satisfied:

v·J.. = - v·Jv = - v·

a:= -:,<v .

D) = -

The displacement current iv through a specified surface is obtained by integration of the normal component of Jv over the surface (just as ic is obtained from J.. ).

iv=1 lv · dS= s

laD· sat

dS=!!J D· dS dts

Here, the last expression assumes that the surfaceS is fixed in space. EXAMPLE 1. Use Stokes' theorem (Section 9.8) to show that ic = i0 Since the two surfaces S, and ~ have the common contour C, ,( H · dl =

'fc

(V X H) • dS =

.51

1 s,

(1c +

in the circuit of Fig. 12-1.

(V X H) • dS

an) ·dS = J. (1r + an) ·dS at at

·'li

192

CHAP. 12)

193

DISPLACEMENT CURRENT AND INDUCED EMF

Assuming the flux is confined to the dielectric between the conducting plates, free charges are in motion within the dielectric, J, = 0 over S2 • Therefore,

. ds J.(sl J · ds = J an at c

over S1 •

And since no

It should be noted that ant at is nonzero only over that part of S2 that lies within the dielectric. EXAMPLE 2. Repeat Example 1, this time using circuit analysis. Refer to Fig. 12-1. The capacitance of the capacitor is C= EA d

where A is the plate area and dis the separation.

The conduction current is then

i = cdv = EAdv c dt d dt

On the other hand, the electric field in the dielectric is, neglecting fringing, aD at

Hence

Edv -ddt

D=EE=-v d

E = v/d.

and the displacement current is (n is normal to the plates)

i = fJ

I an. dS I ~ =

dv dS = EA dv = i d dt d dt c

12.2 RATIO of lc TO JD Some materials are neither good conductors nor perfect dielectrics, so that both conduction current and displacement current exist. A model for the poor conductor or lossy dielectric is shown in Fig. 12-2. Assuming the time dependence eiwt for E. the total current density is

J, =Jc +ln = aE

a at

+-(~:E)=

aE + jwEE

from which

Jc Jn

a WE

As expected, the displacement current becomes increasingly important as the frequency increases.

Fig. U-2 EXAMPLE 3. A circular-cross-section conductor of radius 1.5 mm carries a current ic = 5.5 sin (4 x 1010t) (~-tA). What is the amplitude of the displacement current density, if a= 35 MS/m and Er = 1?

Jc JfJ

Then

JD =

=-W-E= (4

3.50 X 107 101<')( 10 9 /36:r) = 9 "90

(5.5 x 10 6 )/[:r(I.5 x 10- 3 ) 2 ] 9.90 X 107

7.86

107 __ ,

10 - IJA 1m

194

DISPLACEMENT CURRENT AND INDUCED EMF

[CHAP. 12

12.3 FARADAY'S LAW AND LENZ'S LAW The minus sign in Faraday's law (Section 11.3) implicitly gives the polarity of the induced voltage v. To make this explicit, consider the case of a plane areaS, bounded by a closed curve C, where S is cut perpendicularly by a time-variable flux density b (Fig. 12-3). Faraday's law here takes the integral form ,( E Âˇ dl = ic

d dt

B Âˇ dS

in which the positive sense around C and the direction of the normal, dS, are corrected by the usual right-hand rule (Fig. 12-3(a)). Now if B is increasing with time, the time derivative will be positive and, thus, the right side of the above equation will be negative. In order for the left integral to be negative, the direction of E must be opposite to that of the contour, Fig. 12-3(b). A conducting filament in place of the contour would carry a current ic , also in the direction of E. As shown in Fig. 12-3(c), such a current loop generates a flux </>'which opposes the increase in B. Lenz's law summarizes this discussion: the voltage induced by a changing flux has a polarity such that the current established in a closed path gives rise to a flux which opposes the change in flux.

~ ---E

(b)

(a)

(c)

fig. ll-3

In the special case of a conductor moving through a time-independent magnetic field, the polarity predicted by Lenz's law is yielded by two other methods. (1) The polarity is such that the conductor experiences magnetic forces which oppose its motion. (2) As indicated in Fig. 12-4, a moving conductor appears to distort the flux, pushing the flux lines in front of it as it moves. This same distortion is suggested by the counterclockwise flux lines shown around the conductor. By the right-hand rule the current which would result if a closed path were provided would have the direction shown, and the polarity of the induced voltage is+ at the end of the conductor where the current would leave. Figure 12-5 confirms this by comparing the moving conductor and its resulting current to a voltage source connected to a similar external circuit.

Fig. 12-4

CHAP. 12]

DISPLACEMENT CURRENT AND INDUCED EMF

195

(b)

(a)

Fig. J.l..S

U.4 CONDUCTORS IN MOTION THROUGH TIME-INDEPENDENT FIELDS The force F on a charge Q in a magnetic field 8, where the charge is moving with velocity U, was examined in Chapter 10.

F= Q(UXB) A motional electric field intensity, Em, can be defined as the force per unit charge: F

Em= Q=UXB When a conductor with a great number of free charges moves through a field B, the impressed Em creates a voltage difference between the two ends of the conductor, the magnitude of which depends on how Em is oriented with respect to the conductor. With conductor ends a and b, the voltage of a with respect to b is Vab

Em • dl =

(U X B) • dl

If the velocity U and the field B are at right angles, and the conductor is normal to both, then a conductor of length f will have a voltage v=BfU

For a closed loop the line integral must be taken around the entire loop:

(U X B) • dl

Of course, if only part of the complete loop is in motion, it is necessary only that the integral cover this part, since Em will be zero elsewhere. EXAMPLE 4. In Fig. 12-6, two conducting bars move outward with velocities U, = 12.5( -a,.) m/s and U 2 = 8.0a,. m/s in the field B = 0.35a. T. Find the voltage of b with respect to c. At the two conductors, Em,= U 1 X 8=4.38(-a_.)V/m Emz= VzX B =2.8011_. V/m z

196

DISPLACEMENT CURRENT AND INDUCED EMF

[CHAP. 12

and so Vat>=

[~ 4.38( -a.)· dxa.. = -2.19 V

Vdc

= r·~ 2.80a.. • dxa.. = 1.40 V 0

v"" = v,., + v,., + v. =2.19+ 0+ 1.40= 3.59V Since b is positive with respect to c, current through the meter will be in the ay direction. This clockwise current in the circuit gives rise to flux in the -a. direction, which, in accordance with Lenz's law, counters the increase in the flux in the +a. direction due to the expansion of the circuit. Moreover, the forces that B exerts on the moving conductors are directed opposite to their velocities.

U.S CONDUCI'ORS IN MOTION THROUGH TIME-DEPENDENT FIELDS When a closed conducting loop is in motion (this includes changes in shape) and also the field B is a function of time (as well as of position), then the total induced voltage is made up of a contribution from each of the two sources of flux change. Faraday's law becomes

-!!_f.

dts

B • dS =

-f.

aB · dS + J. (U X B) · dl sat j

The first term on the right is the voltage due to the change in 8, with the loop held fixed; the second term is the voltage arising from the motion of the loop, with B held fixed. The polarity of each term is found from the appropriate form of Lenz's law, and the two terms are then added with regard to those polarities. EXAMPLE 5. As shown in Fig. 12-7(a), a planar conducting loop rotates with angular velocity w about the x axis; at t = 0 it is in the xy plane. A time-varying magnetic field, B = B(t)an is present. Find the voltage induced in the loop by using the two-term form of Faraday's law. z

(b)

(a)

FIJ.ll-7

Let the area of the loop be A.

v1 =

The contribution to v due to the variation of B is

-L aB. -L Sat

dS =

dB a • dSa = - dB A cos wt

.~dt

since a. • a,= cos wt. To calculate the second, motional contribution to v, the velocity U of a point on the loop is needed. Fig. 12-7(b) it is seen that U = rwa., = _Y_ wa., cos wt

so that

U X B = _Y_ wa., X Ba. COS

= _Y_ wB sin wt( -a.. ) COS

From

DISPLACEMENT CURRENT AND INDUCED EMF

CHAP. 12]

since •" X •· = sin wt( -•.. ).

197

Consequently,

wBsinwtp y•.. • dl coswt

v2 = (U X 8) • dl = -

Stokes' theorem (Section 9.8) can be used to evaluate the last integral.

py•.. ·dl=

(VXy•.. )·dS=

v2 =-

1berefore

Since V X y•.. = -•. ,

(-•.)·dS•"=-Acoswt

wBsinwt . (-A cos wt)= BAwsm wt coswt

Solved Problems 12.1.

In a material for which o = 5.0 S/m and £, = 1 the electric field intensity is E = 250sin 1010t (V/m). Find the conduction and displacement current densities, and the frequency at which they have equal magnitudes. lc = oE = 1250 sin 1010t

(A/m2 )

On the assumption that the field direction does not vary with time, }0

aD=~ ( £0 £,250 sin 1010t) = 22.1 cos 1010t

at at

(A/m2 )

For lc=lo,

which is equivalent to a frequency

12.2.

5.0 II x _, = 5.65 x 10 rad/s 8 854 10 2

O=W£

w= _

f = 8. 99 x

1010 Hz= 89.9 GHz.

A coaxial capacitor with inner radius 5 mm, outer radius 6 mm and length 500 mm has a dielectric for which £, = 6. 7 and an applied voltage 250 sin 377t (V). Determine the displacement current iv and compare with the conduction current ic. v = 0. Then, from Problem 8.7, the potential at 0.005 ::s

Assume the inner conductor to be at rs0.006m is

From this, E=-Vv=D = £0 £,E = -

t.37 x r

ur sm377t•, .

8 13 X 10- 8

sin 37718, (C/m2 )

lo =aD=- 3.07 X 10_, cos 37718,

(V/m)

(A/mz)

= J0 (2m-L) = 9.63 x 10-s cos 377t

1be circuit analysis method for ic requires the capacitance,

2n:£o£,L =

lnm

02 F n

(A)

198

DISPLACEMENT CURRENT AND INDUCED EMF

Then

ic =

C~~ = (1.02 X 10-9 )(250)(377)(cos377t) =9.63 X 10-scos377t

It is seen that ic = i0

U.J.

[CHAP. 12

(A)

Moist soil has a conductivity of 10-3 S/m and

2.5.

£, =

E=6.0x 10-6 sin9.0x 109 t First, lc = oE = 6.0 X 10- sin 9.0 X 10 t (A/m 9

Find lc and lv where (V/m)

Then. since

= £o£,E,

lo =at= £o£, a,= 1.20 X 10-6 cos 9.0 X 109 t (A/m 2 )

U.4.

Find the induced voltage in the conductor of Fig. 12-8 where B = 0.04a, T U = 2.5 sin lWtaz

and

(m/s)

Em= U X B = 0.10sin lWt( -•x)

(V/m)

0.10 sin 1Wt( -•x) • dx•..

= -0.02 sin 1Wt (V) The conductor first moves in the •· direction. at the z axis for this half cycle.

: "t

The x = 0.20 end is negative with respect to the end

~--8------y X

.-----0.20 I

U.S.

Rework Problem 12.4 if the magnetic field is changed to B = 0.04ax (T). Because the conductor cuts no field lines, the induced voltage must be zero. This may be verified analytically by use of Problem 1.8.

(U X B) . dl =

I u.

(B X dl) = 0

since B and dl are always parallel.

U.6.

An area of 0.65 m2 in the z = 0 plane is enclosed by a filamentary conductor. induced voltage, given that B = 0.05 cos 1Wr(

See Fig. 12-9. v= =

-1 ar s

aB·dS•

8 "

~a.)

•

L50sin1Wt( Y~•·)·dS•. 8

=23.0sin 1Wt (V)

(T)

Find the

CHAP. 12)

DISPLACEMENT CURRENT AND INDUCED EMF

199

Fig. U-9 The field is decreasing in the first half cycle of the cosine function. The direction of i in a closed circuit must be such as to oppose this decrease. Thus the conventional current must have the direction shown in Fig. 12-9.

U.7.

The circular loop conductor shown in Fig. 12-10 lies in the z = 0 plane, has a radius of 0.10 m and a resistance of 5.0 Q. Given B = 0.20 sin 1Wtaz (T}, determine the current.

4> = B Âˇ S= 2 x 10-3 n sin lWt (Wb) v =- dtl> = -2n- cos 1Wt (V) dt

i=~= -0.4n-cos 1Wt (A) At t = 0+ the flux is increasing. In order to oppose this increase, current in the loop must have an instantaneous direction -â&#x20AC;˘. . where the loop crosses the positive x axis.

s y X

Fig. 12-10

u.s.

The rectangular loop shown in Fig. 12-11 moves toward the origin at a velocity U -250.,. m/s in a field B = 0.80e-o.soyaz

(T)

Find the current at the instant the coil sides are at y = 0.50 m and 0.60 m, if R = 2.5 Q. z R

..... 12-11

200

DISPLACEMENT CURRENT AND INDUCED EMF

Only the 1.0-m sides have induced voltages.

Let the side at y =0.50m be 1.

v 1 = B 1lU = 0.80e- s(1)(250) = 155.8 V 0 2 ·

The voltages are of the polarity shown.

[CHAP. 12

v 2 = B 2 tU = 148.2 V

The instantaneous current is

i = 155.8- 148.2 = _ A 3 04 2.5

12.9.

A conductor 1 em in length is parallel to the z axis and rotates at a radius of 25 em at 1200 rev/min (See Fig. 12-12). Find the induced voltage if the radial field is given by B = 0.5a, T. z

Fig. 12-ll The angular velocity is rev) ( 1 min) rad) = 40:rrrad - ( 2:rr( 1200min 60 s rev s Hence

U = rw = (0.25)(40:rr) m/s Em = 10:rra• X 0. 5a, = 5. O:rr( -a.) V /m V

r·OI 5.0:rr( -a,) • dza, = -5.0

10-

2V 11"

The negative sign indicates that the lower end of the conductor is positive with respect to the upper end.

U.IO. A conducting cylinder of radius 7 em and height 15 em rotates at 600 rev/min in a radial field B = 0.20a, T. Sliding contacts at the top and bottom connect to a voltmeter as shown in Fig. 12-13. Find the induced voltage. w = (600)(~)(2n) = 20n rad/s

U = (20:rr)(0.07)a. m/s Em= UX 8=0.88(-a.)V/m

Each vertical element of the curved surface cuts the same flux and has the same induced voltage. These elements are effectively in a parallel connection and the induced voltage of any

T h

j_ Fig.l.l-13

DISPLACEMENT CURRENT AND INDUCED EMF

CHAP. 12]

201

element is the same as the total. rU-15

v = Jo

0.88( -a,) • dza,

(+ at the bottom)

= -0.13 V

12.11. In Fig. 12-14 a rectangular conducting loop with resistance R = 0.20 Q turns at 500 rev/min. The vertical conductor at r 1 = 0.03 m is in a field 8 1 = 0.25ar T, and the v-f+ conductor at r2 = 0.05 m is in a field 8 2 = 0.80ar T. Find the current in the loop.

Mat head

U, = (500){J-,)(2.n')(0.03)a<~> = O.SO.rraop m/s 05(1

v, =

(O.SO.rr~~,p

x 0.25a,) • dza,

= -0.20 V

Similarly, U 2 =

0.83.rr~~,p

m/s and

v 2 = -1.04 V. Then i=

1.04-0.20 0.20

=4.20A

in the direction shown on the diagram.

12.12. The circular disk shown in Fig. 12-15 rotates at w (rad/s) in a uniform flux density B = Ba2 • Sliding contacts connect a voltmeter to the disk. What voltage is indicated on the meter from this Faraday homopolar generator?

Fig. 12-15

One radial element is examined. so that

A general point on this radial element has velocity

wra<~>,

and

L o

= U X B = wrBa, wa 2 B wrBa, ·dra,.. = -2 -

where a is the radius of the disk. The positive result indicates that the outer point is positive with respect to the center for the directions of B and w shown.

12.13. A square coil, 0.60 m on a side, rotates about the x axis at w = 60n rad/s field B = 0.80az T, as shown in Fig. 12-16(a). Find the induced voltage. v-I+

Math cad

in a

202

DISPLACEMENT CURRENT AND INDUCED EMF z

[CHAP. 12

(a)

(b)

Fig. 12-16

Assuming that the coil is initially in the xy plane, a= wt = 60m (rad)

The projected area on the xy plane becomes (see Fig. 12-16(b)]: A =(0.6)(0.6cos60nt)

Then

(m2 )

4> = BA = 0.288 cos 60.nt (Wb) and v = - dtl> = 54.3 sin 60nt (V) dt

Lenz's law shows that this is the voltage of a with respect to b. Alternate Method

Each side parallel to the x axis has a y component of velocity whose magnitude is

IV,. I = lrw sin al = 118.0n sin 60.ntl (m/s) The voltages B t IU,.l for the two sides add, giving lvl = 2(Bt IUJ'I) = 154.3 sin 60.ntl

(V)

Lenz's law again determines the proper sign.

1.2.14. Check Example 5 by means of the original, differential form of Faraday's law. From Fig. 12-7(b) the projected loop area normal to the field is A cos wt, whence

tP = B(t) (A cos wt) and

dt/>

v =- dt = -dtA cos wt + BAw sin wt= v 1 + v2 (It is almost always simpler to use the differential form.)

12.15. Find the electric power generated in the loop of Problem 12.11. Check the result by calculating the rate at which mechanical work is done on the loop. The electric power is the power loss in the resistor: ~

=i R = (4.20) (0.20) = 3.53 W 2

The forces exerted by the field on the two vertical conductors are

=(4.20)(0.50)(0.25)(•. X •,) = 0.525-. N = i(iz X 8 = (4.20)(0.50)(0.80)( -•. X 8,) = -1.68-. N

F1 =i(l1 X B 1) F2

CHAP. 12]

DISPLACEMENT CURRENT AND INDUCED EMF

203

To tum the loop, forces -F1 and -F2 must be applied; these do work at the rate P = (-F1 ) • U 1 + (-F2 ) • U 2 = (-0.525)(0.50.n) + (1.68)(0.83.n) = 3.55 W

To within rounding errors,

P = P• .

Supplementary Problems ll.16. Given the conduction current density in a lossy dielectric as lc = 0.02 sin 109t (A/m2 ), find the displacement current density if 0 = 1W S/m and £, = 6.5. Ans. 1.15 X 10-6 cos 109 1 (A/m2) 12.17. A circular-cross-section conductor of radius 1.5 mm carries a current ic = 5.5 sin 4 x 1010t (I'A). is the amplitude of the displacement current density, if o = 35 MS/m and £, = 1? Ans. 7.frl X w-J I'A/m 2

What

12.18. Find the frequency at which conduction current density and displacement current density are equal in (a) distilled water, where o=2.0x10- 4 S/m and £,=81; (b) seawater, where o=4.0S/m and £, = 1. Ans. (a) 4.44 X let Hz; (b) 7.19 X 1010 Hz

ll.l9. Concentric spherical conducting shells at r1 = 0.5 mm and r2 = J mm are separated by a dielectric for which £, = 8.5. Find the capacitance and calculate ic, given an applied voltage v = 150 sin 5000t (V). Obtain the displacement current i0 and compare it with ic . Ans. ic = io = 7.09 X 10- 7 cos 5000t (A) ll.:ZO. Two parallel conducting plates of area 0.05 m2 are separated by 2 mm of a lossy dielectric for which £, = 8.3 and o = 8.0 x w-• S/m. Given an applied voltage v = 10 sin 107 t (V), find the total rms current. Ans. 0.192 A 12.11. A parallel-plate capacitor of separation 0.6mm and with a dielectric of £, = 15.3 has an applied rms voltage of 25 V at a frequency of 15 GHz. Find the rms displacement current density. Neglect fringing. Ans. 5.32 x 105 A/m2

ll.l2. A conductor on the x axis between x = 0 and x = 0.2 m has a velocity U = 6.0a. m/s in a field B = 0.04ay T. Find the induced voltage by using (a) the motional electric field intensity, (b) df/>/dt, and (c) BtU. Determine the polarity and discuss Lenz's law if the conductor was connected to Ans. 0.048 V (x = 0 end is positive) a closed loop. 12.23. Repeat Problem 12.22 for B = 0.04 sin kzay flux in one direction to the reverse direction.

(T). Discuss Lenz's law as the conductor moves from Ans. 0.048 sin kz (V)

ll.l4. The bar conductor parallel to they axis shown in Fig. 12-17 completes a loop by sliding contact with the conductors at y = 0 and y = 0.05 m. (a) Find the induced voltage when the bar is stationary at x = 0.05 m and B = 0.30 sin 1ifta. (T). (b) Repeat for a velocity of the bar U = 150._. m/s. Discuss the polarity. Ans. (a)-7.5cos1Cft (V); (b) -7.5cosleft-2.25sin1Cft (V)

o.os ..... ll-17

204

DISPLACEMENT CURRENT AND INDUCED EMF

[CHAP. 12

12.25. The rectangular coil in Fig. 12-18 moves to the right at speed U = 2.5 m/s. The left side cuts flux at right angles, where B, = 0.30 T, while the right side cuts equal flux in the opposite direction. Find the instantaneous current in the coil and discuss its direction by use of Lenz's law. Ans. 15 rnA (counterclockwise) 81

c==rl Im

t:::J:_y_ son

Fig. U-18 12.26. A rectangular conducting loop in the z = 0 plane with sides parallel to the axes has y dimension and x dimension 2 em. Its resistance is 5.0 Q. At a time when the coil sides are at 20 em and x = 22 em it is moving toward the origin at a velocity of 2.5 m/s along the x axis. the current if B = 5.0e- 111'a, (T). Repeat for the coil sides at x = 5 em and 7 em. Ans. 0.613 rnA, 2.75 rnA

1 em x= Find x=

12.27. The 2.0-m conductor shown in Fig. 12-19 rotates at 1200 rev/min in the radial field B = 0.10 sin cpa, (T). Find the current in the closed loop with a resistance of 100 Q. Discuss the polarity and the current direction. Ans. 5.03 x 10--> sin 40.m (A)

___ J __ _

,- r = 0.20 m -,,

T ---

-----~1

100.!1

I I

-J

Fig. U-19 12.28. In a radial field B = 0.50a, (T), two conductors at r = 0.23 m and r = 0.25 m are parallel to the z axis and are 0.01 m in length. If both conductors are in the plane cp = 40m, what voltage is available to circulate a current when the two conductors are connected by radial Ans. 12.6 m V conductors?

12.29. In Fig. 12-20 a radial conductor, 3 ::S r ::S 6 em, is shown embedded in a rotating glass disk. Two 11.2, mQ resistors complete two circuits. The disk turns at 12 rev/min. If the field at the disk is B = 0.30an (T), calculate the electric power generated. What is the effect of this on the rotation? Discuss Lenz's law as it applies to this problem. Ans. 46.3 f,!W

Fig. 12-20 12.30. What voltage is developed by a Faraday disk generator (Problem 12.12) with the meter connections at r, = 1 mm and r2 = 100 mm when the disk turns at 500 rev/min in a flux density of 0.80 T? Ans. 0.209 V 12.31. A coil such as that shown in Fig. 12-16(a) is 75 mm wide (y dimension) and 100 mm long (x dimension). What is the speed of rotation if an rms voltage of 0.25 Vis developed in the uniform field B = 0.45ay (T)? Ans. 1000 rev/min

Chapter 13 Maxwell's Equations and Boundary Conditions 13.1 INTRODUCTION The behavior of the electric field intensity E and the electric flux density D across the interface of two different materials was examined in Chapter 7, where the fields were static. A similar treatment will now be given for the magnetic field strength H and the magnetic flux density B, again with static fields. This will complete the study of the boundary conditions on the four principal vector fields. In Chapter 12, where time-variable fields were treated, displacement current density JD was introduced and Faraday's law was examined. In this chapter these same equations and others developed earlier are grouped together to form the set known as Maxwell's equations. These equations underlie all of electromagnetic field theory; they should be memorized.

13.2 BOUNDARY RELATIONS FOR MAGNETIC FIELDS When H and B are examined at the interface between two different materials, abrupt changes can be expected, similar to those noted in E and D at the interface between two different dielectrics (see Section 7.7). In Fig. 13-1 an interface is shown separating material 1, with properties a 1 and !J-,1 , from 2, with o 2 and !J-,2 â&#x20AC;˘ The behavior of B can be determined by use of a small right circular cylinder positioned across the interface as shown. Since magnetic flux lines are continuous,

1 B . dS = L j

Bt . dSI +

end I

L B . dS + L

82 . dS2 = 0

end 2

cyl

Now if the two planes are allowed to approach one another, keeping the interface between them, the area of the curved surface will approach zero, giving

- B,..

dSI

+ 8,.2

end 2

end l

from which

Fig. 13-1

205

dS2 = 0

206

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

[CHAP. 13

In words, the normal component of 8 is continuous across an interface. Note that either normal to the interface may be used in calculating 8,1 and 8, 2 â&#x20AC;˘ The variation in H across an interface is obtained by the application of Am~re's law around a closed rectangular path, as shown in Fig. 13-2. Assuming no current at the interface, and letting the rectangle shrink to zero in the usual way,

O=fHÂˇdi-H~ 1 6f1 -Hnll.f2 H~ 1 =Hn

whence

Thus tangential H has the same projection along the two sides of the rectangle. can be rotated 90" and the argument repeated, it follows that

Since the rectangle

H, 1 =H,2

In words, the tangential component of H is continuous across a current-free interface. The relation --=-

tan 6 2

1-'rt

between the angles made by H 1 and H2 with a current-free interface (see Fig. 13-2) is obtained by analogy with Example 6, Section 7. 7.

Fig.lJ-2

13.3 CURRENT SHEET AT THE BOUNDARY If one material at the interface has a nonzero conductivity, a current may be present. This could be a current throughout the material; however, of more interest is the case of a current sheet at the interface. Figure 13-3 shows a uniform current sheet. In the indicated coordinate system the current sheet has density K = Koay and it is located at the interface x = 0 between regions 1 and 2. The magnetic field H' produced by this current sheet is given by Example 2, Section 9.2, H~ = !K X a, 1 = !Koaz

Hi= !K X a,z = !Ko( -az)

Thus H' has a tangential discontinuity of magnitude IKol at the interface. If a second magnetic field, H", arising from some other source, is present, its tangential component will be continuous at the interface. The resultant magnetic field, H=H' +H"

will then have a discontinuity of magnitude IKol in its tangential component. This is expressed by the vector formula

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

CHAP. 13)

207

0 X

Fig. 13-3

where a, 12 is the unit normal from region 1 to region 2. The vector relation, which is independent of the choice of coordinate system, also holds for a nonuniform current sheet, where K is the value of the current density at the considered point of the interlace.

13.4 SUMMARY OF BOUNDARY CONDmONS For reference purposes, the relationships forE and D across the interlace of two dielectrics are shown below along with the relationships for H and B. Mqnetic Fields

Electric Fields (charge-free) (with surface charge)

H,1 = H,2 { (HI - H2) X anl2 = K tan 61 #Jr'l --=tan 62 llrl

(current-free) (with current sheet) (current-free)

tan 6,

Er2

--=-

These relationships were obtained assuming static conditions. be found to apply equally well to time-variable fields.

(charge-free)

However, in Chapter 14 they will

13.5 MAXWELL'S EQUATIONS A static E field can exist in the absence of a magnetic field H; a capacitor with a static charge Q furnishes an example. Likewise, a conductor with a constant current I has a magnetic field H without an E field. When fields are time-variable, however, H cannot exist without an E field nor can E exist without a corresponding H field. While much valuable information can be derived from static field theory, only with time-variable fields can the full value of electromagnetic field theory be demonstrated. The experiments of Faraday and Hertz and the theoretical analyses of Maxwell all involved time-variable fields. The equations grouped below, called Maxwell's equations, were separately developed and examined in earlier chapters. In Table 13-1, the most general form is presented, where charges and conduction current may be present in the region. Note that the point and integral forms of the first two equations are equivalent under Stokes' theorem, while the point and integral forms of the last two equations are equivalent under the divergence theorem.

208

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

[CHAP. 13

Table 13-1. MoweD's Equatiou, Geuenl Set Integral Form

Point Form

ao VXH=Jc +ar -

_fH·dl= i(Jc+ ~~)·dS

VXE= - -

V·D=p

f.o·dS=

a& at

E . dl =

V·B=O

L(- ~~) .dS [pdv

(Ampere's law) (Faraday's law; S fixed)

(Gauss' law)

(nonexistence of monopole)

fa·dS=O

For free space, where there are no charges (p = 0) and no conduction currents (Jc = 0), Maxwell's equations take the form shown in Table 13-2. Table 13-2. MoweD's EquatioD!i, Free-Space Set Point Form

Integral Form

ao VXH=-

fn·dl=

L(~~) ·dS

VXE=

L(-~~) .

at a& -at

E • dl =

V·D=O

V·B=O

fB·dS=O

D·dS=O

The first and second point-form equations in the free-space set can be used to show that time-variable E and H fields cannot exist independently. For example, if E is a function of time, then D = EoE will also be a function of time, so that 8D/ at will be nonzero. Consequently, V X H is nonzero, and so a nonzero H must exist. In a similar way, the second equation can be used to show that if H is a function of time, then there must be an E field present. The point form of Maxwell's equations is used most frequently in the problems. However, the integral form is important in that it better displays the underlying physical laws.

Solved Problems 13.1. In region 1 of Fig. 13-4, 8 1 = l.2ax + 0.8ay + 0.4az (T). Find H 2 (i.e., Hat ::1 the angles between the field vectors and a tangent to the interface •

z = +0) and

Write H, directly below B, . Then write those components of H 2 and 8 2 which follow directly from the two rules B normal is continuous and H tangential is continuous across a current-free interface.

209

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

CHAP. 13)

+ O.Bay + 0.4

8, = 1.2a..

H, = - (8.0... + 5.33ay + 2.67 P.o

(T)

a.)I0- 2

(A/m)

-2

(A/m)

= - (8.0... + 5.33ay + 1v P.oH. 2a.)10

l'o

82 =

Bx2Bx

By~y

+ 0.4

(T)

Now the remaining terms follow directly: By2

= 5.33

10- 2 (T)

Angle 6, is 90°- a, , where a 1 is the angle between 8 1 and the normal, a• . 8, ·a. cos a,= liiJ = 0.27

whence a,= 74.5° and 6, = 15.5°. Similarly, Check: (tan 6,)/(tan 6 2 ) = p.,2 /1',1 •

13.2. Region 1, for which

l'rl

= 3,

6 2 = 76.5°.

is defined by x < 0 and region 2, x > 0,

has

#-'r2 = 5.

Given

H 1 = 4.0a.., + 3.0a,- 6.0az

show that

6z = 19.7° and that

(A/rn)

Hz= 7.12 A/rn.

Proceed as in Problem 13.1. H1 =

4.0.. + 3.0.y- 6.0..

(A/m)

8, = p. 0 (12.0... + 9.0.y- 18.0..) (T) 8 2 = p. 0 (12.0... + 15.0.,. - 30.0..) (T) H2 =

Now

2.408.. + 3.0.y- 6.0..

= V(2.40f + (3.0) + ( -6.0) 2

(A/m)

= 7.12 A/m

The angle a 2 between H 2 and the normal is given by or

13.3. Region 1, where l'rt = 4, is the side of the plane y + z = 1 containing the origin (see Fig. 13-5). In region 2, l'r2 = 6. 8 1 = 2.0a.., + l.Oay (T), find Bz and Hz.

lit

Mill head

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

210

[CHAP. 13

F1g. 13-5

Choosing the unit normal a.,= (ay + a.)tv2, 8

_ (2.0.x + l.O.y) • (ay +a,)

.. ~-

8.,1 = (: )a.. = 0.5ay + 0.5a. = 8.,2 2 8,1 =81- 8,1 =2.0.x + 0.5ay- 0.5a, 1

U.1 =- (0. 5ax + 0.125ay - 0.125a.) = H,2 P.o 8,2 = P.ollr2Hr2 = 3.0.x + 0. 75ay - 0. 75a, Now the normal and tangential parts of 8 2 are combined. 8 2 = 3.0.x + 1.25ay - 0.25a,

(T)

1 H2=-(0.50.x +0.2lay -0.04a,) P.o

13.4. In region 1, defined by

z < 0,

llr 1 = 3

(A/m)

and

H 1 =-(0.2a.., +0.5ay + l.Oa,)

(A/m)

llo

Find Hz if it is known that

Bz = 45°. cos Q'

Then,

6 1 =61.7°

HI·· =- · = 0.88 IHII

and tan 61. r = Jl.?. tan45° 3

From the continuity of normal 8,

p.,1H.1 = p..?.Hz2 ,

P.r2 = 5.51

and so

P.rl ) =1 (0.2ax + 0.5ay + 0.54a,) (A/m) 1 ( 0.2ax + 0.58y + -1.0., H 2 =Jlo Jl,z P.o

13.5. A current sheet, K = 6.5a. A/m, at x = 0 separates region 1, x < 0, lOay A/m and region 2, x > 0. Find Hz at x = +0.

where

H1 =

Nothing is said about the permeabilities of the two regions; however, since H 1 is entirely tangential, a change in permeability would have no effect. Since 8, 1 = 0, 8,2 = 0 and therefore H,2 = 0.

CHAP. 13)

211

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS (H,-H2)Xan,2=K (lO.y- Hy28y) X a,.= 6.5a, (10- Hy2 )( -a.)= 6.5a. Hy2 = 16.5 (A/m)

Thus, H 2 = 16.5ay (A/m).

13.6. A current sheet, K = 9.0ay A/m, is located at z = 0, the interface between region 1, z<O, with l-'r1 =4, and region 2, z>O, #lr2=3. Given that H 2 =14.5ax+ 8.0az (A/m), find H 1 •

A C8CI

The current sheet shown in Fig. 13-6 is first examined alone.

H; =

H9.0)ay

(-a.)= 4.5( -a.. )

H; = ~(9.0}8y X a,= 4.5a,. z

0 X

K = 9.0ay

Fig. 1.3-6

From region 1 to region 2, H,. will increase by 9.0 A/m due to the current sheet. Now the complete H and B fields are examined. H 2 = 14.5a,. + 8.0.,

(A/m)

82 = ~-& 0(43.5•.. + 24.0.,) (T)

8, = ~-& 0(22.0.,. + 24.0.,) (T) H, = 5.5a,. + 6.0., (A/m) Note that H"' must be 9.0A/m less than H" 2 because of the current sheet. An alternate method is to apply (H 1 - H 2) X 8" 12 = K:

8,. 1 is obtained as lloJJ,1 H, 1

•

(H",a" + Hyt8y + H,,a,) X a,= K + (14.5a,. + 8.0.,) X a. -H.-~ay

+ Hy 18x= -5.5ay

from which H"' = 5.5 A/m and Hy1 = 0. This method deals exclusively with tangential H; any normal component must be determined by the previous methods.

13.7. Region 1, z < 0,

has Jlr 1 = 1.5,

81 = 2.40ax + lO.Oaz

while region 2, (T)

> 0, has #lr2 = 5. Near (0, 0, 0),

=25.75ax -17.7ay + lO.Oaz

If the interface carries a sheet current, what is its density at the origin? Near the origin, 1 1 H, =--8, =-(1.608" +6.67a.) 1-&ollrt I-to 1 H 2 = - (5.15a"- 3.54ay + 2.0..) llo

(A/m)

(T)

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

212

[CHAP. 13

Then the local value of K is given by

13.8. Given E = Em sin ( wt - f3z )a,

1A "'

in free space, find D, 8 and H. D=

The Maxwell equation

.:oE =

E 11 Em

Sketch E and H at t = 0.

sin (wt- fJz)•~

V X E = -a&/ at gives

...

a ay

a az

Em sin ( wt - f3z)

a& - at =

••

·~

a ax

aB at

fJE.., cos ( wt - f3z )•x

Integrating,

B = - f3Em sin ( wt - fJz )ax (1)

where the "constant" of integration, which is a static field, has been neglected.

Then,

fJEm . ~ ) H = --sm(wt-pz • .. WIJ.o

Note that E and Hare mutually perpendicular. At t = 0, sin (wt- fJz) =-sin f3z. shows the two fields along the z axis, on the assumption that Em and fJ are positive.

Figure 13-7

H y

Fig. 13-7

13.9. Show that the E and H fields of Problem 13.8 constitute a wave traveling in the z ~ direction. Verify that the wave speed and E/ H depend only on the properties of free space.

1.11!!!

E and H together vary as sin (wt- fJz).

A given state of E and His then characterized by

wt - f3z = const. = wt0

But this is the equation of a plane moving with speed (1)

c=fJ

(1)

z=-(t-to) fJ

CHAP. 13)

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

213

in the direction of its nonnal, a.. (It is assumed that fj, as well as w, is positive; for fj negative, the direction of motion would be -a•. ) Thus, the entire pattern of Fig. 13-7 moves down the z axis with speed c. The Maxwell equation V X H = aD/ at gives Bx

••

a ax

a ay

a az

--"'sin ( wt - fJz)

-fjE

= :, E 0 Em sin ( wt - fjz )ay

Wl'o

fJ2E

--"'cos ( wt - fJz )ay = E 0 E..,w cos ( wt - fJz )ay Wl'o

1 Eol'o

Consequently,

Moreover, E =Wl'o -= H fJ

= P2

~o -= 120Jt(V/A)= 120JtQ Eo

in free space, find E.

VXH=at

!.._ H e~<wt+P•>a az

"R.H

f!<""+Pz>

D = {jH.., e'(wt+Jiz)B

and E= D/Eo.

l3.ll. Given E = 30JTei<to"t+/lz)a..,

(V /m)

in free space, find Hm and B (/3 > 0). This is a plane wave, essentially the same as that in Problems 13.8 and 13.9 (except that, there, E was in they direction and H in the x direction). The results of Problem 13.9 hold for any such wave in free space:

w 1 - = - - = 3 x Uf(m/s) fj ~

E= H

~o - = 120JrQ Eo

Thus, for the given wave, Uf fj = 3 x

Hf = 3(rad/m)

To fix the sign of H.., , apply

30Jt

= ±120Jt - = ±-(A/m) 4

V X E = -a&/ at:

j{j30Jrf!(lo8t+Jiz)8y

which shows that H... must be negative.

= -jlrf~&oH..,f!(lo"t+Jiz)By

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

214

13.12. In a homogeneous nonconducting region where IJ., = 1,

Ia _

= 30JTei1""-(4/3)yJaz

(V /m)

find

[CHAP. 13

and w if

= I.Oei1""-(4/3)yJax

(A/m)

Here, by analogy to Problem 13.9,

3xUf

fi =YEP.= w:ii, Thus, since

-E = { ; - = 120Jt H E

(m/s)

~, E,

(Q)

p., = 1,

.-= 3xUf .c

1 30Jt = 120Jt-

w 3

which yield

= 16,

Vi.

V€,

l(f rad/s. In this medium the speed of light is c/4.

Supplementary Problems 13.13. Region 1, where p., 1 = 5, is on the side of the plane 6x + 4y + 3z region 2, fJ,2 = 3. Given 1 H 1 = - (3.0a_.- 0.5ay) P.o Ans.

12.15a.. +0.60ay + 1.5811,

= 12

that includes the origin.

(A/m)

(T), 56.6°

13.14. The interface between two different regions is normal to one of the three cartesian axes.

8 1 = fJ 0 (43.5a.. + 24.0a,)

0.506

Ans.

what is the ratio (tan 6 1)/(tan 6 2 )?

13.15. Inside a right circular cylinder, fJ, 1 = 1000. The exterior is free space. the cylinder, determine 8 2 just outside. Ans. 2.5a., (mT)

If 8 1 = 2.5a.,

13.16. In spherical coordinates, region 1 is r <a, region 2 is a < r b. and 3 are free space, while p.,2 = 500. Given 8 1 = 0.20a, (T), find H in each region. o.2o 4 x w- 4 0.20 Ans. -(A/m), (A/m), -(A/m) P.o P.o P.o

(T)

inside

Regions 1

13.17. A current sheet, K=(8.0/p. 0 )ay(A/m), at x=O separates region 1, x<O and IJ,1 =3, region 2, x > 0 and P..-?. = 1. Given H 1 = (10.0/ IJo)(ay +a,) (A/m), find H 2 • 1 Ans. - (10.0ay + 2.0a,) (A/m)

from

fJo

13.18. The x = 0 plane contains a current sheet of density K which separates region 1, x < 0 and 2, from region 2, x < 0 and p.,2 = 7. Given

8 1 = 6.0a.. + 4.0ay + IO.Oa, find K.

Ans.

13.19. In free space,

1 - (3. 7211,.- 9.288,) P.o

82 = 6.0.. - 50. 96ay

(T)

+ 8. 96a, (T)

(A/m)

D = Dm sin (wt + fh)a... B=

Using Maxwell's equations, show that -WtJoDm .

sm(wt+fjz)ay

p.,, =

CHAP. 13]

MAXWELL'S EQUATIONS AND BOUNDARY EQUATIONS

Sketch the fields at I= 0 along the z axis, assuming that D.., > 0,

{J > 0.

propagation

y-------~1

13.21. In free space,

Show that

13.21. In a homogeneous region where fA, = l and Find wand H,.. if the wavelength is 1.78 m.

= 50,

An.r.

1.5 X Uf rad/s, 1.18 A/m

An.r.

215 See Fig. 13-8

Chapter 14 Electromagnetic Waves 14.1 INTRODUCfiON Some wave solutions to Maxwell's equations have already been encountered in the Solved Problems of Chapter 13. The present chapter will extend the treatment of electromagnetic waves. Since most regions of interest are free of charge, it will be assumed that charge density p = 0. Moreover, linear isotropic materials will be assumed, with D = £E, B = #lH, and J = oE.

14.2 WAVE EQUATIONS With the above assumptions and with time dependence equations (Table 13-1) become

eiru' for both E and H, Maxwell's

VXH= (o+ jc.o£)E

(1)

V X E = -jwllH

(2)

V·E=O

(3)

V·H=O

(4)

Taking the curl of (1) and (2), V X (V X H) = (o + jw£ )(V X E) V X (V X E)= -jwll(V X H) Now, in cartesian coordinates only, the Laplacian of a vector

satisfies the identity V X (V X A) = V(V • A) - V2A Substitution for the ..curl curls" and use of (J) and ( 4) yields the vector wave equations V2 H = jwll( o + jw£ )H = y 2 H V2E = jwll( o + jw£ )E = y2 E The propagation constant y is that square root of y 2 whose real and imaginary parts are positive: y= a+ jf3

w~~£( ~1 + (:J -t)

{3=

(l)~~£ ( ~1 + (:J2 + 1)

with

216

(5)

(6)

CHAP. 14)

ELECI'ROMAGNETIC WAVES

217

14.3 SOLUTIONS IN CARTESIAN COORDINATES The familiar scalar wave equation in one dimension, cf-F

1 cf-F

8z 2

=u 2 af

has solutions of the form F = f(z- ut) and F = g(z + ut), where f and g are arbitrary functions. These represent waves traveling with speed u in the +z and -z directions, respectively. In Fig. 14-1 the first solution is shown at t = 0 and t = t 1 ; the wave has advanced in the +z direction a distance of ur1 in the time interval 11 â&#x20AC;˘ For the particular choices f(x) =

Ce-;ruxtu

and

g(x) =

De+imxtu

/(zo) ~----1111 - - - - - - - - . . j

Fig. 14-1

harmonic waves of angular frequency w are obtained:

F = cei<wt-fJz>

and

= Dei(rot+/Jz)

in which f3 w/u. Of course, the real and imaginary parts are also solutions to the wave equation. One of these solutions, F = C sin (rot- f3z), is shown in Fig. 14-2 at t = 0 and t = n/2w. In this interval the wave has advanced in the positive z direction a distance d = u(n/2w) = n/2/J. At any fixed t, the waveform repeats itself when x changes by 2nI f3; the distance

is called the wavelength.

The wavelength and the frequency f J.f=u

).=

=w/2n enjoy the relation

where T 1// = 2n/w is the period of the harmonic wave. The vector wave equations of Section 14.2 have solutions similar to those just discussed. Because the unit vectors ax, ay, and az in cartesian coordinates have fixed directions, the F

......., It=

2: I

Fig. 14-2

218

ELECI'ROMAGNETIC WAVES

(CHAP. 14

wave equation for H can be rewritten in the form ~H

c?H

2 -+ 2 -+ 2 - =2y H

Of particular interest are solutions (plane waves) that depend on only one spatial coordinate, say z. Then the equation becomes d 2H

-=y2H 2 dz

which, for an assumed time dependence e'ru', is the vector analog of the one-dimensional scalar wave equation. Solutions are as above, in terms of the propagation constant y.

H(z, t) = H 0 e±yzeirutaH The corresponding solutions for the electric field are

E(z, t) = E 0 e±Yzf!rutaE. The fixed unit vectors aH and a£ are orthogonal and neither field has a component in the direction of propagation. This being the case, one can rotate the axes to put one of the fields, say E, along the x axis. Then from Maxwell's equation (2) it follows that H wiU lie along the ±y axis for propagation in the ±z direction. EXAMPLE 1. Given the field E = E0 e-,··~ (time dependence suppressed), show that E can have no component in the propagation direction, +a•. The cartesian components of •~ are found by projection: E = E 0 e -y~[(a~ • a_.)a_. +

<•~

·ay)a,. +(a~ • a.)a.]

From V· E=O,

which can hold only if •~ ·a. = 0.

Consequently, E has no component in a•.

The plane wave solutions obtained above depend on the properties #J, because these properties are involved in the propagation constant y.

£,

and o of the medium,

14.4 SOLUTIONS FOR PARTIALLY CONDUCI'ING MEDIA For a region in which there is some conductivity but not much (e.g., moist earth, seawater), the solution to the wave equation in E is taken to be E=E0 e-yza:r

Then, from (2) of Section 14.2, H =

~0 +. jw£ E 0 e -yz a,. JWIJ

The ratio E I H is characteristic of the medium (it is also frequency-dependent). More specifically for waves E = E:rax, H = Hyay which propagate in the +z direction, the intrinsic impedance, 'I· of the medium is defined by

Thus

219

ELECfROMAGNETIC WAVES

CHAP. 14)

where the correct square root may be written in polar form, I'11/...!!._, with

Vii/i

I'11 = 41

+(:J

tan 20 =

0" < 0 < 45°

and

W£

(If the wave propagates in the -z direction, other square root used.) Inserting the time factor t~m• and writing fields in a partially conducting region:

E(z, t) = H

<z,

ExiHY =

In effect, y is replaced by -y and the

-1].

y = a + jf3 results in the following equations for the

E0 e-"'zt~<m~-fJz>ax

t) = Eo

I'll e

-az...i(M-/Jz-8)

The factor e-az attenuates the magnitudes of both E and H as they propagate in the +z direction. The expression for a, (5) of Section 14.2, shows that there will be some attenuation unless the conductivity o is zero, which would be the case only for perfect dielectrics or free space. Likewise, the phase difference 0 between E(z, t) and H(z, t) vanishes only when o is zero. The velocity of propagation and the wavelength are given by

.A.=21r=

w~IJ2£ ( ~1 +

(:£r

+ 1)

If the propagation velocity is known, .A./ = u may be used to determine the wavelength .A.. The term (o/w£) 2 has the effect of reducing both the velocity and the wavelength from what they would be in either free space or perfect dielectrics, where o = 0. Observe that the medium is dispersive:

waves with different frequencies w have different velocities u.

14.5 SOLUTIONS FOR PERFECf DIELECfRICS For a perfect dielectric,

o = 0,

and so

fJ=wv'/ii

a=O

Since a = 0, there is no attenuation of the E and H waves. The zero angle on '1 results in H being in time phase with E at each fixed location. Assuming E in ax and propagation in az , the field equations may be obtained as limits of those in Section 14.4: E(z, t) =

Eoei<m~-IJz>ax

H(z, t) =Eo t~<m~-tJz>ay

The velocity and the wavelength are

u=f3=\fiii

.A.-2n_

- f3-

wv;i

220

ELECfROMAGNETIC WAVES

[CHAP. 14

Solutions in Free Space. Free space is nothing more than the perfect dielectric for which

w-9

1J = #Jo = 4n X 10- 7 H/m

For free space,

= '1o = 120nQ

= £ 0 = 8.854 X 10- 12 F/m = -- F/m 36n

= c =3 x

and u

lOS m/s.

14.6 SOLUTIONS FOR GOOD CONDUCfORS; SKIN DEPTH Materials are ordinarily classified as good conductors if o~ w£ in the range of practical frequencies. Therefore, the propagation constant and the intrinsic impedance are y= a+ jf3

~~cz= v--z-=

a={J=

J]=

V:rr/#JO

-y-:;f45°

It is seen that for aU conductors the E and H waves are attenuated. Numerical examples will show that this is a very rapid attenuation. a will always be equal to (3. At each fixed location H is out of time phase withE by 4SO or n/4 rad. Once again assuming E in ax and propagation in az, the field equations are, from Section 14.4,

H(z t) =Eo e-azei<wr-fJz-.rrt4>a ' 1'11 :r 2n A= {i =

Moreover,

V:rrf#JO = 2nlJ

The velocity and wavelength in a conducting medium are written here in terms of the skin depth or depth of penetration, 1

{) =

- V:rrf#JO EXAMPLE 2. Assume a field E = l.Oe-•..ei<""-ll•>a% (V /rn), with f = w/2tr = 100 MHz, at the surface of a copper conductor, a= 58 MS/m, located at z > 0, as shown in Fig. 14-3. Examine the attenuation as the wave propagates into the conductor. X

~.......

u=O

_.., E) ,...____:: Propagation direction

Fig. 14-3

At depth z the magnitude of the field is

lEI= l.Oe ... = l.Oe-''6 where

6= -

1 -

VrifJla

= 6.61

Jlm

CHAP. 14)

ELECfROMAGNETIC WAVES

221

Thus, after just 6.61 micrometers the field is attenuated to e-• = 36.8% of its initial value. micrometers, the magnitude is 0.67% of its initial value-practically zero.

At 5b or 33

14.7 INTERFACE CONDmONS AT NORMAL INCIDENCE When a traveling wave reaches an interface between two different regions, it is partly reflected and partly transmitted, with the magnitudes of the two parts determined by the constants of the two regions. In Fig. 14-4, a traveling E wave approaches the interface z = 0 from region 1, z < 0. E; and Er are at z = -0, while E' is at z = +0 (in region 2). Here, i signifies "incident," r "reflected" and t "transmitted." Normal incidence is assumed. The equations for E and H can be written

E;(z, t) =

E~-y,zeiw'ax

Er(z, t) = E(jeY•zeiw'a.. E;(z, t) = E~-y2zeiro'a.. H;(z, t) = H~-y,zeicu'ay ur(z, t) =

H(jeY•Zei<»~ay

H'(z, t) =

H~-y2 zeiwTay

One of the six constants-it is almost always E:r-may be taken as real. Under the interface conditions about to be derived, one or more of the remaining five may turn out to be complex.

Fig. 14-4

With nominal incidence, E and H are entirely tangential to the interface, and thus are continuous across it. At z = 0 this implies E~+E(i=E~

H~+H'Q=H~

Furthermore, the intrinsic impedance in either region is equal to

±ExlHy

(see Section 14.4).

E~ H~=r~z

The five equations above can be combined to produce the following ratios in terms of the intrinsic impedances: E(i

E~ E~ -=

'12- '11 '11 + '12 2'12 '1• + '12

H(i H~ H~

'1•- '12 '11 + '12 2'11 '1• + '12

ELECfROMAGNETIC WAVES

222

[CHAP. 14

The intrinsic impedances for various materials have been examined earlier. They are repeated here for reference.

partially conducting medium:

7]=

'J~

conducting medium:

7]=

..f¥/45°

perfect dielectric:

7]=~

free space:

T]o

= ~ = 120.dl

EXAMPLE 3. Traveling E and H waves in free space (region 1) are normally incident on the interface with a perfect dielectric (region 2) for which £, = 3.0. Compare the magnitudes of the incident, reflected, and transmitted E and H waves at the interface. '} 1

= '1o = 120.n 0

120.n =--=217.70

'12=

v;,

E~ = '1 2 - 'I• = -0.268 E~ '1• + '12

H~ = '1•- 112 =0.268 H~ 'I• + '12

£~=~=0732

H~=~=1.268

'I• + '12

'1• + '12

14.8 OBLIQUE INCIDENCE AND SNELL'S LAWS An incident wave that approaches a plane interface between two different media generally will result in a transmitted wave in the second medium and a reflected wave in the first. The plane of incidence is the plane containing the incident wave normal and the local normal to the interface; in Fig. 14-5 this is the xz plane. The normals to the reflected and transmitted waves also lie in the plane of incidence. The angle of incidence 8;, the angle of reflection fJ,, and the angle of transmission 8,-all defined as in Fig. 14-~bey Snell's law of reflection, 0;

= B,

and Snell's law of refraction,

Reflec1ed

wa,·e normal

Fig. 14-S

223

ELECfROMAGNETIC WAVES

CHAP. 14)

~.!!eXAMPLE 4. A wave is incident at an angle of

30" from air to teflon, transmission, and repeat with an interchange of the regions. Since IJ• = /J2,

sin 8, sin 30" --=--= sin 8, sin 8,

~r2 • ,;;;-. -=v2.1 E.1

•

sin 30" 1 sine,= V2.T

£.

= 2.1.

Calculate the angle of

8, =20.18"

From teflon to air, 8, =46.43°

Supposing both media of the same permeability, propagation from the optically denser medium (£1

> £ 2 ) results in 8, > 8;. As 8; increases, an angle· of incidence will be reached that results

in 8, = 90°. At this critical angle of incidence, instead of a wave being transmitted into the second medium there will be a wave that propagates along the surface. The critical angle is given by

it:i

~.!!!EXAMPLE 5. The critical angle for a wave propagating from teflon into free space is 8 =sin-•:-b=43.64o c v2.1

14.9 PERPENDICULAR POLARIZATION The orientation of the electric field E with respect to the plane of incidence determines the polarization of a wave at the interface between two different regions. In perpendicular polarization E is perpendicular to the plane of incidence (the xz plane in Fig. 14-6) and is thus parallel to the (planar) interface. At the interface,

''12 cos 0; - '11 cos 0,

E~ = '1 2cos 0; + '1 1cos 0, and Note that for normal incidence 14.8.

27] 2 cos B; + 1] 1 cos 0,

-= E~ 1] 2 cos 0;

8; = 8, = 0" and the expressions reduce to those found in Section

Fig. 14-6

224

ELECfROMAGNETIC WAVES

[CHAP. 14

It is not difficult to show that, if 1J 1 = 1J 2 , 'h cos 0;- 'It cos 0,

for any 0;

Hence, a perpendicularly polarized incident wave suffers either partial or total reflection.

14.10 PARALLEL POLARIZATION For parallel polarization the electric field vector E lies entirely within the plane of incidence, the xz plane as shown in Fig. 14-7. (Thus E assumes the role played by H in perpendicular polarization.)

At the interface,

Eo 1] 2 cos B,- 'It cos 0; E~= 'It cos B; + rJ 2 cos B, E~ -=

and

21] 2 cos B; 'It cos 0;- '12 cos 0,

CD.T 0 ~H'

~Âˇ

ew- z

E'/ E'\ Fig. 14-7

In contrast to perpendicular polarizations, if 1J 1 = 1J 2 there will be a particular angle of incidence for which there is no reflected wave. This Brewster angle is given by

.!~!!ExAMPLE 6. The Brewster angle for a parallel-polarized wave traveling from air into glass for which E.,= 5.0 is

~ rf.l!!t4.11 STANDING WAVES When waves traveling in a perfect dielectric (o 1 = a 1 = 0) are normally incident on the interface with a perfect conductor ( o 2 =co, 11 2 = 0), the reflected wave in combination with the incident wave produces a standing wave. In such a wave, which is readily demonstrated on a clamped taut string, the oscillations at all points of a half-wavelength interval are in time phase. The combination of incident and reflected waves may be written E(z, t) = Since

rJ 2 = 0,

[E~(rot-llzl

+ E;;ei(wt+llz>]a,. = eiw'(E~-illz + E(ieillz)a..

Eo/ Eh = -1 and E(z, t) = ei""(E~-illz- E~llz)a.. = - 2jE~ sin {Jzeiw'ax

225

ELECTROMAGNETIC WAVES

CHAP. 14)

or, taking the real part, E( z, t) = 2E;J sin {3z sin wtax

The standing wave is shown in Fig. 14-8 at time intervals of T /8, where T = 2n I w is the period. At r = 0, E = 0 everywhere; at t = l(T /8), the endpoints of the E vectors lie on sine curve 1; at t = 2(T /8), they lie on sine curve 2; and so forth. Sine curves 2 and 6 form an envelope for the oscillations; the amplitude of this envelope is twice the amplitude of the incident wave. Note that adjacent half-wavelength segments are 180° out of phase with each other.

.......--..,.. .4.

2£1

,. .'.......... _.,.,."''1 \

.....,__

....

Fig. 14-8

14.12

POWER AND THE POYNTING VECTOR

Maxwell's first equation for a region with conductivity a is written and then E is dotted with each term.

aE ot

VXH=aE+E~

aE ot

E· (VXH)= a£-+ E· E-

where, as usual, £ 2 = E • E. The vector identity V ·(A X B)= B · (V employed to change the left side of the equation. H • (V X E) - V • (E

aE at

H) = a£2 + E • E -

By Maxwell's second equation, H. (V X E)= H.

(-Ji oH) =-!!:. oH2 ar 2 at

Eo£ 2

E·E-=-0( 2 ac

Similarly, Substituting, and rearranging terms, a£2

o£2

oH 2

= - - - - -11- - V · (Ex H) 2 ar 2 ar

A)- A· (V

ELECfROMAGNETIC WAVES

226

(CHAP. 14

Integration of this equation throughout an arbitrary volume v gives

oE 2 dv =-

£ a£ - +"'aH - -)

I(v2at

2at

dv-

f. (EXH) · dS s

where the last term has been converted to an integral over the surface of v by use of the divergence theorem. The integral on the left has the units of watts and is the usual ohmic term representing energy dissipated per unit time in heat. This dissipated energy has its source in the integrals on the right. Because ££ 2 /2 and ~-&H2 /2 are the densities of energy stored in the electric and magnetic fields, respectively, the volume integral (including the minus sign) gives the decrease in this stored energy. Consequently, the surface integral (including the minus sign) must be the rate of energy entering the volume from outside. A change of sign then produces the instantaneous rate of energy leaving the volume: P(t) =

(EX H) · sS =

!1' • dS

where !1' = E X H is the Poynting vector, the instantaneous rate of energy flow per unit area at a point. In the cross product that defines the Poynting vector, the fields are supposed to be in real form. If, instead, E and H are expressed in complex form and have the common time-dependence eiw', then the time-average of !1' is given by

where H* is the complex conjugate of H. This follows the complex power of circuit analysis, S = !VI•, of which the power is the real part, P =iRe VI*. For plane waves, the direction of energy flow is the direction of propagation. Thus the Poynting vector offers a useful, coordinate-free way of specifying the direction of propagation, or of determining the directions of the fields if the direction of propagation is known. This can be particularly valuable where incident, transmitted, and reflected waves are being examined.

Solved Problems 14.1. A traveling wave is described by y = 10 sin (fJz - rot). Sketch the wave at t = 0 and iJ:i at t = 11 , when it has advanced )./8, if the velocity is 3 x 10" m/s and the angular !II!! frequency ro = 106 rad/s. Repeat for ro = 2 x 106 rad/s and the same t 1 • The wave advances A in one period,

T = 2tr/w.

Hence

lt=-=8 4w

8= ctt = (3 x 10") 4(Hf') =236m The wave is shown at t = 0 and t = t 1 in Fig. 14-9(a). At twice the frequency, the wavelength A is one-half, and the phase shift constant fJ is twice, the former value. See Fig. 14-9(b). At t 1 the wave has also advanced 236m, but this distance is now A/4.

227

ELECfROMAGNETIC WAVES

CHAP. 14)

t=0 )0

(b)

(a)

Fig. 14-9

14.2. In free space,

E(z, t) = lW sin (wt- f3z )ay

(V /m).

Obtain H(z, t).

Examination of the phase, wt- f3z, shows that the direction of propagation is +z. must also be in the +z direction, H must have the direction -a.. Consequently,

-~. =J] 0 =120JrQ

and

H(z, t) = -

H"=-

Jrsin(wt-f3z) 120

~~~Jr sin ( wt- f3z )a.

Since EX H

(A/m)

14.3. For the wave of Problem 14.2 determine the propagation constant y, given that the frequency is f = 95.5 MHz. In general, y

= Vjw~J(u + jw£).

In free space,

u = 0,

so that

2Jr(9S.S X lif) . • ~ ·(21rf\ y=JWVIJo€o=J ~J=i 3xHf =j(2.0)m-t

a=0

Note that this result shows that the attenuation factor is f3 = 2.0 rad/m.

and the phase-shift constant

14.4. Examine the field

il:i_ .II!!,

E( z, t) = 10 sin ( wt + f3z )ax + 10 cos ( wt + f3z )ay in the

z = 0 plane, for wt = 0, ;r/4, ;r/2, 3;r/4 and 1r.

The computations are presented in Table 14-1.

Table 14-1 wt

E.= 10sin wt

Ey =coswt

E=E.a. +Eyay

10ay

-Jr

-10

10(··~ay) lOa.

-Jr

3Jr 4

-10

10(a·:2ay) 10{-ay)

228

ELECfROMAGNETIC WAVES As shown in Fig. 14-10, E(x, t) is circularly polarized. direction.

(CHAP. 14

In addition, the wave travels in the -a,

Fig. 14-10

14.5. An H field travels in the -az direction in free space with a phaseshift constant of 30.0 rad/m and an amplitude of (l/31r) A/m. If the field has the direction -ay when t = 0 and z = 0, write suitable expressions for E and H. Determine the frequency and wavelength. In a medium of conductivity u, the intrinsic impedance J], which relates E and H, would be complex, and so the phase of E and H would have to be written in complex form. In free space this restriction is unnecessary. Using cosines, then 1 H(z, t) =- Jr cos (rot+ f3z)ay 3

For propagation in -z,

Ex Hy

- = -J] 0 =

-120Jr Q

E.. = +40 cos ( wt + f3z)

E(z, t)=40cos(wt+f3z)a..

Thus

(V /m)

(V/m)

Since f3 = 30 rad/m, 2Jr Jr A=-=-m f3 15

c 3 X Hf 45 /=-=--=-xl(fHz A Jr/15 1r

14.6. Determine the propagation constant y for a material having f.l.r = 1, 0.25 pS/m, if the wave frequency is 1.6 MHz.

= 8,

and

In this case, (1

(J)£

o.zs x w-

2Jr(1.6 X Hf')(8)(10- 9 /36Jr) = l0- =0

so that a=O

f3 = wV/ii = 2rif ..r,;:£, = 9.48 X 10- 2 rad/m c

and y =a+ jf3 =j9.48 x 10-2 m- 1 • The material behaves like a perfect dielectric at the given frequency. Conductivity of the order of 1 pS/m indicates- that the material is more like an insulator than a conductor.

14.7. Determine the conversion factor between the neper and the decibel. Consider a plane wave traveling in the + z direction whose amplitude decays according to E=E0 e-"''

229

ELECfROMAGNETIC WAVES

CHAP. 14)

From Section 14.12, the power carried by the wave is proportional to £ 2 , so that

Then, by definition of the decibel, the power drop over the distance z is 10 log10 (P0 /P) dB.

But

Po 10 Po 20 10 logto p = _ In p = _ ( az) = 8. 686( az) 2 3026 2 3026 Thus, az nepers is equivalent to 8.686(az) decibels; i.e.,

1 Np = 8.686 dB

14.8. At what frequencies may earth be considered a perfect dielectric, if o to-' S/m, p.- I, and 8? Can <r be assumed zero at these frequencies?

<.-

•

Assume arbitrarily that

u 1 WE-100

-<-

marks the cutoff. Then

w 100u / = - 2 : - = 1.130Hz 2Jr 2JrE For small a/wE,

w~T( ~1 + (~f -1)

-I'E[1( - -o ) 2 2 WE

2 ]

u~, =-u~ -=-(120Jr)=0.333Np/m 2 E 2 E,

Thus, no matter how high the frequency, a will be about 0.333 Np/m, or almost 3 db/m (see Problem 14.7); a cannot be assumed zero.

14.9. Find the skin depth 6 at a frequency of 1.6 MHz in aluminum, where and 1-'r = 1. Also fin: y and the wave velocity u.

~ 38.2 MS/m

lJ = - - = 6.44 X 10- 5 m= 64.4 I'm

Yrifi'U

and

u =-= wlJ =647(m/s)

14.10. A perpendicularly polarized wave propagates from region 1 ( Er1 = 8.5, 1-'rt = 1, o 1 = 0) to region 2, free space, with an angle of incidence of 15°. Given £~= 1.01-'V/m, find: £~, E~, H~, H~, and

Yo.

The intrinsic impedances are J}o 120 'It=--=--= 129 Q

v'i:t V8.S

and

TJz = J}o = 120Jr Q

and the angle of transmission is given by sin15° _ sin8,

{£ v~

8,=48.99°

230

ELECTROMAGNETIC WAVES

[CHAP. 14

Then E~ = 7] 2 COS 8,- 7] 1 COS 8, = E~

7] 2 COS

2'lz cos 8;

E 'o = _ E~

Finally,

8; + 7] 1 COS 8,

7] 2 COS 8;

+ 7] 1 COS 8,

H~=E~/7J 1 =7.75nA/m,

_ 0 623 623

H~=4.83nA/m,

£~=0.6231-'V/m

and

= 1.623 I'V/m

H~=4.3lnA/m.

14.11. Calculate the intrinsic impedance fJ, the propagation constant y, and the wave velocity u for a conducting medium in which o =58 MS/m, f.l.r = 1, at a frequency f = 100 MHz.

VwiJU /45° = 2.14 x lOS /45° m'1 = /45° =3.69 x w-l /45° Q

..Ji¥

1 lJ =-= 6.61 I'm a

a= f3 = 1.51 x lOS

u = wlJ = 4.15 x tfrl m/s

14.12. A plane wave traveling in the +z direction in free space (z < 0) is normally incident at z = 0 on a conductor (z > 0) for which o = 61.7 MS/m, f.l.r = 1. The free-space E wave has a frequency f = 1.5 MHz and an amplitude of 1.0 V/m; at the interface it is given

by E(O, t) = 1.0 sin 21rftay

(V /m)

Find H(z, t) for z > 0. For z > 0,

and in complex form, E(z, t) = l.Oe-""'ei<Znf•-ll•lay

where the imaginary part will ultimately be taken.

(V /m)

In the conductor,

a= fJ = Vrifl'u= VIr(1.5 x Hf')(41r x 10 7 )(61. 7 x Hf') = 1.91 x 104 '1 =

Then, since

Ey/( -H,) =

..Ji¥/45°

4.38 X 10

ei"14

7],

H(z, t) = -2.28 X 1We-'"'ei12""- 11' "'4 >a,

(A/m)

or, taking the imaginary part, H(z, t) = -2.28 x 1We-"' sin (2rift- {Jz- Ir/4)a,

where f, a, and

(A/m)

f3 are as given above.

14.13.In free space E(z,t)=50cos(rot-Pz)ax (V/m). circular area of radius 2.5 m in the plane z = const.

Find the average power crossing a

In complex form, E = 5()eilwr-llzla, and since '1 = 1201r Q

and propagation is in +z, H=..2_ei(wr-llzla

121f

Then

(V /m)

91'.v8 = ~ Re (EX u•) =

(A/m)

~(SO)(l~1r)a, W/m

231

ELECfROMAGNETIC WAVES

aiAP. 14]

The flow is normal to the area, and so

14.14 A voltage source, v, is connected to a pure resistor R by a length of coaxial cable, as shown in Fig. 14-ll(a). Show that use of the Poynting vector !1' in the dielectric leads to the same instantaneous power in the resistor as methods of circuit analysis.

_C1R u

~t~........ D

u_·.....,.. .. .._.;;..;.;.;........._....__....J {a)

{b)

Ftg_ 14-11

From Problem 7.9 and

Am~re's

law,

v E

rln(b/a)

•

i H=-a•

and

21fT

where a and bare the radii of the inner and outer conductors, as shown in Fig. 14-ll(b).

Then

s-=EXH

21r1'2 ln(b/a) 8 '

This is the instantaneous power density. The total instantaneous power over the cross section of the dielectric is

P(t)=

z.rl" vi Lo ., 2m' In (b/a )a,·rdrdt/Ja.=vi 2

which is also the circuit-theory result for the instantaneous power loss in the resistor.

14.15. Determine the amplitudes of the reflected and transmitted E and H at the interface shown in

232

ELECfROMAGNETIC WAVES

(CHAP. 14

Fig. 14-12, if E~ = 1.5 X 10- 3 v /m in region 1, in which 0. Region 2 is free space. Assume normal incidence. 'It=

r;;;:. = 129 Q 'J';;;:;

1} 2 =

120Jr Q

Ert

== 8.5,

f.l.rl

= 1,

and

= 377 Q

E~ = 'lz- 'It E~= 7.35 x l0- 4 V/m 'lz +'It

Eo=~E~=2.24x to

'lz + 'It

V/m

= - = 1.16 X 10- A/m

'It

H~ = 'It - 'lz H~ = -5.69 x 10-6 A/m 'It+ 'lz

H~=~H~=5.91 x 10-

'It+ 'lz

A/m

-!I'

., f.·o

Fig. 14-12

14.16. The amplitude of E; in free space (region 1) at the interface with region 2 is 1.0 V /m. If H 0= -1.41 X 10- 3 A/m, Er2 = 18.5 and o 2 = 0, find f.l.r2· From E~ H~ =

E~ = H~

Then

-120Jr Q = -377 Q

1.0 -1.41 X 10 1234 =

and

-377(377 + 71 2 ) 1} 2 - 377

r;;;:;-

Eo or

1} 2 - 377 377 + 'lz

71 2 = 1234 Q

'J~

14.17. A normally incident E field has amplitude E~ = 1. 0 VIm in free space just outside of .,.... seawater in which Er = 80, f.l.r = 1, and o = 2.5 S/m. For a frequency of 30 MHz, at 1 what depth will the amplitude of E be l.OmV/m?

ced

Let the free space be region 1 and the seawater be region 2. 'It= 377 Q

J] 2

Then the amplitude of E just inside the seawater is or

= 9. 73 /43S Q

. E~ =

5.07

10- 2 V/m

233

ELECTROMAGNETIC WAVES

CHAP. 14) From

y = Vjwll(a

+ jwE) = 24.36 f46.53o m

a= 24.36 cos 46.53° = 16.76 Np/m Then, from 1.0 = 10

= (5.07

10 2 }e- 1676z

z =0.234m.

14.18. A traveling E field in free space, of amplitude 100 V /m, strikes a sheet of silver of thickness ..,... 1+ 5J.Lm, as shown in Fig. 14-13. Assuming a=61.7MS/m and a frequency f= ~ 200 MHz, find the amplitudes l£2 1, l£ 3 1. and l£4 1. Math cad

• I.!

Fig. 14-13

For the silver at 200 MHz, 2(5.06

£2

71 = 5.06 x 10- 3 /4SO Q.

377 + 5.06 X 10-' {45°

IE2I = 2.68 X 10- 3 V/m

whence

Within the conductor,

a= f3 =

Vn:flla =

2.21

105

Thus, in addition to attenuation there is phase shift as the wave travels through the conductor. Since IE3 1and 1£4 1represent maximum values of the sinusoidally varying wave, this phase shift is not involved.

and

£4 E,

2(377) 377 X 5.06 X 10

{45°

whence

1£41 = 1. 78 X 10- 3 V /m

Supplementary Problems 14.19. Given

E(z, t) = 103 sin (6 x l0 8 t- f3z)a,

(V /m)

in free space, sketch the wave at t = 0 and at time t 1 when it has traveled A/4 along the z axis. t., {3. and A. Ans. t 1 = 2.62 ns, f3 = 2 rad/m, ). = .nm. See Fig. 14-14.

Find

(CHAP. 14

ELECTROMAGNETIC WAVES

234

...... 14-14 14.20. In free space,

H(z, t) =

(A/m)

t.()ei<t.5><tolt+P•la..

Obtain an expression for E(z, t) and determine the propagation direction. Ans. E 0 =377V/m, -a, 14.21. In free space,

H(z, t) = 1.33 x 10- 1 cos (4 x l07 t)- fJz)a.. Obtain an expression for E(z, t).

Find

f3 and A.

Ans.

(A/m) (~) rad/m,

Eo= 50 V /m,

lS.n m

14.22. A traveling wave has a velocity of llf' m/s and is described by y

= 10 cos (2.5z + rot)

Sketch the wave as a function of z at t = 0 and t = t 1 = 0.8381JS. traveled between these two times? Ans. t See Fig. 14.15.

What fraction of a wavelength is

>.. = 2.51 m

...... 14-15

14.23. Find the magnitude and direction of

E(z, t) = 10 sin (rot- fJz)a.. -15 sin (rot- fJz}ay at t =0,

z =3A/4.

Ans.

18.03 V /m,

(V/m)

0.555a_. - 0.832ay

14.24. Determine y at 500kHz for a medium in which '-'• = 1, £, = 15, o = 0. At what velocity will an electromagnetic wave travel in this medium? Ans. j4.06 x 10- 2 m-t, 7.74 x 107 m/s 14.25. An electromagnetic wave in free space has a wavelength of 0.20 m. When this same wave enters a perfect dielectric, the wavelength changes to 0.09 m. Assuming that IJ, = 1, determine £, and the wave velocity in the dielectric. Ans. 4.94, 1.35 x lOS m/s 14.16. An electromagnetic wave in free space has a phase shift constant of 0.524 rad/m.

The same wave has a

235

ELECfROMAGNETIC WAVES

CHAP. 14]

phase shift constant of 1.81 rad/m upon entering a perfect dielectric. and the velocity of propagation. Ans. 11.9, 8.69 x 107 m/s

Assuming that ,_,, = 1,

14.27. Find the propagation constant at 400 MHz for a medium in which 0.6 S/m. Find the ratio of the velocity v to the free-space velocity c. Ans. 99.58f60.34om-•, 0.097 14.28. In a partially conducting medium, E, = 18.5, ,_,, = 800, and o velocity u, for a frequency of 109 Hz. Determine H(z, t), given E(z, t) = 50.0e_ .., cos (rot- fJz)a,

Ans.

= 16,

= 1 S/m.

,_,, = 4.5,

find

and

Find a, f3, fl, and the

(V/m)

1130 Np/m, 2790rad/m, 2100 {22.1° 0, 2.25 x lO"m/s, 2.38 X 10- 2e-..• cos (rot- 0.386- f3z)( -ax) (A/m)

14.29. For silver, o = 3.0 MS/m. Ans. 84.4 kHz

At what frequency will the depth of penetration 6 be 1 mm?

14.30. At a certain frequency in copper (o = 58.0 MS/m) lOS rad/m. Determine the frequency. Ans. 601 MHz

the

phase shift constant

3.71 x

14.31. The amplitude of E just inside a liquid is 10.0 V /m and the constants are ,_,, = 1, E, = 20, and o = 0.50 S/m. Determine the amplitude of E at a distance of 10 em inside the medium for frequencies of (a) 5MHz, (b) 50MHz, and (c) 500MHz. Ans. (a) 7.32V/m; (b) 3.91 V/m; (c) 1.42V/m 14.32. In free space, E(z, t) = 1.0 sin (rot- fJz)a.., (V /m). disk of radius 15.5 m in a z = const. plane is 1 W.

Show that the average power crossing a circular

14.33. In spherical coordinates, the spherical wave 100 . E =-sm 8cos (wt- f3r)a 8 r

(V/m)

0.265 H=--sin8cos(wt-f3r)a. r

(A/m)

represents the electromagnetic field at large distances r from a certain dipole antenna in free space. Find the average power crossing the hemispherical shell r = 1 km, 0 ~ 8 ~ :rr/2. Ans. 55.5 W 14.34- In free space, E(z, t) = 150sin (rot- fJz)a.., (V/m). Find the total power passing through a rectanguAns. 13.4 mW lar area, of sides 30 mm and 15 mm, in the z = 0 plane. 14.35. A free space-silver interface has E~ = 100 V /m on the free-space side. The frequency is 15 MHl and the silver constants are E, = ,_,, = 1, o = 61.7 MS/m. Determine E~ and E~ at the interface. Ans. -100 V /m, 7.35 X 10- 4 /45° V /m 14.36. A free space-conductor interface has H~ = 1.0 A/m on the free-space side. The frequency is 31.8 MHz and the conductor constants are E, = ,_,, = 1, o = 1.26 MS/m. Determine H~ and H~ and Ans. 1.0 A/m, 2.0 A/m, 80 ~-tm the depth of penetration of H'. 14.37. A traveling H field in free space, of amplitude 1.0 A/m and frequency 200 MHz, strikes a sheet of silver of thickness 5 ~-tm with o = 61.7 MS/m, as shown in Fig. 14-16. Find H~ just beyond the sheet. Ans. 1. 78 x 10-s A/m 14.38. A traveling E field in free space, of amplitude 100 V /m, strikes a perfect dielectric, as shown in Fig. Ans. 59.7V/m 14-17. Determine E~. 14.39. A traveling E field in free space strikes a partially conducting medium, as shown in Fig. 14-18. frequency of 500 MHz and E~ == 100 V /m, determine E~ and H~. Ans. 19.0 V /m, 0.0504 A/m

Given a

236

ELECfROMAGNETIC WAVES

~--:t

(CHAP. 14

Eo,J.Io

~--1

t'o, Po

20 u=O

t', =

J.!,"' I

tE'

1-lmm~ FIJ. 14-17

£,=55 . :·-,: '. ,:~,=1

. :-

o:=l -

,: .. ;4......· . I'

~30mm4 f'i&. 14-18 14.40. A wave propagates from a dielectric medium to the interface with free space. is the critical angle of 20", find the relative permittivity. Ans. 8.55

If the angle of incidence

14.41. Compute the ratios E~/ E~ and E~l E~ for normal incidence and for oblique incidence at 8; = 10". For region 1, £, 1 = 8.5, 1J,1 = 1, and o 1 = 0, region 2 is free space. Ans. For normal incidence, E~/£~=0.490 and Eo/E~= 1.490. At 10", E~/£~=0.539 and E~/ E~ = 1.539. 14.42. A parallel-polarized wave propagates from air into a dielectric at Brewster angle of 75°. E,. Ans. 13.93

Find

Chapter 15 Transmission Lines (by Milton L. Kult)

15.1 INTRODUCDON Unguided propagation of electromagnetic energy was investigated in Chapter 14. In this chapter the transmission of energy will be studied when the waves are guided by two conductors in a dielectric medium. Exact analysis of this two-conductor transmission line requires field theory. However, the performance of the system can be predicted by modeling the transmission line with distributed parameters and using voltages and currents associated with the electric and magnetic fields. Only uniform transmission lines will be considered; that is, the incremental distributed parameters shall be assumed constant along the line.

15.2 DISTRIBUTED PARAMETERS The incremental distributed parameters per unit length of line are inductance and capacitance as determined in Chapters 7 and 11, the resistance of the conductors, and the conductance of the dielectric medium. It was seen that the parameters depend on the geometry of the configuration, the characteristics of the materials, and in some cases the frequency. In the following summary list the dependence on geometry is represented by a geometrical factor GF.

Capacitance. [Ed =permittivity of dielectric] Conductance. [ad= conductivity of dielectric]

Inductance (external). L, = #Jd (oFL)

(H/m)

[#Jd = permeability of dielectric = #Jo]

DC Resistance (useful for operation up to 10kHz). 1 Rd = -(oFRd)

(!J/m)

[oc =conductivity of conductors]

Oc:ff

Ac Resistance (for frequencies above 10kHz). 1

21f0cu

Ra = ~(GFR0 ) (!J/m)

[ lJ =

~ 7if#Jc0c

skin depth]

Inductance (internal).

L- = {Ra127if (H/m) ' #Jo/4n (H/m)

237

for for

f> 10kHz f <10kHz

TRANSMISSION LINES

238

[CHAP. 15

lndudallce (total). L, = Lr + L; = Lr

For three common line configurations the geometrical factors are as follows:

Coaxial Line (inner radius a. outer radius b, outer thickness t). 2 GFC =In

1 a

GFL=GFC

(b/a)

GFRd=2+

1 (b t +t)

for

GFR = - + a a b

t~ {J

Parallel Wires (radius a. separation d). 1

GFL =

GFC=GFL

cosh- 1 - = In2a a

for

d~a

2 a

GFRa=-

GFRd=2

Parallel Plates (width w. thickness t. separation d). w ;rd

GFC=-

21r

GFRJ=-

GFL=GFC

4;r GFRa=W

for

15.3 INCREMENTAL MODEL; VOLTAGES AND CURRENI'S

The model in Fig. 15-1, where R, L, G. and Care as given in Section 15.2, permits analysis of the line using voltages and currents. For within a cell of length Ill the voltages across the line at points a and b differ by . ai(x t) 6.v(x. t) = (R fll)l(X, t) + (L Ill) ' at

In the limit as I l l - 0,

this becomes av(x, t) ax

R"(

l X'

) ai(x, t) t + L ---''---'at

G.lx

C.lx

(1)

C.lx

Â·I

1.. f'i&.15-l

239

TRANSMISSION LINES

CHAP. 15)

Likewise, the current at point c differs from that at b by l1i(x, t) = (G l1x)v(x, t) + (C l1x)

av(x t) ' at

from which di(x, t) G ( ) C av(x, t) - - - = v x,t + ax at

(2)

The first-order PDEs (1) and (2) imply a single second-order PDE,

a:zt(x,2 t) = RGf(x ax

t) + (RC

+ LG) af(x, t) + LC a:zt(x, t) at 2

(3)

for either v(x, t) or i(x, t). Now, (3) is an equation of hyperbolic type, very similar to the wave equation. Indeed, for a lossless line (R = G = 0), (3) is precisely the one-dimensional scalar wave equation studied in Chapter 14. Thus it is known in advance that transmission lines support voltageand current-waves which can be reflected and/or transmitted at discontinuities (sites of abrupt parameter changes) in the line.

15.4 SINUSOIDAL STEADY-STATE EXCITATION

When the transmission line of Fig. 15-1 is driven for a long time by a sunusoidal source (angular frequency ro ), the voltage and current also become sinusoidal, with the same frequency: v(x, t) = Re [t?'(x)ei""]

i(x, t)

= Re [i(x)ei.,1

Here, the phasors t?'(x) and i(x) are generally complex-valued; often they are indicated in polar form (with the x-dependence suppressed) as

where Â˘ denotes the angle between the complex vector and the real axis. Steady-state analysis of the transmission line is much simplified when all voltages and currents are replaced by their phasor representations. Figure 15-2 models in the phasor domain a uniform line of length f that is terminated in a (complex) load ZR at the receiving end and is driven at the sending end by a generator with internal impedance Z 8 and voltage ~ = V8 m /.!!_ â&#x20AC;˘ The per-unit-length series impedance and shunt admittance of the line are given by

= R + jroL

Y=G +jroC

Distance from the receiving end is measured by the variable x; from the sending end, by d.

Sending End

Receiving End (Load)

v.clf:'': X

d=O - + d

IR-

Z.Y

+_, __

x=O

240

TRANSMISSION LINES

[CHAP. 15

Equations (1), (2), and (3) of Section 15.3 become ODEs for the phasors lf(x) and i(x). dV(x)

~=Zl(x)

(1 bis)

di(x) = YV(x)

(2 bis)

dx 2

d F(x)= dx2

2 p.() r,. x

(3 bis)

with y = v'ZY = a + j{J, the square root being chosen to make a and fJ nonnegative. Equation (3 bis) is identical in form to the equation of plane waves (Section 14.3); it has the traveling-wave solutions V(x) = v+eyx + l!-e-yx = t?inc(x)

+ Yrefl(x)

f(x) = J+eyx + J-e-yx = l;nc(x) + fren(x)

The coefficients

tf+, etc., are phasors independent of x that are interrelated by the r Râ&#x20AC;˘ defined as

characteristic

impedance Z 0 and the boundary reflection coefficient

It is easy to express

j~ = ~

Zo =

l!- __ j- _ lf.en(O)

R--v+-

i+-v, <o> InC

r R in terms of the characteristic and

load impedance:

_ZR-Zo r R-

ZR+Zo

Then, if a pointwise reflection coefficient is defined by f(x)

~ren(x) V;nc(X)

it follows that r(x)

= r Re- 2 = ZR- Zo e- 2 yx

ZR+Zo

Similarly, if Z(x) = lf(x)/i(x) (x =0), then

is the pointwise impedance looking back to the receiving end Z(x) = Zo 1 + r(x) 1- f(x)

The conditions at the sending end [rerotate r(t), etc., as fs, etc.) are

z -z s-

1+rs o1-fs

lfs= lfB Z Zs Z

_lfs

ls-

Average power received at the load and average power supplied to the sending end are

241

TRANSMISSION LINES

CHAP. 15}

calculated as

PR =! Re (t?'RIR) =! liRI Re (ZR) 2

= Pine(X = 0) - Prefl(x = 0)

Ps =! Re (-.?'sf:>=! lisl 2 Re (Zs) Simp6fications for High-Frequency or Lossless Lines. For frequencies such that R .q;; wL and G .q;; roC (e.g., above 1 MHz), Zo =

~ R + ~wL = G

+ JWC

g = Ro

'J(;

y = V(R + jwL)(G + jroC) = (~ + GRo) + jwVLc =a+ jf3 2R0 2 1

u.p =v'LC --

2Jl 1 .A=-=-{3 tVLc

and

where, as always, uP and A denote phase velocity and wavelength. For the ideal lossless line with R = 0 and G = 0, the reflection coefficient is of constant magnitude. r(x)

r Re-i2fJx

where <fJR is the polar angle of r R·

ZR- Rol/tPR- 2{3x ZR+Ro

The voltage is given by V(x) = v+(l + r R J-2{3x)

which implies Adjacent maxima and minima are separated by {3x = 90°, or one-quarter wavelength. resulting wave the voltage standing-wave ratio, VSWR, is defined as

For the

For the small-dissipation line the VSWR can st'ill be used if a correction is made for the attenuation (see Problems 15.2, 15.9, 15.41).

15.5 THE SMITH CHART The Smith Chart (Fig. 15-3) is a graphical aid in solving high-frequency transmission line problems. The chart is essentially a polar plot of the reflection coefficient in terms of the normalized impedance r + iX· Z(x) I+ r(x) --=z(x)=r(x) + ix(x) =----'-....::... Rv 1- r(x) r(x) = r R J-2{3x =

ro+ixo-11 . ro+Jxo+ 1

/tPR- 2{3x = r, + jr;

(for a= 0)

where r0 = r(O) and Xo = x(O). In the complex r plane the curves of constant r are circles, Fig. 15-4(b), as are of course the curves of constant !rl, Fig. 15-4(a). The curves of constant x are arcs of circles, Fig. 15-4(c). Some important correspondences are listed in Table 15-1.

242

TRANSMISSION LINES

[CHAP. 15

Radial scales 1 II

II H II. 'I

II fl II 7

11.2

II 4 II~

II 3

Reflectron coefficient magnitude Linear d1\.llto 1.11 for r

Ji1&. 15-3

U II I

I 2 1.11

1.6 I4

2.11 I.H

Ill

3.11

Voltage standing-wave ratio (VSWR) R > 1 scale for VSWR

Smith Chart: Normalized Resistance and Readallce Coordin8tes

211

CHAP. 15}

243

TRANSMISSION LINES

nso· r,

-I- j

1-90· o

lfl

I - j

I- j

-1-j

= const.

o ,.

(a)

-1-j

const.

1-j ox= const. (c)

(b)

J.i1&. 15-4 Table 15-1

Condition

Open-circuit Short-circuit Pure reactance Matched line

1f!!.. 1/180" 1/±90" 0

oo (arbitrary) 0 0 1

arbitrary (oo) 0 ±1 0

The complete Smith Chart of Fig. 15-3 is obtained by superposing Figs. 15-4(b) and (c). The circles of constant lfl are not included; instead the value of lfl corresponding to a point (r, X) is read off the left-hand external scale. The value of the VSWR is read from the right-hand scale. The two circumferential distance scales are in fractions of a wavelength. From r = 0, x = 0, the outer scale goes clockwise toward the generator (i.e., measures x/A.), and the inner one counterclockwise toward the load (i.e. measures d/A.). Once around the chart is one-half wavelength. The third circumferential scale gives ¢r = ¢R - 2/Jx. The chart can be used for normalized admittances, Y(x)

Go == y(x) == g(x) + jb(x)

where r-circles are used for g, x-arcs are used forb, the angle of r for a given y is 1SOO + ¢r, and the point y = 0 + jO is an open-circuit. 15.6 IMPEDANCE MATCHING

At high frequencies it is essential to operate at mmtmum VSWR (ideally, at VSWR = 1). Several methods are used to match a load ZR to the line, or to match cascaded lines with different characteristic impedances. Matching networks can be placed at the load (x = 0) or at some position x =x 1 along the line, as in Fig. 15-5. The two sets of normalized conditions are as follows: (a) Before match: z(O)=zR=r0 +jx0 ; y(O)=g0 +jb0 ; VSWR> l After match: z(O) = 1 + jO; y(O) = 1 + jO; VSWR = 1

244

TRANSMISSION LINES VSWR= I

VSWR =I

(CHAP. 15

:Jz,

MATCH. "fT.

VSWR> I

MA f< H. "I'T.

---lx=x

---l.r=O

--l.r=ll

(b)

(a)

Fig. 15-5

(b) At load: z(O) = r0 + jx 0 ; y(O) = g 0 + jb 0 ; VSWR(O) > 1 Before match: z(x 1 ) = r 1 + jx 1 ; y(x 1 ) = g 1 + jb 1 ; VSWR = VSWR(O) After match: z(x 1 ) = 1 + jO; y(x 1 ) = 1 + jO; VSWR = 1 The matching networks at lower (radio) frequencies can be made with lumped low-loss reactive components; one lumped L-C network is shown in Fig. 15-6. If ZR has a reactive component, a reactance of opposite sign is added in series so that Z~ = R + jO. Then, for a match, Y. = jroC2 + m

1 L 1 =-VR(R0 -R)

(J)

+ jwL 1 =R-0 and

If R > R 0

, the capacitor should be connected to the other end of the inductor. To minimize dissipation losses at higher frequencies a length of open- or short-circuited line is used for matching, in either a single-stub or double-stub configuration.

-R,. Fig. J.S-6

15.7 SINGLE-STUB MATCHING

The configuration shown in Fig. 15-7 uses one shorted stub, of length from the load. To accomplish matching:

fso

placed at a distance x 1

(1) Determine x 1 such that y(x.) = 1 + jb â&#x20AC;˘. (2) Determine f. such that y( f.)= 0- jb â&#x20AC;˘ . After matching, y(x 1 ) = 1 + jO and

VSWR = 1 from

to f.

EXAMPLE 1. The above two steps may be accomplished on the Smith Chart (Fig. 15-8).

(i) Plot YR and trace the jrRl [or VSWR(O)J circle. (ii) Mark the intersections of the 1rRl circle and the circle g = 1. (iii) From YR move toward the generator to the first intersection, read y, distance x 1 as a fraction of J... (or read off angle 2fJx.).

= 1 + jb

and note the

CHAP. 15)

245

TRANSMISSION LINES

--x=f Fig.IS-7

l+j

-I-

(iv)

Mark the point y = 0- jb 1 on the WI= 1 circle. From the short-circuit position y = co, move toward the generator to the point y = -jb 1 • Note the distance t. as a fraction of J.... If the first intersection is not accessible, the second one can be used by readjusting the stub length for the susceptance at the new position.

For matching two cascaded lines with different characteristic impedances the above procedure is used at the connection point where the equivalent load is the input impedance to the second line.

15.8 DOUBLE-STUB MATCHING A double stub "tuner" has two shorted stub lines separated by a distance ds on the main line, as shown in Fig. 15-9. Stub 1 is nearest the load and frequently is connected at the load (x = 0). Common separations for the two stubs are ).J4 and 3).J8, hence the names "quarterwavelength tuner," etc. The Smith Chart solution for the two-stub matching problem involves the construction of the tuner circle for the given ds. This is the circle gT =g(ds) = 1, which plays the same role for stub 2 as the g = 1 circle plays for the main line. The tuner circle is obtained by clockwise rotation of g = 1, about the center of the chart, 1 + jO: a rotation of 180° gives the )./4 tuner circle, rotation of 90° gives the 3)./8 tuner circle, etc. See Fig. 15-10.

246

TRANSMISSION LINES

[CHAP. 15

1~Âˇ ="'

=II

flg.IS-9

-I+ I

-I - I

I + j

flg. 15-10 EXAMPLE 2.

Assuming that d.= A/4,

a five-step sequence is used for two-stub matching, as shown in Fig.

15-11. Plot the A/4 tuner circle. 2. Mark the intersection(s) of the tuner circle and the 8R circle through the entry point YR = 8R + jbR. Read bT at this point, which may be either intersection. 3. Stub 1 at x = 0 is used to change the susceptance bR to bT. 4. From YT = 8R + jbT move on the JrRI circle a distance d. = J.../ 4 toward the generator onto the g = 1 circle and read y = 1 + jb 2 â&#x20AC;˘ 5. Cancel the susceptance b 2 by adjusting stub 2 to produce y = 1 + jO, the matched condition. 1.

A problem arises when using the M4 tuner to match a load with gR > 1, since the conductance circle does not intersect the tuner circle. The 3)./8 tuner works for some values of gR > 1. In any case, a tuner can be displaced from the load by a distance x 1 to put g in the range for matching. Should g = 1 at the displaced position, the single-stub condition holds and stub 2 must be set for b = 0.

247

TRANSMISSION LINES

CHAP. 15)

I+ j

Fig. 15-11

Note the standing waves on the various sections of line. The shorted stubs each for 0 < x < ds, VSWR is determined have VSWR = oo. For ds < x < l, VSWR = 1; by y = gR + ibr; and if line is added between the load and stub 1, the VSWR is determined by YRÂˇ

15.9 IMPEDANCE MEASUREMENT A slotted line is used with high-frequency coaxial lines to measure VSWR and to locate voltage minima on the line. With the aid of the Smith Chart, the impedance of an unknown termination can be easily found from the VSWR and the shift of a voltage minimum from a short-circuit reference position. In Fig. 15-12 the slotted line is inserted at a convenient terminal. With the ZR in place, a probe is moved along the line to locate and measure maximum and minimum voltages. A suitable amplifier/indicator converts the probe output to a VSWR reading. ZR is replaced by a short circuit, and the reference minima are located for the high-VSWR condition. As would be expected, maxima and minima alternate at intervals of )./4. Max I

GENERATOR

---t----+-' Fig. 15-ll

~z.

EQUIVALENT LOCATION

248

TRANSMISSION LINES

[CHAP. 15

To find zR with the Smith Chart draw the measured VSWR circle as in Fig. 15-13 and locate the voltage-minimum line (from 0 to 1 on the x = 0 line). Convert the measured tu to wavelengths and mark the points on the VSWR circle that are tu from the V min line. The correct z, is capacitive; a rotation through tu toward the generator takes it into a Vmin point. (If zR were inductive, tu would be greater than a quarter wavelength and a V max point would occur before the Ymin point.)

Fig. 15-13

15.10 TRANSIENTS IN LOSSLESS LINES In switching applications and pulse operation a change in voltage is suddenly applied to the line. An analysis of this transient condition generally requires recourse to the time PDEs of Section 15.3 or to their Laplace transforms. However, in the special case of a lossless line (R = G = 0, R0 = VfJC, uP= 1/ LC), a simple graphical method is available, based on superposition of multiply reflected waves. Figure 15-14 shows a model for the lossless system, in which the exciting voltage v 8 (t) is switched on at t = 0 and where R8 is the source resistance. Now, an abrupt change at one end of the line has an effect at the other end only after one delay time, tv= t/uP, has elapsed. Reflection will occur at the receiving end if the load is not matched to the line (RR R 0 ); at the sending end if the source is not matched (R8 R 0 ).

Rel·einng End

Sending End

R. ~~:.-----------------------·~

v.:(-1)--------'-R-oU-P--'-----\-H

v--(t-)______

.t =I d=ll-+d

~:r·

-X=Il

d=l

Fig. 15-14

EXAMPLE 3.

For the case where Vg(t) is a 10-V step at t = 0 (i.e., a de voltage), and where the line is matched at both ends (RR = R8 = Ro), the transient voltage conditions are displayed in the time-distance plots of Fig. 15-15. Because the source voltage is constant and there is no reflection at the receiving end, the system reaches a steady state, v(d, t) = 5 V, after only one delay time. In Fig. 15-15(a) and (b) different boldface numerals indicate different space-time combinations corresponding to the steady-state condition. For instance, 5 signals v(O.St, 0.510 ) = 5 V.

CHAP. 15]

I d=O I

: : : : ~ :~ 1[ : .. J ~=r

•=Vl II

1d =

o.st

21 0

II<

\'~ //

211,

(

I d=l I

II=

21 1,

(a) o

J : II

= 111

(h) 0 1"\"~.cl

t·v~.l

u·~~

2rn /

lr=O.lr 0

'"l

249

TRANSMISSION LINES

Fig. 1.5-15

EXAMPLE 4. Assume that everything is as in Example 3, with the exception of the load, which is now an open-circuit (RR = oo). Figure 15-16 gives the time-distance plots. Because of the single reflection at the load, a uniform steady state of 10 V is attained after 210 •

.. 'T

Id = o I

(Ill

'·:t

ld= o.s1

·:

II~/

ll=r0

Ill

]

REFL--

I d=li

1"11

2r0

I'=

Ir = 0.5r I

21n

If>

Ill

IN(" II

f'"' " [)

II II

2r"

'I>

(a)C t·n.l

11.~/

I I

--I

2rn

-----1 I

(f\ = 0)

I I

II~~

I I

d (c) o t•d plot

(b)I'VS. d

Fig. 1.5-16

EXAMPLE 5.

A line is excited as in Examples 3 and 4; parameter values are Yg=20Vdc

-! r B-2

fR = 1 (open-circuit)

The voltage transient is described in Fig. 15-17. With reflections occurring at both ends of the line, an infinite time is needed for the attainment of a uniform steady state of 20 V.

250

TRANSMISSION LINES

(CHAP. 15 d rlro

l'al0.5/

0 (I'R

= I)

16.25 12.5

(r.

•

= V2)

0.5

1.0

1.5 2.0

5+5=10 10

2.5

10 + 2.5 = 12.5

3.0 12.5 3.5 12.5 + 2.5 =15.0 4.0 15.0

•

15.0 + 1.25 = 16.25

4.5

(b) o t·d plot

(0) C 1\ VS. (

Fig. 15-17

Solved Problems 15.1.

A paraUel-wire transmission line is constructed of #6 A WG copper wire ( dia. = 0.162 in., a11 =58 MS/m) with a 12-inch separation in air. Neglecting internal inductance, find the per-meter values of L, C, G, the de resistance, and the ac resistance at 1MHz. The four geometrical factors for the parallel-wire line involve the conductor radius, a = 206 X 10- 3 m, and the separation d = 0.305 m. Since d~ a

GFL =In (;) = 5.0

GFC = GFL = 0.20

2 GFRd =2=4.72 x Ht'm- 2

2 GFR, =-=971 m- 1

For air dielectric, L=

/Jd

/Jd (GFL)

= /Jo and

£d

= 2.0 pH/m

C = IrEd{GFC) = 5.56 pF/m

G=OS/m

15.1.

£0;

for copper, /Jc = /Jo·

Hence:

1 Rd = - (GFRd) = 2.59 x 10-3 rl/m nac 1 {J = =66pm Ylif!Jcac 1 Ra = - 2 J: (GFR,) = 4.04 X w-z rl/m Irvcac

The specifications for rigid air-dielectric coaxial line used in a radar set operating at 3 GHz are: copper material, stub-supported at intervals to maintain the air dielectric; outside diameter, i inch; wall thickness, 0.032 inch; inner-conductor diameter, 0.375 inch; characteristic impedance, 46.4 Q; attenuation, 0.066 dB/m; maximum peak power, 1.31 kW; operating peak power, 200 kW; lowest safe wavelength, 5.28 em. Determine the per-meter values of L, C, G, and R. for the line, neglecting internal inductance.

CHAP. 15}

251

TRANSMISSION LINES

The inner radius a is 4.76mm and the outer radius b is 10.3 mm. 0.771, GFL=0.386, GFC=2.59. L

1'C

{) = 1.2 pm.

1 1 1 GFR =-+-=307m" a b

and

In (b/a) =

C = 1'C£0 (GFC) = 71.9 pF/m

= /l() (oFL) =0.154 pH/m

For copper and a frequency of 3 GHz,

Then

Then, 1

R, =--.., (GFR,)=0.7020./m 2nacu

For air dielectric, G = 0 S/m.

15.3.

Show that the voltage v(x, t) =A cos (cot+ O)eiflr (3), for a uniform lossless line, if fJ = co..;LC. For the lossless line, R = G = 0,

satisfies the transmission line equation

so that the equation reduces to

ifv(x, t) = LC ifv(x, t)

For the given voltage, this requires

- {3 2 v = LC(- olv)

15.4.

fJ=wVLC

For the parallel-wire line of Problem 15.1, find the characteristic impedance, propagation constant (attenuation and phase shift), velocity of propagation, and wavelength, for operation at 5 kHz. At 5kHz the de resistance may be used. Z

=Rd + jwL = 2.59 X 10-

+ j2n(5 X

1~)(2 X

10- 6 )

= 6.289 X

10- 2 /87.6° 0./m

Y = G + jwC = j2n(5 X 1~)(5.56 X 10- 12) = l. 747 X 10-7 /9ff>S/m Zo=

~=600J-1.ro.

r= YZY= 1.048 x w-• /88.8°= (2.19 x w-6 ) + j(t.048 x w-•) m- 1 Then a=2.19 X 10- 6 N/m, 59.96km.

.. 15.5.

f3 = 1.048 X

10- 4 rad/m,

uP= w/{3 = 2.998 X 10Sm/s,

A.= 2n/{J =

A 10-km parallel-wire line operating at 100kHz has Z 0 = 557 Q, a= 2.4 x w-s Np/m, and fJ = 2.12 x 10-3 rad/m. For a matched termination at x = 0 and VR = 10 /!!_ V, evaluate V(x) at x-increments of )./4 and plot the phasors. The

y+ eax l..fk.

line

matched

the

receiving

end,

that

rR=O

and

V(x)=

But

V(O) = y+ = VR = 10/!!._V V(x) =tOe,..~ (V)

whence For x = n(A/4)

(n = 0, 1, 2, ... , 13, 13.48),

fJx

where n = 13.48 corresponds to the 10-km length,

=n(irad) = n(WO) a

ax =~(fJx) =n0.0178 Np)

By use of these increments, Table 15-2 is generated. Fig. 15-18.

A polar plot of the tabulated results is given in

252

TRANSMISSION LINES

[CHAP. 15

Table 15-l

f3x = fl>v,

Quarter Wavelengths from Load 0 1 2 3 4 5 6 7 8 9 10 11 12 13 13.48

rv(x)l.

deg.

a:r,Np

0.0 0.0178 0.0356 0.0534 0.0711 0.0889 0.1067 0.1245 0.1423 0.1601 0.1779 0.1956 0.2134 0.2312 0.24

10.00 10.18 10.36 10.55 10.74 10.93 11.13 11.33 11.53 11.74 11.95 12.16 12.38 12.60 12.71

180 270 360 450 540

630 720 810 900 990 1080 1170 1215

27(lc

Fig. 15-18

15.6.

Repeat Problem 15.5 if a mismatched load results in the same.

= 0.41!!_;

all other data remain

In contrast to Problem 15.5, the voltage is now given by the superposition of an incident and a reflected wave: V(x) = V;nc(x) + v.en(x) = v+eax Lf!!_ + r Rv+e-""' L-f3x

The boundary condition at x = 0 gives (omitting physical units)

Thus

10/!!_=l.4V+f!!_ or y+=7.14/!!._ V(x) = 7.14e"'" /1!!_ + 2.86e_,.. L-f3x

CHAP. 15)

TRANSMISSION LINES

253

Table 15-3

A/4from Load

ax, Np

e""

0 1 2 3 4 5 6 7 8 9 10

0.0 0.0178 0.0356 0.0534 0.0711 0.0889 0.1067 0.1245 0.1423 0.1601 0.1779 0.1956 0.2134 0.2312 0.24

1.0 1.018 1.036 1.055 1.074 1.093 1.113 1.133 1.153 1.174 1.195 1.216 1.238 1.260 1.271

12 13 13.48

fJx ()"

90" 180" 270"

360" 450" 54QO

630" 720" 810" 9000 990" 1080" 1170" 1215°

V;nc(x)

V...,,(x)

7.14/!!_ 7.27/90° 7.40/180" 7.53/270" 7.67 l3fff 7.80/450" 7.941540° 8.09/630" 8.23/720" 8.38/810" 8.53/900° 8.68/990" 8.84/1080° 9.00 /1170" 9.08J12l5°

2.86/!!_ 2.81/-90" 2.76f-180" 2. 71f-270" 2.66/- 360" 2.62/-450" 2.57/-540" 2.52J-630" 2.48J-720" 2.44/-810" 2.39J-900° 2. 35/-990" 2.31/-1080" 2.211 -1170" 2.25 L-1215°

IV(x)l

10.00/!!_ 4.46/90° 10.16/180" 4.821-90" 10.33 /!!.. 5.18190" 10.51/180" 5.57/-90" 10.71 /!!.. 5.94/WO 10.92/180" 6.33/-WO 11.15/!!_ 6.73 JWO 9.34[148S

where a and f3 are as specified in Problem 15.5. The required calculations are presented in Table 15-3 and Fig. 15-19.

4xi"A =

Fig. 15-19

15.7

Measurements are made at 5kHz on a 0.5-mile-long transmission line. The results show that the characteristic impedance is 941-23.2° Q, the total attenuation is 0.06 Np, and the phase shift between input and output is 8". Find the R, L, G, and C per mile for the line; the phase velocity on the line; and the power lost on the line when the sending-end power is 3 W and the load is matched. The measured attenuation is at= 0.06 Np, whence 8° = 0.14 rad, so that f3 = 0.28 rad/mi. Hence,

YZY= y = 0.12 + j0.28=0.305 f66.8°miFrom this and the measured value Z

a= 0.12 Np/mi,

ZY =0.093f133.6°mi- 2

VZfY = 94 f- 23.2° 0.:

= 28.67 [43.6° = 20.8 + j19.8 0./mi = R + j21ifL

Y = 3.24 X 10- 3 f90" = j3.24 x 10- 3 S/mi = G

which imply:

= 20.8 0./mi;

L = 630 ~JH/mi; G = 0;

the phase shift is {Jt =

+ j21ifC

C = 0.103 ~JF/mi.

[CHAP. 15

TRANSMISSION LINES

254

The phase velocity is uP =2:rif I {J = 2Jr(5 x L£t1)10.28 = 1.12 x lOS mi/s. For a matched load (no reflections), the received power is given by

PR = Pse- 2"'' = 3e- 0 . 12 = 2.66 W and the power lost as the result of attenuation is 0.34 W.

15.8.

A 600-Q transmission line is 150m long, operates 10-3 Np/m and fJ = 0.0212 rad/m, and supplies a load Find the length of line in wavelengths, r R• r s. and Zs. 50/!!_ V, find Vs. the position on the line where the voltage

at 400kHz with a= 2.4 x impedance ZR = 424.3/4~ D. For a received voltage R = is a maximum, and the value of

lVI max· Because J... = 2Jr I {J = 296.4 m,

ZR - Zo ZR + Zo

t = 150m = 0.5U.. At x = 0, 300 + j300- 600 0 300 + j 300 + 600 = 0.45/116.6

•

= -0.2 + ]0.4

Therefore, at x = t, rs = jrRl e- 2 "'' /1/JR- 2fJf= 0.45e-0."12/116.6°- 363° =0.22/113.6°= -0.09+j0.20 1 + rs) (0.91 + j0.2) Zs = Zo( l _ r.~ = 600 l.09 _ jO.Z

and From

VR=50LQ:=V+(t+rR),

= 5(}2. 7/22.8" 0.

"V+=56.2/-26.6°V. Then,

Ys = (V .. e"'' I.J!!)[l + rs] = (56.2e0 · 36 J-26.6° + 18t.S0)[0.9t + j0.2) = 75.0 Jt67.3°V To find x where the voltage is a maximum, construct the phasor diagram Fig. 15-20. At x = 0 the incident and reflected voltages are separated by an angle of ll6.6°. When vane rotates 58.3° counterclockwise and V.-e~~ rotates the same angle clockwise the two phasors add together. The distance x for which {Jx = 58.3° is 48.2 m, the position of the maximum. The magnitude is

IYimax = 56.2e0 116 + (0.45)(56.2)e_ 0 _116 = 85.5 V

'""~/ /

5!\.30 / /

F'i&· 15-ZO

.. 15.9.

For the coaxial line specified in Problem 15.2, determine the actual characteristic impedance and attenuation, and compare the values with the specifications. Determine the length of the shorted stub required to support the center conductor at the operating frequency of 3 GHz and calculate the highest "safe" frequency of operation for this line from the specifications. The characteristic impedance for the high-frequency low-loss line is

/L /t.54 x w"Vc= \h.I 9 x 10 =46.33 0. 7

Ro =

(specification is 46.40.)

CHAP. 15]

255

TRANSMISSION LINES

The per-meter attenuation is

a= Ro 2

0.702 I 3 ( _ ) = 7.58 X 10- Np/m = 0.0658 dB m 2 46 33

(specification is 0.066 dB/m)

where the conversion 1 Np =8.686 dB has been used. A stub to support the center conductor must be A/4 long so that the short circuit reflects to an open circuit at the point of connection to the main line. At 3 GHz the length should be

1 1(3 x 109 Hf) =0.025m X

~=4y=4

or 2.5cm

The "safe" highest frequency of operation is determined by the specification for lowest ..safe" wavelength. Up

fm =A,""'

3 X lOS 5.28 X 10-2 = 5.68 GHz

At frequencies above this value, propagation modes other than the TEM could exist.

15.10. A 70-Q high-frequency lossless line is used at a frequency where l = 80 em with a load at X= 0 of (140 + j91) Q. Use the Smith Chart to find: r R' VSWR, distance to the first voltage maximum from the load, distance to the first voltage minimum from the load, the impedance at vmax ' the impedance at vmin ' the input impedance for a section of line that is 54 em long, and the input admittance.

â&#x20AC;˘

On the Smith Chart plot the normalized load ZR/ R 0 = 2 + jl.3, as shown in Fig. 15-21. Draw a radial line from the center through this point to the outer A-circle. Read the angle of r R on the angle scale: tPR = 29". Measure the distance from the center to the z-point and determine the magnitudes of r R and VSWR from the scales at the bottom of the chart. VSWR=3.0

and

Draw a circle at the center passing through the plotted normalized impedance. Note that this circle intersects the horizontal line at 3 + jO. This point of intersection could be used to determine the VSWR instead of the bottom scale, because the circle represents a constant VSWR. Locate the intersection of the VSWR circle and the radial line from the center to the open-circuit point at the right of the z-chart. This intersection is the point where the voltage is a maximum (the current is a

0.21 >..

()(..

0 sc 0.5 >..

0.25 >..

0.38Sk - -

Fil-15-11

TRANSMISSION LINES

256

(CHAP. 15

minimum) and the impedance is a maximum. The normalized impedance at this point is 3 + jO, whence Zm.. = 210 + jO Q. To find the distance from the load to the first Ym.. use the outer scale (wavelengths toward the generator). The reference position is at 0.2U and the max. line is at 0.25 ).; so the distance is 0.04). toward the generator, or 3.2 em from the load. From the Vmax point move 0.25). toward the generator and locate the Vmin point. The normalized impedance is 0.33 + jO, and Zmin = 23.1 + jO Q. The distance from the load to the first minimum is

0.25). + 0.04). = 0.29). = 23.2 em To find the input impedance, move ~ = 0.675 wavelengths from the load toward the generator, and read the normalized impedance. Once around the circle is 0.5J., so locate the point that is 0.175). from the load on the outer scale. The point is at 0.2U + 0.175). = 0.385).. Through this point draw a radial line and locate the intersection with the VSWR circle. The normal impedance is 0.56- jO. 71 and Z; 0 = 39.2- j49. HL The normalized input admittance is located a diameter across on the chart, which corresponds to the inversion of a complex number. For z = 0.56- jO. 71, y = 0.68 + j0.87; therefore, Y;n = ~ = (9. 71 + j12.4) mS

15.11. The high-frequency lossless transmission system shown in Fig. 15-22 operates at 700 MHz with a phase velocity for each line section of 2.1 x lOS m/s. Use the Smith Chart to find the VSWR on each section of line and the input impedance to line #1 at the drive point. (There are three distinct transmission line problems to be solved.)

zR =

+j 1n n

=son j

ZH=41l+j0H

Fig.

ts-n

For the three lines the wavelength is ). = (2.1 x 1rf)/(7 x 1rl) = 30 em. For line #2 the length is (43.5/30)). = 1.45). and the normalized load is (0 + j70)/70 = jl. Plot this value as point 1 in Fig. 15-23. Note the reference position, 0.125>.. and VSWR = oo. Move on the VSWR circle 1.45). toward the generator to point 2 and read the value Z;n = 0 + j0.5l. The input impedance to line #2, Zin2 = Z.nRm = 0 + j35. 7 Q

is one part of the load on line # 1. For line #3 the length is jA=0.7A and the normalized load is (40+j0)/90=0.44+j0. Plot this value as point 3 and note the reference position of OJ. and the VSWR = 2.25. Move on the VSWR circle 0. 7A toward the generator to point 4, and read off Z; 0

= 1.62 + j0.86

This is the second part of the load on line # 1.

Zin3

Z;nRm =

145.8 + j77.4 Q

257

TRANSMISSION LINES

CHAP. 15)

0.125 I

()(

0.5

11.:!57

Fig. 15-23

For line # 1: the length is 1.25/0.30 = 4.167A and the load is the parallel combination of Z;.,2 and Z;.,3 • Normalize each impedance to the 50-Q line, find each admittance, add the admittances for YR• and then find zR. ~=j (

35.7) =0+j0.714 (pomt 5 50 0

)

and

145.8 + j77.4 2 92 5 SO = . + J1. 5 (point 7) 0

YR = 0.27- jl.55

(point 9)

with

Y2 = 0- jl.41

and

(point 6)

YJ = 0.27- j0.14

(point 8)

VSWR=14

Invert by moving a diameter across to point 10 for zR = 0.1 + j0.63 at the reference position 0.09).. Now move 4.167A toward the generator from zR on the VSWR = 14 circle to point 11, and read z;., = 9.5- j6.3. The input impedance to line #1 is 50(9.5- j6.3) = 475- j315 Q

15.12. (a) A high-frequency 50-Q lossless line is 141.6 em long, with a relative dielectric constant E, = 2.49. At 500 MHz the input impedance of the terminated line is measured as Z;n = (20 + j25) Q. Use the Smith Chart to find the value of the terminating load. (b) After the impedance measurement an 8-pF lossless capacitor is connected in parallel with the line at a distance of 8.5 em from the load. Find the VSWR on the main line. (a)

For

£, =

2.49,

3xlOS

uP= • hffi =

v2.49

1.9 x lOS m/s

). =up=

1.9 x lOS= 3Bcm 5 X lOS

and the length of line is (141.6/38)). = 3. 726).. The normalized input impedance is Z;n = (20 + j25)/50 = 0.4 + j0.5. Plot this value on Fig. 15-24 as point I, measure the VSWR, draw the VSWR = 3.2 circle, and note the reference position at 0.418). toward the load. From z,., move 3.726). toward the load on the VSWR circle (a net change of 0.226).) and read the normalized load impedance zR = 0. 72 - jO. 98 at point Z. The load impedance is ZR = (36 j49) Q at 500 MHz. (b) Since the capacitor is connected in parallel it is convenient to work on the y-chart, Fig. 15-25. In Fig. 15-24 read the value diametrically opposite zR: YR = 0.48 + j0.67. Plot YR as point 3 in Fig. 15-25 and draw the VSWR = 3.2 circle. The reference position is 0.105). toward the generator, corresponding to x = 0. Move 8.5 em, or (8.5/38)). = 0.224). toward the

258

TRANSMISSION LINES

zoe

0.5 0

(CHAP. 15

()

ysc

0.5

0.2~

Fig. 1.5-14

generator on the VSWR circle and read y(x 1) at point 4. Before the capacitor is added, y(x 1) = 1.04- jl.22. The normalized admittance of the capacitor is Yc = (j2tifC)Ro = j2n(5 X Hf)(8 X 10- 12)(50) = 0 + j1.26 and the new admittance at x 1 is Yc + y(x 1) = 1.04 + j0.04. Plot this new admittance as point 5 and measure VSWR = 1.04 (a significant reduction from 3.2).

15.13. A 4-m-long, stub-supported, lossless, 300 Q, air-dielectric line (Fig. 15-26) was designed for operation at 300 MHz with a 300-Q resistive load, using shorted (l/4)-supports. With no changes in dimensions or load, the line is operated at 400 MHz. Use the Smith Chart to find the VSWR on each section of line, including the supports, and the input impedance at the new frequency. I

m---•--+1..,.•. . .___ 2.~ m---•-+-1••0.5 m--J

Fie· 15-16 The wavelength is A= uP/!= (3 x Uf}/(4 x Uf) = 75 em and distances in terms of A are Total length = 4 m = 5.333A Load to stub 1 = 50 em = 0.667A Stub length = 25 em = 0.333A Stub separation= 2.5 m = 3.333A Stub 2 to input = 1 m = 1.33A Since the stubs are in parallel, use a y-chart, Fig. 15-27, for the solution. At x = 0, YR = R 0 /ZR = 1 + jO (point 1) and VSWR = 1. The line is flat to the point of connection of the first stub, 0.667A from the load, with y(x 1 ) = 1 + jO in the absence of stubs.

259

TRANSMISSION LINES

CHAP. 15]

<=> 1------+--=-±----~--l--+--~

·;

ysc

Fig. 15-17

To find the admittance of the shorted stubs, plot Ysc at point Z, move 0.333A. toward the generator on the VSWR = oo circle to point 3, and read y = 0 + jO. 58. This value must be added to y = 1 + jO to get the admittance at x 1 with stub 1 connected; thus, y(x 1) = 1 + j0.58 (point 4), VSWR= 1.75, and the reference position is 0.148A toward the generator. Draw the VSWR circle through point 4; move 3.333A toward the generator from 4 to 5; and read the admittance at x 2 without stub 2 in place: y(x 2 ) = 0.57- j0.08. At this point add the second stub to get y(x 2 ) = 0.57 + j0.50 (point 6) draw the VSWR = 2.3 circle. The reference position is 0.092A toward the generator. From point 6 move 1.333A toward the generator on the VSWR = 2.3 circle to point 7 and read the normalized input admittance Y;n = 0.52- j0.38. Invert this value by moving across a diameter and read z;n = 1.23 + jO. 92 (point 8). The input impedance to the line at 400 MHz is Z; R 0 = (369 + j276) Q. 0

15.14. The lossless lumped-parameter network shown in Fig. 15-28 is used to match a 50-Q line to the input of an RF transistor operating at 1 GHz. The input reflection coefficient for the transistor is r=0.6f-150°, measured for a 50-Q system. Find the values of Land C for the conjugate matched condition.

Transislor

f11.1S-Z8 Normalizing the reactances of the matching network to the 50-Q line gives x = wL/50 and b = 50wC. The normalized impedance looking back to the network from the transistor is

1 Zm

Now, the matching criterion is

r R = r•, Zm

=ix+ 1 +jb

(1)

1+r* _ r* =0.27 + j0.25 1

(2)

260

TRANSMISSION LINES together (1) and (2) yield b = Âą1.64. corresponds to a capacitance. Then

15.15.

x = +0.70 where the positive sign on b

For b = +1.64,

b C=-=5.2pF wRo

[CHAP. 15

xRo L=-=5.6nH

and

(J)

A 15 m length of 300-Q line must be connected to a 3 m length of 150-Q line that is terminated in a 150-Q resistor. Assuming the lossless condition for the air-dielectric lines and operation at a fixed frequency of 50 MHz, find the R 0 and the length for a quarter-wave section of line (quarter-wave transformer) to match the two lines for a VSWR = 1 on the main line. If no transformer is used, what is the VSWR on the main line? A model for the system is shown in Fig. 15-29. For f =50 MHz and uP = 3 x lOS m/s, wavelength is ). = 6 m; a )./4 section of line must be 1.5 m long.

Ru~-

lx ~ 0

>J4

1~~

150 fl

the

Fig. ts-29 With no transformer in place the termination of the 300-Q line is 150 Q, since line 2 is R0 -terminated. The reflection coefficient on line 1 is r = 150- 300 150+300

and

VSWR=l+G}=2

l-<D

With the transformer inserted as shown, the reflection coefficient at the load RR is

r R= and the input impedance at

fJx = 90",

RR - RoT RR +RoT

which, as the load on line 1, must be 300 Q, is t+rR~

Z;n=300Q=RoT1-rR/-180" Substitution of (1), with RR = 150 Q,

(1)

1-rR RoTl+rR

(2)

in (2) gives

300(150 +RoT+ 150- RoT)= RoT(150 +RoT- 150 +RoT) or RoT= V300 X 150 = VR01 R02 = 212.1 Q.

15.16. A generator at 150 MHz drives a 10-m-long, 75-Q coaxial line terminated in a composite load consisting of the parallel connection of two 50-Q lines of lengths 0.5 m and 1 m, each terminated in a 50 Q resistance. All lines are lossless with ER = 2.2. With reference to Fig. 15-30, determine the length ls and connection point x 1 of a parallel-connected 75-Q stub that will produce minimum VSWR on the feed line. The stub should be as close as possible to the load. Phase velocity, uP= 3 x 10S/v'2.2 = 2.02 x lOS m/s; wavelength, ). =uP/!= 1.35 m. The input impedance to each of the 50-Q lines is 50 Q for R0 -termination, the composite load on the 75-Q line is 25 Q or, when normalized, zR = 0.333 + jO. Plot zR on the Smith Chart and through this point draw the IfRl circle and the radial line to the angle scale, as shown in Fig. 15-31. Read tPR = 180" and measure and

VSWR=3

CHAP. 15]

TRANSMISSION LINES

261

I~= 10m

X= X1

Fig. 15-30

Fig. 15-31

Since the matching stub is in parallel, locate YR = 3 + jO by projecting a diameter across from zR. Locate the intersections of the g = 1 and VSWR = 3 circles and note the distances from the load at x =0: y y

= 1 + jl.15

=1-

at a distance 0.25 + 0.166 = 0.416l toward the generator j1.15 at a distance 0.3335 - 0.25 = 0.0835l toward the generator

Locate the 75-D stub at x, = 0.0835l = 11.3 em. The stub length ~ is that which makes y. = 0 + jl.15, for a net y = 1 + jO, (matched condition). Thus, plot y. and determine the distance from the short-circuit condition y = oo to this point. The length of stub is (0.25 + 0.1345)l = 51.9cm. The VSWR on lines t, and f 2 is 1.0 for the matched loads. On the shorted stub the VSWR is infinite. From x =0 on the main line to x =x 1 the VSWR is 3.0 and from x =x 1 to x = l 3 the VSWR is 1.0.

15.17. Find the shortest distance from the load and the length (both in centimeter) of a shorted stub connected in parallel to a 300-Q lossless air-dielectric line in order to match a load ZR = (600 + j300) Q at 600 MHz. The matching stub is the same type of line as the main line.

262

TRANSMISSION LINES

(CHAP. 15

For both the line and the stub, uP = 3 x lOS m/s and A= 0.5 m. Plot zR = (600 + j300)/300 = 2 + jl on the y-chart, Fig. 15-32. Draw the VSWR = 2.6 circle, move diametrically across to YR = 0.4- j0.2, and read the reference position 0.464A toward the generator. Move from YR on the VSWR circle to the first intersection with the g = 1 circle and read y(x 1) = 1 + jl at the reference position 0.162A.. The stub location is x 1 = [(0.5- 0.464) +0.162)A=0.198A =9.9em

from the load. To match the line for VSWR == 1, the admittance of the shorted stub must be y, = 0- j1 to cancel the susceptance at point x 1 • The required length of stub is 0.125A = 6.25cm.

Fig. J.5..3Z If this position is not accessible, the second intersection with the g = 1 circle may be used, where y(x 2 ) = 1 - jl and

x 2 = ((0.05- 0.464) + 0.338)A = 0.374A = 18.7 em The stub would have to be adjusted to give y, = +jl,

for a length of 0.375A = 18.75 em.

15.18. A high-frequency lossless 70-Q line, with E, = 2.1, is terminated in ZR =50 jJfJ' Q at 320 MHz. The load is to be matched with a shorted section of 50-Q line, with £, = 2.3, connected in parallel; the stub must be at least 5 em from the load. If such matching is possible, find the distance from the load and the length of the stub. For the main line, uP = 3 x lOS /W = 2.07 x lOS m/s, A= uP//= 64.7 em; for the stub line, u,.. = 3 x lOS/v'2.3 = 1.98 x lOS m/s, .A.= 61.9cm. The normalized load is zR = (50/3ff')/70= 0.62 + j0.36, with VSWR = 1.92, and the admittance is YR = 1.20- jO. 70 at reference position 0.327A, toward the generator; see Fig. 15-33. Move on the VSWR circle from YR toward the generator to the first intersection, y(x 1 ) = 1- j0.66, at 0.350A, or a distance of 0.023).. = 1.49 em. This point can not be used due to the 5-cm limitation. Continue on the VSWR circle to y(x 2 ) = 1 + j0.66 at position 0.15 U, ; the distance x 2 = (0.5- 0.327) + 0.151 = 0.324A.., = 21.0 em

gives the point of connection for the stub. As the stub has a different Ro , it is necessary first to "denormalize" y(x 2): Y(x 2 ) = l

;g·

+ 66 (1.4 + jO. 94) X 10-2 S

263

TRANSMISSION LINES

CHAP. 15]

0.~

I /

0.43

y, = - j0.47

I, = 0.43- 0.25 = 0.18).

Fig. 15-33

which shows that for cancellation of susceptance we must have y, = ( -j0.94 X 10- 2 )(50) = -j0.47

The length of shorted stub is then

(0.43 - 0.25))., = 0.18)., = 11.1 em.

15.19. A complex load is measured with a VHF bridge at 500 MHz; the impedance is 29/3f1' Q. This load is connected to a 50-Q air-dielectric line, with a 50-Q 3)./8 tuner between the load and line. Find the lengths of each shorted stub to produce a VSWR of 1.0 on the main line. Show both solutions if they exist. The model for the system is shown in Fig. 15-9.

For the air line,

uP = 3 x

m/s and

). =

uP// = 60 em. The normalized load impedance is ZR

= Ro

=0.58/30° =0.5 + j0.29

On the Smith Chart draw the 3)./8 tuner circle, plot zR , draw the VSWR = 2.25 circle, locate YR = 1.52- j0.88, and find the intersections of the tuner circle and the gR =

Fie· 15-34

'··

(b)

264

TRANSMISSION LINES

(CHAP. 15

1.52 circle. There is a solution for each intersection; first consider y = 1.52- j1.82 (Fig. 15Here, the first stub must be adjusted to change the susceptance from -0.88 to -1.82 (point 1); thus y•• = 0- jO. 94 at point Z. The stub length for this b is read on A scale from y = oo toward the generator: t.. = (0.380- 0.25)A = 7.8 em 34a).

From point 1 move 3A/8 toward the generator to point 3, where y = 1 + j1.53. add y = 0- j1.53 (point 4); and the stub length is (. 2

Stub 2 must

= (0.342- 0.25)A = 5.52 em

Figure 15-34b presents the second solution, which follows the same pattern. At 1': y = 1.52- j0.16 At Z': y, 1 = 0 + j0.72 and (. 1 = (0.25 + 0.099)A = 21.6 em At 3': y = 1 + j0.45 At 4': y, 2 = 0- j0.45 and f, 2 = (0.433 - 0.25)). = 11.4 em The first solution is preferred because the total length of the stubs with infinite VSWR is 12.63 em, which will introduce lower losses in a practical system.

15.20. Use a two-stub quarter-wave tuner (50-Q, shorted stubs) located 7.2 em from the load of Problem 15.19 in order to match the load to the line. The normalized load admittance in Problem 15.19 is YR = 1/zR = 1.52- j0.88, and the wavelength is A = 60 em. At 7.2 em or 0.12). from the load, y 1 = 0. 54 - jO. 36; this value is to be matched to the line with the tuner. Two solutions exist. Solution I

<Fie· 15-35)

y 1 = 0.54- j0.36 1: Yr = 0.54- j0.50 Z: y•• = -j0.14 (. 1 = (0.478- 0.25)A = 13.68cm 3: Move A/4 toy= 1 + j0.95 4: y.2 = -j0.95 (. 2 = (0.3795- 0.25)A = 7. 77 em

Total stub length = 21.45 em

Fig. 15-35

(preferred)

Solution 2 (Fig. 15-36) Y2 = 0.54- j0.36 1': Yr = 0.54 + j0.50 2': y•• + j0.86 (. 1 = (0.25 + 0.113)A = 21.78 em 3': Move A/4 toy= 1- j0.95 4': y.2 = + j0.95 (. 2 = (0.25 + 0.1205)). = 22.23 em

Total stub length = 44.01 em

Fig. 15-36

ts.Zl

265

TRANSMISSION LINES

CHAP. 15]

A 70-Q double-stub tuner is used to match a load YR = ( 4. 76 + jl. 43) mS at 600 MHz to a 70-Q Iossless air-dielectric line. The first stub is located at the load and the separation between the stubs is 10 em. Find the shorted-stub lengths for the matched condition. For the air line, A= (3 x Hf)/(6 x 10") = 0.50 m and the stub separation is 10 em= A/5. Draw the A/5 tuner circle as shown in fig. 15-37. Plot the normalized load YR = YRRo = 0.33 + jO.lO, which determines the VSWR = 3.0 circle. Two solutions exist, one for the intersection with the tuner circle at y, = 0.333- j0.18 (VSWR = 3.2) and the other at y2 = 0.333 + j0.84 (VSWR=4.9).

0 0.5

ysc 0.25

Fig. IS-37 First solution. y,, = -j0.28 to change YR to y., for a length of 0.207A = 10.35 em. Move on the VSWR = 3.2 circle 0.2A. toward the generator from YR , to y = 1.0 + jl.23. The second stub must be adjusted to give y, 2 = -jl.23 for a net y = 1 + jO, the matched condition. The length is 0.109A. = 5.45 em. Second solution. y;, = +jO. 74, for y2 = 0.33 + j0.84 and a length 0.35U = 17.55 em. Move on the VSWR = 4.9 circle 0.2A toward the generator from YRâ&#x20AC;˘ to y' = 1.0- jl. 75. The second stub must be adjusted for y ; 2 = -jl. 75 to produce 1 + jO, the matched condition. The length is 0.417A = 20.85 em.

ts.zz.

A 50-Q slotted line that is 40 em long is inserted in a 50-Q lossless line feeding an antenna at 600 MHz. Standing-wave measurements with a short-circuit termination and with the antenna in place yield the data of Fig. 15-38; the scale on the slotted line has the lowest Min Jcm

Min 28cm

-----4

I ----i

HighVSWR Max

20.5cm I VSWR = 2.2

Fig.IS-38

Min

I I

'8 5cm

~:R

266

TRANSMISSION LINES

[CHAP. 15

number on the load side. Find the impedance of the antenna, the reflection coefficient due to the load, and the velocity of propagation on the line. For the short circuit, minima are separated by a half-wavelength, so A= 50 em. For a frequency of 600 MHz the phase velocity is uP= fi = 3 x 10" m/s (air dielectric). With the antenna in place the minimum shifts 5 em = 0. U toward the generator. On the Smith Chart draw the VSWR = 2.2 circle and identify the voltage minimum line as in Fig. 15-39. Locate zR on the VSWR circle O.lA toward the load from the V min position: ZR

= 0.64- j0.52

and

The load r R is read from the chart: 4>R = -108Â° and center to zR, as read off the external scale.

IfRl = 0.375 is the distance from the

Fig. 15-39

15.23. A 40-m length of lossless 50-0 coaxial cable with a phase velocity of 2 x 10" m/s is connected at t = 0 to a source with v8 (t) = 18 V de and R8 = 100 C. If the receiving end is short-circuit terminated, sketch the sending-end voltage v5 (t) from t = 0 to t = 2.5ps. The delay time of the line is t 40 tn=-=--=0.2ps uP 2 x lOS

The incident voltage at the sending end at t = 0 is Vs(O)=

V6 R 0 = 18x 50 = 6 V R6 + Ro 100 + 50

The reflection coefficients at the two ends are

Figure 15-40 shows the t = d plot over a total time of 2.5ps = l2.5t0 â&#x20AC;˘ From this, the desired v. = t plot, Fig. 15-41, is easily derived. At any time v. is the sum of all incident and reflected waves present at d = a, up to and including the last-created incident wave. For example,

2 2

v.(4.0lt0 ) =6- 6-2 +2 +3=3 On account of

= -1

the waves preceding the last incident wave cancel in pairs.

267

TRANSMISSION LINES

CHAP. 15)

rR =

-1

Fig.lS-40

Ill

-2

Fig. 1.5-41

15.24. A well-designed, lossless, 100-0, 100-ps delay line produces a good 10-ps pulse at the output 100 ps after it is driven at the input, at t = 0, by a 10-ps rectangular pulse recurring with a period of 2 ms. The generator has a 9 V peak open-circuit output and an internal resistance of 50 Q. Sketch v 5(t) and vR(t) from t = 0 to t = 650 ps if the termination is a 50-Q resistor. At t = 0 the sending-end incident voltage is pulse of

duration with a peak value of

= (9)(100)

'""

10-J~S

50+ 100

The sending-end and receiving-end reflection coefficients are

R - Ro 50 - 100 R6 + R0 50+ 100

= 6- - = B

1 3

r _R R -

-Ro= 50-100 =--1 + R0

50 + 100

Since the pulse period is 2000 JlS, only the pulse sent out at t = 0 need be considered in the t = d plot of Fig. 15-42, which covers only the first 650 JlS. Applying to Fig. 15-42 the summation technique described in Problem 15.23, one obtains the required voltage plots, Fig. 15-43. For proper operation the delay line must be terminated in Ro = 100 !J.

''~

Fig. 1.5-42

268

TRANSMISSION LINES

[CHAP. 15

l] v

Fig. 15-43

Supplementary Problems 15.25. A coaxial cable with the dimensions a= 0.5 mm, b = 3 mm, and t = 0.4 mm is filled with a dielectric material having E, = 2.0, ad= 10 p.S/m. The conductors have a,.= 50 MS/m. Calculate the per-meter values of L, C. G, Rd, and R,. at 50 MHz. Neglect internal Am. 0.358p.H/m, 62.0 pF/m, 35.1p.S/m, 0.030 Qjm, 0.743 Q/m inductance. 15.26. Find the per-meter values of L, C, G, and R for a parallel-wire line constructed in air of #12 A WG copper wire (dia. = 0.081 in., ac = 52.8 MS/m) with a 4-inch separation. Operation is at 100kHz. Ans. 1.84p.H/m, 6.05 pF/m, 0, Ra = 0.0268 Q/m 15.27. In a ·•twin-lead" transmission line, two parallel copper wires (ac =50 MS/m} are embedded 0.625 in. apart in a low-loss dielectric with E, = 2.4. Neglecting losses, determine the diameter of the conductors for a characteristic impedance of 300 Q. For this size of conductor, find the de resistance and the ac resistance at 100MHz. Ans. 0.026in. =0.66mm, 0.117Q/m, 2.72Q/m 15.28. A high-frequency application uses a coaxial cable with copper conductors, where the diameter of the inner conductor is 0.8 mm and the inside diameter of the outer conductor is 8.0 mm. The dielectric material has E, = 2.35, and the thickness of the outer conductor is much greater than the depth of penetration at the operating frequency. The engineer wants to use a new cable having the same R 0 , but with a larger outer conductor such that b 2 - a = 1. 5( b 1 - a). Find E, for the new cable and calculate R 11 and the capacitance per meter for each cable. Ans. 3.18, 90Q, 56.8pF/m(old},66.lpF/m(new) 15.29. For the coaxial cable of Problem 15.28, calculate the line characteristics Zu , a, {3, uP, ). for operation at 10kHz. Ans. 32.7 fl5.2° Q, 1.04 x 10 3 Np/m; 4.2 x 10 ·4 rad/m, 1.49 x 108 m/s, 14.9 km 15.30. Find the characteristic impedance, propagation constant, velocity of propagation, and wavelength for the parallel-wire line of Problem 15.26. Ans. 552.5/-0.65° Q, (2.4 + j210}10- 5 m 1, 2.99 x 108 m/s, 2.99 km 15.31. A transmission line is 2 miles long, operates at 10kHz, and has parameters R = 30 Q/mi, C = 80 nF/mi, L = 2.2 mH/mi, and G = 20 nS/mi. Find the characteristic impedance, attenuation per

CHAP. 15]

TRANSMISSION LINES

269

mile, phase shift per mile, phase velocity, and wavelength. What is the received power to a matched load when the sending-end power is 1.2 W? Ans. 167.7/-6.1° D, 0.0896Np/mi, 0.838 rad/mi, 7.5 x tltmi/s, 7.5 mi; 838.6mW

15.32. A transmission line 250m long operates at 2 MHz with a load impedance of 200 D. The line characteristics are Zo = 300 /!!_ D, 11' = 4 x 10- 4 Np/m, {J = 0.06 rad/m. If the sending-end voltage is 30 /!!_ V, find the receiving-end voltage, power to the load, sending-end current and power, and the reflected powerfrom the load. Ans. 22f-l30°V; 1.21 W; 105.4f18.3°mA, 1.5W; 50mW 15.33. One method of determining the characteristics of a line is to measure the input impedance at x = t (line disconnected from source) when the receiving end is opened for Zoe , and when it is shorted for Zsc . From the product ZocZsc and the ratio Zoe/Zsc the characteristic impedance and the propagation constant per unit length can be calculated. If measured values at 5 kHz are Zoe= 141.9 j-84.1° D and Zsc = 62.0 j37.7o D for a 2-mile length of line, use the equation for Zs to find Z 0 , 11' (per mile), and {J (per mile). Ans. 93.8 j-23.2° D, 0.12 Np/mi, 0.28 rad/mi 15.34. A 200-m length of 300-D transmission line has 11' = 2.5 x 10- 3 Np/m and f3 = 0.02 rad/m when operating at 200kHz. If VR = 20 /!!_ V and ZR = 350 f20o D, find the distance from the receiving end to the first impedance minimum. What is the value of this Zm;n? Ans. 107.2 m, 239.4 D 15.35. A 500-D line is connected to a 10-kHz generator rated at 80 /!!_ V open-circuit with an internal resistance of 600 D. The line is 3 miles long, with 11' = 0.05 Np/mi and f3 = 0. 9 rad/mi at 10kHz. For a matched load at x = 0, find the sending-end power, the receiving-end power, and YR. If the line is opened at the receiving end, what is the sending-end power? Ans. 1.32W, 0.98W, 31.3f-154.7°V; 0.65W

15.36. For the line of Problem 15.35, find the receiving-end current and the sending-end power if the line is shorted at x = 0. Ans. 0.12/-157.6° A, 0.55 W 15.37. The rigid coaxial line of Problems 15.2 and 15.9 would be classified as a low-loss line. (a) What are the reflection coefficient at the load and VSWR if the load is a 40-D resistor? (b) Determine the maximum and minimum load resistance for VSWR = 1.5. (c) Calculate the reflection coefficient 3 em from the load, if ZR =(55+ jO) D (consider attenuation). Ans. (a) -0.073, 1.16; (b) 69.45 D, 30.89 D; (c) 0.0856/-216° 15.38. A 90-D, lossless, high-frequency, coaxial line, with £, = 2.1, operates at 150 MHz. Of interest is the sensitivity of the VSWR to small changes in terminating resistance. (a) Tabulate r R and VSWR against RR = (90 ± 2n) D, for integral values of n from 0 to 5. (b) If the specifications for an application limit the maximum VSWR to 1.025, find the maximum and minimum values of terminating resistance. Ans. (a) See Table 15-4. (b) 92.24 D, 87.81 D

15.39. For the transmission line of Problem 15.38, (a) find the phase velocity, wavelength, and the phase shift per meter. (b) If the terminating resistance is 100 D, find the input impedances for line lengths A/2, "A/4, and A/8. Ans. (a) 2.07x 10"m/s,1.38m, 4.55rad/m; (b) 100/-00D, 81/-00D, 90f-6°D 15.40. Use the Smith Chart to find (a) rR, (b) VSWR, and (c) y, for the following (ZR, R 0 )-pairs, in ohms: (100 + j150, 50); (28- j35, 70); 90/-300' 90); (120 /9f.f' 50); (0, 70); (50+ j5, 50). Ans. (a) 0.75/26S, 0.53/-121°,0.27/-90, 1.0/45°, -1.0, 0.05/90° (b) 7, 3.3, 1.75, co, co, 1.1 (c) 0.16-j0.225, 0.97+j1.23, 0.87+j0.52, -j0.415, co, 1-j0.1

15.41. Find (a) YR• (b) VSWR, and (c) YR (in mS) for the following (rR , R 0 )-pairs: (0.5/600, 50 D); (1/-SOO, 90 D); (0.1/!!_, 70 D); ( -0.6 /-300,50 D); (0.8 + j0.4, 70 D). Ans. (a) 0.44- j0.495, 0 + j0.84, 0.83 + jO, 2.0- jl.85, 0.06- j0.238 (b) 3.0, oo, 1.2, 4.0, 17 (c) 8.8- j9.9, 0 + j9.3, 11.9 + jO, 40- j37, 0.86- j3.4

Z70

TRANSMISSION LINES

[CHAP. 15

Table 1.5-4

0.0

92 88

0.0110 -0.0112

1.022 1.023

0.0217 -0.0227

1.044 1.046

0.0323 -0.0345

1.067 1.071

98 82

0.0426 -0.0465

1.089 1.098

100 80

0.0526 -0.0588

1.111 1.125

VSWR 1.0

15.42. A lossless high-frequency line 3 m long, with Ro = 50 Q and £, = 1. 9, is operated at 350 MHz. The VSWR on the line is 2.4 and the first voltage maximum is located 7 em from the load. Use the Smith Chart to find the load impedance, the reflection coefficient at the receiving end, the location of the first voltage minimum, and the input impedance. Ans. (40 + j39) Q, 0.42/81°, 22.6 em, (22.3- jll) Q 15.43. A 70-Q lossless line, with £, = 2.2, is 2.5 m long and operates at 625 MHz. The VSWR on the line is 1. 7 and the first voltage minimum is located 5 em from the load. Use the Smith Chart to find the load admittance, the reflection coefficient at x = 0, and the input admittance to the Ans. (10.4 + j5.4) mS, 0.27 /-68" , (16- j7. 9) mS line. 15.44. An air-dielectric line with R0 = 150 Q is terminated in a load of (150- j150) Q at the operating frequency, 75 MHz. (a) Use the Smith Chart to find the shortest length of line for which the input impedance is (150 + j150) n. (b) What are the VSWR on the line and the reflection coefficient at the load? What is the shortest length of line for which Zin = R + jO, and what is the value of R? Ans. (a) 1.3m; (b) 2.6, 0.47/-64°; (c) 64.8cm, 57C 15.45. Two lines are connected in parallel at the input to a 250-MHz source. Each line is 2m long and is terminated in a 70-C resistance. Line #1 has R 0 =50 C, £, = 1. 9, line #2 has Ro = 90 C, £, = 2.3. Use the Smith Chart to find the input impedance to the parallel combination. (Be careful in combining the two input impedances/admittances.) Ans. (24.6 + j3.5) Q 15.46. A lossless 50-Q line, with a phase velocty 2.5 x lOS m/s, is 105 em long and is terminated in a load YR = (20- j16) mS at 500 MHz. A short-circuited line, 17.85 em long and also having R 0 = 50 Q, is connected across Y,. as shown in Fig. 15.44. Use the Smith Chart to find the VSWR on the main line and the input impedance. What is the equivalent capacitance (or inductance) of the short-circuited line? Ans. 1.0, 50 C; 5.1 pF

I·

,, = ws

-----~·*r--

Fig. 1.5-44

17.S5cm

-1

CHAP. 15)

TRANSMISSION LINES

271

15.47. In the line of Problem 15.46 the short-circuit on line two is inadvertently changed to an open circuit. Use the Smith chart to find the VSWR on the main line and the input impedance. Ans. 6.2, (26.5 + j72.5) !J 15.48. A parallel-wire line of the type in Problem 15.1 is operated at 20 MHz to supply a resistive load of 500 !J through a quarter-wave matching transformer connected at the load. Neglect losses on the main line and the transformer section. (a) For the transformer calculate the length of line and characteristic impedance required for matching. (b) If the same sized wire is used for the main line and the transformer, find the separation d (in inches) required for matching. Ans. (a) fT=3.75m, RoT=547.7!J; (b) dT=7.77in.

15.49. A lossless 70-Q line is terminated in ZR 60.3/30. 7o Q at 280 MHz. Use the Smith Chart to find the value of the inductance or capacitance to connect in parallel with the load for minimum VSWR on the line. What length (in centimeter) of shorted line would give the desired value, if ÂŁ, = 2.1? Ans. C = 4.8 pF (VSWRm;n = 1.0); 24.8 em 15.50. A 200-!J air-dielectric line is terminated in YR = (3.3- jl.O) mS at 200 MHz. (a) Find the VSWR and the position nearest the load where the real part of the normalized admittance is unity, using the Smith Chart. (b) What value of susceptance (in millisiemes) should be connected at this point to make VSWR= 1 on the line? Ans. (a) 1.65, 29.3cm; (b) -2.55 (inductive) 15.51. Two 72-!J resistive loads are connected in parallel as the termination for a 120-!J air-dielectric lossless line at 150 MHz. Find the location nearest the load and the length (both in centimeters) of a single Ans. 16 em, 78.6 em shorted parallel-connected stub to match the line for a VSWR = 1.0. 15.52. (a) In Problem 15.51, if the maximum length of the adjustable shorted stub is 50 em, can the load be matched to the line? (b) If the answer to (a) is Yes, find the position and length of stub for the matched condition. (c) If the stub were left at its original position and set to the 50-cm maximum, what would be the VSWR on the line? Ans. (a) Yes; (b) 83.8cm, 21.4cm; (c) 3.3. 15.53 A 90-Q lossless line with ÂŁ, = 1.8 operates at 280 MHz and is matched to the termination with a single shorted stub that produces VSWR = 1.0. The stub is located 15.8 em from the load and is of length 10 em. Find the ohmic value of the terminating impedance. (Hint: Remove y. , find the VSWR, and move back toward the load.) Ans. 201.2/26.6Â°!J 15.54. A 50-!J air-dielectric lossless line has ZR = (25 - j30) !J at 120 MHz. An adjustable shorted stub is located 45 em from the load (fixed in position). Find the length of stub for the best match on the line. What is the minimum VSWR on the line? Ans. 96 em, 1.08 15.55. (a) An air-dielectric lossless 70-Q line is matched at 200MHz to a 140-!J load by means of a shorted parallel stub. Find the position nearest the load and the length of the stub (both in centimeter) for the matched condition. (b) The line is now used at 220 MHz without changing the position or length of the stub. Find the VSWR on the main line at the new frequency. Ans. (a) 22.65 em, 22.80 em; (b) 1.22 15.56. The termination on a 90-Q lossless air-dielectric line is ZR = (270 + jO) !J at 600 MHz. A doublestub 0.25A tuner is connected with the first stub at the load for matching. Find the lengths for the shorted stubs (both solutions). Which solution is preferred? Ans. 9 em and 4.95 em (preferred); otherwise, 16 em and 20.05 em. 15.57. A 3A/8 tuner is connected at the load to match ZR = (50- j50) !J at 400 MHz to a 50-Q lossless air-dielectric line. Find the lengths of the shorted stubs for both solutions and indicate the preferred solution. Ans. 4.5 em and 4.3 em (preferred); otherwise, 12.1 em and 26.1 em. 15.58. A 50-Q air-dielectric line, with a load YR = (0.024- j0.02) S at 470 MHz, has a A/4 tuner with the first stub located 7 em from the load. Find both solutions for the lengths of the shorted stubs to match the load to the line. Indicate the preferred solution. Ans. 14.5 em and 7.6 em (preferred); otherwise, 23.1 em and 24.3 em.

272

TRANSMISSION LINES

(CHAP. 15

15.59. A two-stub 3A/8 tuner is constructed of 70-!J line with £, = 2.0 for use at 272 MHz on the same type of line and a certain load. For the matched condition the shorted stub at the load is 4. 76 em long and the other shorted stub is 4.60 em long. Find the ohmic impedance of the load at the operating Ans. (58.1 - j58.1) !J frequency. 15.60. A 225-Q resistive load is matched to a 90-!J air-dielectric line at 300 MHz. The matching is via two shorted stubs separated by 30 em, with the first stub connected at the load. Find the lengths of the stubs (both solutions) and indicate the preferred solution. Ans. 13.6 em and 8.5 em (preferred); otherwise, 28.1 em and 37.5 em. 15.61. A 50-!J slotted line is used to determine the load impedance at 750 MHz on a lossless 50-Q line. When the line is terminated in a short circuit, the· high VSWR has adjacent minima at 30 em and 10 em (the scale has the low numbers on the load side). With ZR connected the VSWR is 3.2, a minimum is located at 13.2 em and the adjacent maximum is at 23.2 em. Find the ohmic value of ZR at the operating frequency. Ans. 31.24/-50.2o!J 15.62. Find the load impedance and the operating frequency for a 90-!J air-dielectric system that has the following slotted-line measurements: With load:

VSWR= 1.6 and a voltage minimum at to em (high numbers on load side).

With short circuit:

Ans.

VSWR > 100, minimum at 40 em, maximum at 10 em.

144 !J; 250 MHz

15.63. A 50-!J slotted line is used to measure the load impedance at 625 MHz on a 50-!J lossless coaxial line. Adjacent voltage minima are found at 10 mm and 250 mm (high numbers at the load side) when the termination is a short circuit. With the load connected, VSWR > 100 and a minimum occurs at 172.7mm. Find the ohmic value of the load impedance. Ans. 80!J (capacitative) 15.64. A tuner is connected at the load to match the load to a 50-!J lossless air-dielectric line at 517 MHz. To check the quality of the matching a slotted line is inserted in the system. With the tuner and load connected, VSWR = 1.15, with a Vmin at 253.4 mm. When the tuner and load are both removed and replaced by a short circuit, adjacent minima are found at 40 mm and 330 mm (low numbers on the scale are at load side). Find the residual normalized admittance on the line that results from the "best match." Ans. 0. 98 - j0.14 15.65. A 60 m-long, lossless, 50-!J coaxial cable, with a phase velocity of 2 x lOS m/s, is terminated with a short circuit. The line is connected at t = 0 to a 30-V de source having internal resistance 25 !J. Plot the sending-end voltage from t = 0 up to the time when the voltage drops below 0.1 V. Ans. See Fig. 15-45.

,.,. v !lit-------.

6.67

"~n-----,J~.t.:--____)11!~=2:.2:2=~,~;::::::=0=.7=4=~2~4=~0,;.2;,5=:::::::-:Jn;,--~0~.~~-- ~s '·

Fig. 1..5-45

15.66. A 90-!J lossless line, with £, = 2. 78, is connected at t = 0 to a 70-V de source with an internal resistance of 120 !J. If the line is 135m long, find the time when the open-circuit voltage at the receiving end is 97% of the steady-state value. When is the voltage 99.95% of the steady-state Ans. 2.25ps, 5.25ps value? 15.67. A pulse generator with internal resistance 150 Q produces a 20-Jls pulse with an open-circuit amplitude

273

TRANSMISSION LINES

CHAP. 15)

of +8 V. The generator is connected to a 50-!J, lossless, 200-J~s delay line that is terminated in a 100-!J resistance. If the period of the recurring pulses is 4 ms, sketch the voltage at the input to the delay line Ans. See Fig. 15-46. from pulse onset at t = 0 to t = 1.4 ms

v,.

v 20

1.0

Fig. 15-46 15.68. Sketch v5 and vR versus time, from t = o+ to t = 300 JlS, when a 70-!J, lossless, 50-Jls delay line is terminated with a 30-Q resistor and driven by a pulse generator. The generator has an internal resistance of 70 !J and produces a 2-J~s pulse with a peak open-circuit voltage of + 10 V at a repetition Ans. See Fig. 15-47. rate of 1000 pulses per second.

.... v

)II

-2.0

Fig. 1.5-47

d=O

d =I

Fig. 15-48 15.70. Sketch Vs and VR versus time, from t = o+ to t = 30 JlS, when a 220-m-long, 90-!J, lossless line, with £, = 3.65, is terminated in a 50-Q resistance and driven by a pulse generator. The generator has an internal resistance of 90 !J and produces a 5-J~S pulse with open-circuit peak value + 140 V, at a repetition rate of 100 pulses per second. Ans. See Fig. 15-49.

,·,.Vb

l"R·V~

'"L___J_L__.LI _ 11 2 N "-fl U1 11

,....----+.:'-·

___JL__,_J.,I

211

341 1. f.LS

II I 4

Fig. 15-49

I 10

-"'

1. JLS

Chapter 16 Waveguides (by Milton L Kult)

16.1 INTRODUCTION The electromagnetic waves of Chapter 14 can be guided in a given direction of propagation using several different methods. For instance, the two-conductor transmission line, supporting what are essentially plane waves at megahertz frequencies, was considered in Chapter 15. The present chapter is restricted to single-conductor (hollow-pipe) waveguides, of rectangular or circular cross section, which operate in the gigahertz (microwave) range. These devices too support "plane waves"-in the sense that the wavefronts are planes perpendicular to the direction of propagation. However, the boundary conditions at the inner surface of the pipe force the fields to vary over a wavefront.

16.2 TRANSVERSE AND AXIAL FIELDS The waveguide is positioned with the longitudinal direction along the z axis. In general the guide walls have ac = ~ (perfect conductor) and the dielectric-filled hollow has a= 0 (perfect dielectric), tJ- = tJ-olin and € = € 0 €,. It is further supposed that p = 0 (no free charge) in the dielectric. The dimensions for the cross section are inside dimensions. In Fig. 16-l(a) the a X b rectangular waveguide is shown in a cartesian coordinate system, Fig. 16-l(b) shows the circular or cylindrical waveguide of radius a in a cylindrical coordinate system.

Fig. 16-1

As in Chapter 14 the time dependence e 1w' will be assumed for the electromagnetic field in the dielectric core; this time-factor will be suppressed everywhere in the analysis (as in phasor notation). Thus we have the following expressions for the field vector F (which stands for either E or H), assuming wave propagation in the + z direction.

rectangular coordinates

F = F(x, y)e-ikz

where

F(x, y) = F_.(x, y)ax + Fy(x, y)ay + f;;(x, y)az

== F 7 (x, y) + F;;(x, y )az cylindrical coordinates

F = F(r, cJ> )e-il"

where

F(r, cJ>) = F,.(r, tP )a,+ Fq,(r, tP )aq, + F;;(r, == F T(r, tP) + f'x(r, tP )az

274

tP )az

CHAP. 16]

WAVEGUIDES

275

Because the dielectric is lossless (a= 0), the wave propagates without attenuation; hence the wave number k = 2tr/ A (in rad/m) is constrained to be real and positive. Note:

In the other chapters of this book, unbounded dielectric media are considered, for which the wave number, notated {3, depends on frequency and dielectric properties only. However, as will soon appear, the wave number in a bounded dielectric depends additionally on the geometry of the boundary. This important distinction is emphasized by the employment of a new symbol, k, in the present chapter.

The reason for decomposing the field vector into a transverse vector component F r and an axial vector component F.:az is two-fold. On the one hand, the boundary conditions apply to Er and Hr alone (see Problems 16.1 and 16.2). On the other hand, as will now be shown, the complete E and H fields in the waveguide are known once either cartesian component Dz or Hz is known. Transverse Components from Axial Components. Assume a rectangular coordinate system. Maxwell's equation (2) of Section 14.2 yields the three scalar equations

.kE aEz -jwpHX=} }' + ay

(la)

aEz -jwpH,.= -jkEx- ax

(lb)

-jwpll = aE.,. _ aEx z ax ay

(lc)

Maxwell's equation (1) of Section 14.2, with

a=O,

gives three additional scalar equations:

. E x=J.kH.,.+aHz }WE ay

(2a)

. E .kH aHz }WE.,.=-] X - ax

(2b)

. aH.,. aHx }WEE=--z ax ay

(2c)

Now eliminate Hx between (la) and (2b), and H.,. between (lb) and (2a), to obtain

E = - jk aEz + jwp aHZ y k~ay k~ ax E = - jk aEz- jwp a HZ x k~ax k~ ay

(3b)

in which k~ == w 2 p.E - k 2. The parameter kc (also in rad/m) functions as a critical wave number; see Problem 16.3. Finally, slide (3b) and (3a) back into (2a) and (2b), to find

H _ jk aHZ jwE aEZ }' - - k~ ay - k~ ax

(3c)

H _ _ ik aHz jwEaEz X- k~ ax + k~ ay

(3d)

By exciting the waveguide in suitable fashion it is possible to force either Ez or Hz (but not both) to vanish identically. The nonvanishing axial component will then determine all other components via Equations (3). See Problems 16.4 and 16.5 for the analogous results in cylindrical coordinates.

1:16

WAVEGUIDES

[CHAP. 16

16.3 TE AND TM MODES; WAVE IMPEDANCES The two types of waves found in Section 16.2 are referred to as transverse electric (TE) or transverse magnetic (TM) waves, according as Ez 0 or Hz= 0. When carrying such waves, the guide is said to operate in a TE or TM mode. For any transverse electromagnetic wave, the wave impedance (in ohms) is defined as

1J =

(compare Chapter 14).

IErl IHrl

(4)

For a waveguide in aTE mode, (la) and (lb) imply 2

1Erl2 = 1Ex12 + 1Eyl2 = (kwp)\1Hyl 2 + 1Hxl2 ) = ( wp) 1Hrl2 TE

kTE

(5)

1JTE=kTE

Because ( 4) only involves lengths of two-dimensional vectors, 1J must be independent of the coordinate system. Problem 16.6 confirms the value of 1JTE by recalculating it in cylindrical coordinates. In Problem 16.7 it is shown (using rectangular coordinates) that k™

(6)

1JTM=-(I)£

16.4 DETERMINATION OF mE AXIAL FIELDS All that remains for a complete description of the TE and TM modes is the determination of the respective axial fields: f'z =Hz forTE; ~ = Ez for TM. The good word is that ~e-ikz, being a cartesian component ofF (in either rectangular or cylindrical coordinates), must satisfy the scalar wave equation found in Section 14.2, (7)

together with appropriate boundary conditions which are inferred from the boundary conditions on the components of F T • [Warning: Transverse components such as H<t>e -ikz are not cartesian components and do not obey a scalar wave equation.] ExpUdt Solutions for TE Modes of a Rectangular Guide. The wave equation ( 7) becomes

Solving by separation of variables (Section

where, as previously defined, 8.7),

(8)

where k~ + k~ = k;TE . The separation constants kx and ky are determined by the boundary conditions (review Problem 8.19). Consider first the x-conditions Ey(O, y) = Ey(a, y) = 0; in view of (Ja) and Ez = 0 these translate into

-aHZ ax Applying these conditions to (8) gives

x=O

-3Hz ax

-o

x=a

Bx = 0 and or

k x =m:n (m = 0, 1, 2, ... ) a

277

WAVEGUIDES

CHAP. 16)

By symmetry, the boundary conditions in y force

By = 0 and

n:n

ky=b (n=O,l,2, ... ) Each pair of nonnegative integers (m, n }-with the exception of (0, 0) which gives a trivial solution-identifies a distinct TE mode, indicated as TE, . This mode has the axial field

m:nx n:ny Hzmn(x, y) = Hmn cos-;;-cosb from which the transverse field is obtained through (3).

(9)

The critical wave number forTE, is

in terms of which the wave number and wave impedance for TEmn are (11)

1JTEmn =

(12)

V(I) 2PÂŁ - k2cTEmn

See Problem 16.9 for the TMmn modes of a rectangular waveguide; it is shown there that kcTMmn = kcTEmn . Consequently, the subscripts TE and TM can be dropped from all modal parameters of rectangular guides save the wave impedance. This is not the case with cylindrical guides; see Problem 16.12.

16.5 MODE CUTOFF FREQUENCIES In practice one deals with frequencies, not wave numbers; it is then desirable to replace the concept of critical wave number (kc) by one of cutoff frequency (fc). This is accomplished in the definition (see Problem 16.3) Uo

/c =-2 kc = :n

~ ,..--: kc

In terms of the cutoff frequency fc and the operating frequency (12) become

lcmn = kmn

~o ~(:r + (~f 2

yp- f~mn

(13)

2:n v p.ÂŁ

= w/2:n > fc

(rectangular waveguide)

(10), (11), and

(lObis) (11 bis) (12bis)

where Ao = u0 /f is the wavelength of an imaginary uniform plane wave at the operating frequency and where 1Jo = Vii/i. is the plane-wave impedance of the lossless dielectric. The second form of ( 11 bis) exhibits the relation between the operating wavelength A.o and the actual guide wavelength A,. For TMmn waves, (12bis) is replaced by (see (6)] 1JTMmn = 1Jo

~,--1--(~.-c7--=-f

(14)

278

WAVEGUIDES

[CHAP. 16

The phase velocity of a TEmn or TMmn wave is given by (15)

If (lObis) is replaced by a similar expression involving a Bessel function (see Problems 16.10 and

16.11), all formulas remain valid for cylindrical guides. The meaning of cutoff is made particularly clear in (15). As the operating frequency drops down to the cutoff frequency, the velocity becomes infinite-which is characteristic, not of wave propagation, but of diffusion (instantaneous spread of exponentially small disturbances).

16.6 DOMINANT MODE The dominant mode of any waveguide is that of lowest cutoff frequency. Now, for a rectangular guide, the coordinate system may always be oriented to make a 2:: b. Since (Problem 16.9)

for either TE or TM, but neither m nor n can vanish in TM, the dominant mode of a rectangular guide is invariably TE 10 , with Uo

fcw=

From (9),

1Jto =

Ezto =0,

Aao

1Jo

and the equations of Section 16.2:

.nx HziO = H 10 COS-

Exto=O

Âˇ(2a) SID--;; . .nx

Hxto =1 Aw Hao

(16)

H, 10 =0 For H 10 real, the three nonzero field components have the time-domain expressions Hzao= Hwcos (';;)cos (rot- k 10z) Hxto =

-(~)Hao sin(';;) sin (rot- k 10z)

(17)

E,to= 1Jo(~)H10 sin (':)sin (rot- k 10z) Plots of the dominant-mode fields (17) at t = 0 are given in Figs. 16-2 and 16-3. Both IE,I and IHxl vary as sin (.nx/a). This is indicated in Fig. 16-2 by drawing the lines of E close together near x = a/2 and far apart near x = 0 and x =a. The lines of H are shown evenly spaced because there is no variation withy. This same line-density convention is used to indicate the local value of lEI= IE,I in Fig. 16-3(a) and of

IHI=VH~+H; in Fig. 16-3(b). Observe that the lines of Hare closed curves (divH=O); the H field may be considered as circulating about the perpendicular displacement current density Jn (Section 12.1).

CHAP. 16)

WAVEGUIDES

279

hr-------------------. ----------------

--------

---------

---------------------

u L - - - - - - - - - - - -a- - ! -.\ u

Fig. 16-2. Tnnsverse cross sec:tion

(-k1.z = n/2)

-~.14

-4

u'';!..'-==~-~lf----:~::-:::-:JÂą-=:-::=--,-t-1-,_--::::-==--

......

:=!!fl-!!L- -1- -r[,=ll ~ ~=ll.0

-- - ,...... \

,. f ~Ill

-4-

~Ill 2

\ ::I

I II I I II I

/ /

__ ., //

J~ln

---- '

I , ...._ \ I / /". \ I I I ~- - - "- \ I 1 1 / I I I 1 11 I I I

' , __ .--~

\ \..

I II I I II I

'-.

I \ \'-

'-.....:.::~~~-/ ''--=-~

-4-

(a)Cx=af2

(b) G _,. = const.

Fig. 16-3. Longitudinal cross sections

Figure 16-4 illustrates how the TE 10 mode can be initiated in a rectangular waveguide by inserting a probe halfway across the top wall (y = b, x = a/2), at a distance z = A10/4 from the end of the guide. Higher-order modes are present in the vicinity of the probe, but they will not propagate if the frequency-size condition is selected correctly. See Problem 16.13 for the dominant mode of a cylindrical waveguide. C"OAX

COAX

o~----------------

t--~u/4--l

Fig. 16-4

16.7 POWER TRANSMITIED IN A LOSSLESS WAVEGUIDE The time-average power transmitted in the + z direction is calculated by integration of the z component of the complex Poynting vector over a transverse cross section of the guide (cf. Section 14.12):

pz =

i II Re

cross section

Er X

u;.. 8z dS

(18)

WAVEGUIDES

280

[CHAP. 16

Substituting the field components from (16) and writing A 8 = ab, mode of a lossless rectangular waveguide:

- = Pzw

(2a) (Aw 2a)A

2 'lo 4IHwl Ao

'lo = 4IH1ol

f ) A 8 ( fcw

we obtain for the dominant

~1 - (/cf 10)

(19)

As expected for a lossless system, Pz 1o is independent of z; moreover, the power is proportional to the square of the field amplitude and to the cross-sectional area of the guide. Since the excitation of a guide is commonly specified through the electric field amplitude,

IEaol =

'lo(~) IHwl

it is useful to rewrite (19) as

Pzw = IE1ol

A8 4'1o

1 _ (fcto)

= IE 1ol

A8 4'1TEto

(W)

(19bis)

Relations similar to (19) and (19bis) exist for the higher-order modes. For the lossless cylindrical guide, see Problem 16.15.

16.8 POWER DISSIPATION IN A LOSSY WAVEGUIDE When the conductivity of the guide dielectric is nonzero (but small) and/or the conductivity of the guide walls is noninfinite, the wave in any propagating mode will be attenuated, and transmitted power will decrease exponentially with z. An approximate treatment of these dielectric and wall losses is possible on the assumptions that the two types may be analyzed separately and that the fields which interact with the walls are those which would be present if the dielectric were lossless. To keep the mathematics as simple as possible, only the TE 10 mode of a rectangular waveguide will be treated.

Dieledrk Loss. Maxwell's equations (1)-( 4) of Section 14.2 are unchanged if a= ad, the dielectric conductivity, is replaced by zero and E = Ed , the dielectric permittivity, is replaced by its complex permittivity

Therefore, the field equations for the lossy dielectric may be obtained from those for the lossless dielectric by formal substitution of £ for Ed . In particular, the z dependence of the field vectors in the lossy TE 10 mode is exp(-rwz), where, by (11), 2 2 rw= jklo(c) = rvw pd€- k;w= fV(ro pdEd- k;w)- jwpdad

= .13 (t-jwp.dad) I

p210

112

= (wp.dad) 2/3 10

+ J·p 1o

(20)

In (20), f3w

=VW 21J.dEd- k;1o = kw( Ed)=~ Vl - (fcao//)2

(21)

and the binomial approximation presumes that ad and w are small enough to make wpdad <( {3~0 • To this order of approximation, then, the wave number-the imaginary part of y 10-in the lossy dielectric equals the wave number in the perfect dielectric; while the attenuation factor, ad=

CHAP. 16)

Re y 10 ,

WAVEGUIDES

281

which governs the power loss in the dielectric, is given by

= wpdad _ (V#ldl ÂŁd) ad

a d

_ 1

2fJw - 2Vl- (fcw//)2-

z1JTEto d

(22)

(Np/m)

Wall Loss. The attenuation factor aw governing the wall loss may be determined indirectly, as follows. Because power varies as the square of the field strength, the time-average transmitted power in the TE 10 mode must obey -2a,.z P. ( ) _ P.-

z -

where the entrance power z-length is thus

Pzto

ztoe

is as given in (19).

aw=

whence

Paoss( Z) 2Pav(Z)

The power dissipated in the walls per unit

Paoss(O)

=--2Pzl0

(23)

All that remains is to calculate P1055(0), the power flowing into the first 1m of wall inner surface. Now, it is not hard to show that, at a wall surface, tangential H-which by hypothesis can be obtained from (16)-sets up a Poynting vector, of time-average magnitude (24) and directed into the wall. Here, Rs = Re 1Jw = Vnfp.wl aw (Section 14.6) is the surface resistance (Q) of the wall material at the given frequency f Integrating the appropriate expression (24) over the first I m of each wall surface and adding the results yields finally Paoss(O) = Rs IHwl

2[b + 2a (f lfcw)2]

(25)

(W /m)

From (23), (19), and (25),

aw = RsclO ( 1Jo

Jt)

"V~

a+ 2b(fcwlff (Np/m) abV1- (fcwlf> 2

in which Rsc1o is the surface resistance at the cutoff frequency of TE 10 and plane-wave impedance of the (lossless) dielectric. Combined Losses. The total attenuation factor is dB/m, see Problem 14.7.

(26) 1Jo = VI'd/ ÂŁd

is the

a 101 = aw + ad . To convert from Np/m to the more usual

Solved Problems 16.1.

Give the boundary conditions on E and H at each perfectly conducting wall of the waveguide of Fig. 16-1(a). At a perfect conductor tangential E and normal H must vanish.

Therefore:

top waD

E.(x, b)= Ex(x, b) =0 and

Hy(x,b)=O

left waD

E.(O, y) = Ey(O, y) = 0 and

Hx(O,y)=O

right waD

Ez(a, y) = Ey(a, y) = 0 and

Hx(a,y)=O

and

Hy(X, 0) =0

bottom waD

E.(x, 0) = Ex(x, 0) = 0

282 16.2.

WAVEGUIDES

[CHAP. 16

Repeat Problem 16.1 for the guide of Fig. 16-1(b). At the single cylindrical wall, E,(a, ~)=E.( a, ~) = 0 and

16.3.

H,(a, ~) = 0

What is "critical" about the number kc? For propagation through a lossless dielectric, the wave number k must be real. But k

=Vail'£- k; = Vk~- k;

where k 0 is the wave number of a uniform plane wave in the unbounded dielectric at the given w. Thus kc is a critical wave number in the sense that a guided wave's same-frequency "twin" must have a wave number exceeding kc . Stated otherwise, the frequency f of the guided wave must exceed the quantity (u 0/2~r)k.,, where u 0 = 1/Vii£ is the wave velocity in the unbounded dielectric.

16.4.

Express Maxwell's equations (1) and (2) of Section 14.2 in scalar form in a cylindrical coordinate system. For the curl in cylindrical coordinates, see the Appendix. Equation ( 1) yields (a= 0): . 1 aH. }Wt=E, =--a + jkH• r

jwe= E

(i)

aH.

. H

(ii)

-Jk , - -

1a 1 aH, jwe=E. = - - (rH ) - - -

(iii)

. 1 aE. -JW"H =--+J"kE ,... r r a4J •

(iv)

rar

•

r a4J

Equation (2) yields:

(v) (vi)

16.5.

Using the equations of Problem 16.4, find all cylindrical field components in terms of Ez and Hz. From (i) and (v), with kc as previously defined, E = r

_iWIJ! aH._jk aE, k~

r a4J

(1)

From (ii) and (iv), (2)

From ( 1) and (i), H

•

jwe= aE. jk 1 aH. =------

k; ar k; r a4J

(3)

From (2) and (ii), (4)

16.6.

283

WAVEGUIDES

CHAP. 16)

Calculate 7JTE from the field components in cylindrical coordinates. With E, so 0,

(iv) and (v) of Problem 16.4 yield 2

2 k IErl k ) IE 12 + (k --"!!: ) IE,l =--"!!: ( --"!!: wp • wp wp

whence

16.7.

Calculate 1JTM from the field components in rectangular coordinates. With H."" 0,

(.?il) and (2b) give 2

IE.. l

+ lE,.l 2 =

(::r(IH,l

IErl

whence

16.8.

+ IH..I 2 )

or k™

'1TM"" IHrl =

W€

Show that E and H are mutually perpendicular in any TE or TM wave (as with ordinary plane waves). For either type of wave E.. = f]H,.

and

E,. = -f]H_.;

therefore, since '1 is real,

Er • Hr = Re (ExH: + E,.H;) = Re ( f]H,.H: - f]H.. H;) = '1 Re (H,.H: - H_.H;) = 0 Because E.H: also vanishes,

16.9.

E • H = 0.

Obtain the analogues of ( 9)-( 12) for TMmn . Analogous to (8),

+ D,. sin k,.y)

E,(x, y) =(C.. coskxx +D.. sin k..x)(C,. cosk,.y

where But now the boundary conditions, E.(O, y) = E,(a, y) = 0

E.(x, 0) = E,(x, b)= 0

and

require that C_.=O

k ... =~ a

C,.=O

="Jr b

where m, n = 1, 2, 3, . . . . Note that neither m nor n is zero in a TM mode. The required formulas are . mJrX. nlr)' E """ (X, y ) = E'"" SID -a- SID b

(1)

(2) kTMmn = kTEmn

(3)

k™ '1TMmn=-

(4)

W€

16.10. Determine. the TM modes of a lossless cylindrical waveguide.

284

WAVEGUIDES

[CHAP. 16

The Laplacian in cylindrical coordinates is given in the Appendix; the wave equation ( 7) for E,(r, 4>) becomes erE. 1 aE. 1 erE. 2 -a 2 +--a +2-a z +kc™E• =0 r r r r 4>

subject to the boundary conditions (i) E,(r, 4> + 2.n") = E,(r, 4>); (iii) E,(a, t/J) = 0. Following Section 8.8, one solves by separation of variables to find

(ii)

E,(O, 4>)

bounded;

E.,,p(r, tP) = E.,pf.,(kc™"Pr) cos nt/J

where n = 0, 1, 2,. . . and where x.,P"" kc™"Pa is the pth of J.,(x) = 0. (The first few such roots are listed in Table 16-1.)

( 1)

positive

root

= 1, 2, ... )

Table 16-1. Roots x.., of J'.,(x) = 0

p=l p=2 p=3

n=O

n=l

n=2

n=3

2.405 5.520 8.645

3.832 7.016 10.173

5.136 8.417 11.620

6.380 9.761 12.015

The expression ( 1), together with H, 0, determines all transverse field components in TM via Problem 16.5. The cutoff frequency of TM.,P is given by Uo

(2)

fcTM..p= -x.,P 21fQ

When (2) is used, all rectangular-guide formulas also apply to cylindrical guides; for example, '1TM..p=

'1o~l- (~;r

<J>

16.11. Determine the TE modes of a lossless cylindrical waveguide.

•

In aTE mode the axial field H,(r, t/J) obeys the wave equation and the conditions (i) and (ii) of Problem 16.10. As a consequence of (2) of Problem 16.5, condition (iii) must be replaced by ... ,. (Ill

aH. ar

r-a

The solution by separation is therefore: (1)

H,.,P(r, tP) = H.,pl.,(kc-rE"Pr) cos nt/J

where n = 0, 1, 2, . . . and where of J~(x) = 0. See Table 16-2.

x~P

=kc-rE>op

Table 16-:Z. Roots x:O, of

p=1 p=2 p=3

the

pth

J'~(x) = 0

n=O

n=l

n=2

n=3

3.832 7.016 10.173

1.841 5.331 8.536

3.054 6.706 9.969

4.201 8.015 11.346

positive

root

(p = 1, 2, ... )

WAVEGUIDES

CHAP. 16]

285

The analogues of (2) and (3) of Problem 16.10 are: /.cTEnp

f}TEnp

2.na

(2)

Xnp

--,-----=1- (Aox~p)2 f}o

2.na

16.12. Discuss the relative magnitudes of fcTEnp and fcTMnp

For each fixed n, the zeros Xnp of ln(x) and the stationary points x~-where J"(x) is a maximum or a minimum-alternate along the x axis; this sine-wave-like behavior is clear in Fig. 8-3(a). For n > 0, the function starts at 0, and the first stationary point precedes the first positive zero; thus, x~ < x"P , whence and For n = 0,

/cTEnp

< /cTMrtp

the function starts at a maximum, and the ordering is reversed: and

/cTF.Op

> /cTMOp

16.13. (a) What is the dominant mode of a lossless cylindrical waveguide? modes in order of increasing cutoff frequency.

(b) List the first five

By Problem 16.12, the dominant mode is either TMo1 or the TE" 1 with the lowest cutoff. Tables 16-1 and 16-2 indicate (and analysis establishes) that the winner is TE 11 • (b) TE11 , TMo 1 , TE21 , TE01 , and TM 11 (a tie). [The first column of Table 16-2 is identical to the second column of Table 16-1 because J~(x) = -J1(x).] (a)

16.14. Obtain the transverse fields for the TE 11 (dominant) mode of a cylindrical waveguide.

For m = p = 1,

Equation (I) of Problem 16.11, E.= 0,

and (J)-(4) of Problem 16.5 yield (1) (2)

(3) (4)

16.15. Calculate the time-average power transmission in the TEu mode of a lossless cylindrical guide. Follow Section 16.7, with the transverse fields as given by Problem 16.14. ~ ET X H~ • 8,

= ~(£, 11 H:u- E•nH:n) =

WIJ~~:~~~~I2 {[ Jl~v)r sin2 q, + [J;(v)]2 cos2 q,}

which the integration variable v = kcTF.ur has been introduced.

(1)

In the integration of (1) over the

286

WAVEGUIDES

[CHAP. 16

the sin2 and cos2 both integrate to .1r; therefore,

cross section 0 :5; q, :5; 2Jr and 0 :5; v :5;x;,,

fil {[J;(v)f + [ J,~v)f}v dv

P.u = .1rWIJ~!:.I,Hu12

(2)

There is a general rule for evaluating an integral like the one in (2): Go back to the ordinary differential equation arising from the separation of variables. In this case that equation is (see Section 8.8)

(

J~+;;J:+

1 1- v2

)1,=0

(3)

Thus, using integration by parts, (3), and the end conditions J,(O) = J ;(x;,) = 0,

f. ;.

2 2 [(J :} + (J,/v) )v dv

,..il [J; + (J~ )]2v dv- .

J..

r· [J; r + (~)

(J~+~J;- :

d(J~)

d(~) -J~(x;,)

=HvJ;+J,)21~i•-

we have:

xj,

[

vzJ;+~

1,) dv

-J~(x;,)

=-~J~(x;,)- f'v 2 (J;+~)(-J1 )dv

Substituting this result in (2), and replacing k-."Eu and k,TE 11 by their respective expressions in x;, = (2Jra/u0 )fcTE 11 , one finds after some algebra:

fJo

P.u = 4IHul A 11 in which A« =

.1ra

(

f )

2 2 [(x;,) -1 V1- (fcTEu) - (x;,)2 J ,(xu) 1

.fcTEu

]

(4)

is the cross-sectional area.

16.16. Compare the rectangular and cylindrical waveguides as power transmitters when each operates in its dominant mode. The two power formulas, (19) of Section 16.7 and (4) of Problem 16.15, show identical dependence on H-amplitude, cross-sectional area, and normalized frequency. The only difference lies in a geometrical factor, which has the value 1.0 for the rectangular guide and the value (1 841) 2 - 1 . (0.5814) 2 = 0.239 (1.841) 2 for the cylindrical guide.

16.17. (a) Define the notion of cutoff wavelength. (b) Is the cutoff wavelength an upper limit on the guide wavelength, just as the cutoff frequency is a lower limit on the guide frequency? (a)

The cutoff wavelength J.., is the wavelength of an unguided plane wave whose frequency is the cutoff frequency; i.e., J..cfc = U 0 •

(b)

No; in fact, the formula ),

VF-/~mn

shows that an (m, n) mode can propagate with any guide wavelength greater than J...

22,7

WAVEGUIDES

CHAP. 16)

16.18. A lossless air-dielectric waveguide for an S-band radar has inside dimensions a = 7.214cm and b = 3.404cm. For the TM 11 mode propagating at an operating frequency that is 1.1 times the cutoff frequency of the mode, calculate (a) critical wave number, (b) cutoff frequency, (c) operating frequency, (d) propagation constant, (e) cutoff wavelength, (f) operating wavelength, (g) guide wavelength, (h) phase velocity, (i) wave impedance. (a) By (10), kc 11 = V(.n/0.07214)2 + (.n/0.03404)2 = 102.05 rad/m. (b) By (13), feu= [(3 X 10S)/2.n))(l02.05) = 4.87 GHz. (c) I= l.l.fc 11 = 5.36 GHz. (d) By (11 bis), Yu

c.:-=;:-;---:-:-::::::::. 1 =iku = j 3 X2.nlOS V(5.36f(4.87) (10~ = j46.8 m2

(e) len = U0 /fc11 = (3 X lOS)/(4.87 X 109 ) = 6.16 em. (/) Ao = uo/f = (3 X lOS)/(5.36 X 109 ) = 5.60 em. (g) A11 =2.n/k 11 =2.n/46.8= 13.4em. (h)

(i)

By ( 15), u 11 = (0.134)(5.36 X 109 ) = 7.18 X lOS m/s. For air, = l20.n !J and (14) gives

'lo

'1TMII =

120.n~l- (/1

= 157.5 !J

16.19. A lossless, air-dielectric cylindrical waveguide, of inside diameter 3 em, is operated at 14 GHz. For the TMn mode propagating in the +z direction, find the cutoff frequency, guide wavelength, and wave impedance. By (2) of Problem 16.10, along with Table 16-1, U0 3 X lOS fcTMII = 2.1ta X 11 = .1t(3 X 10 2 ) (3.832) = 12.2 GHz

Then, by (11 bis) and (14), A II -

vr- ,~TMII

3xlOS =4.36em vU4) - (12.2)2 (109 ) 2

'lTMu = 'loV_I1- ('c-TMU) 1

.--=---:;2

= 120.nvI1- (12.2) M

= 185 !J

16.20. Find the inside diameter of a lossless air-dielectric cylindrical waveguide so that a TEn mode propagates at a frequency of 10 GHz, with the cutoff wavelength of the mode being 1.3 times the operating wavelength. The condition is leu = 1.3Ao,

Uo Uo -=1.3-, /.cTEU

fcTEu

= 1.I 3 = 7.6fll. GHz

But, by Problem 16.11, U0 , 0.3 fcTEu = Z.na Xu= nd (1.841) (GHz)

Equating the two expressions yields d = 2.28 em.

16.21. Represent the E field of Problem 16.14 in the time domain, using as variables p r/a, cp, and ~ = kTE 11 z.

~pace

288

WAVEGUIDES

[CHAP. 16

In terms of the lumped constants K ~ wpH11 p k;TElla

which are presumed real, we have (x;. = 1.841): EP(p. <fJ,

C. t) = Re [ÂŁ, 11 ei1"'H 1] =

E<J>(p. <fJ,

C. t) = Re [ÂŁ. 11 ei

1 1 '"H ]

-~J1 (1.841p) sin <P sin (wt- C)

= -K~;(l.841p) cos <P sin (wt-

16.22. For the E field obtained in Problem 16.21, calculate and plot the field lines. (without calculation) the lines of the transverse H field.

Also plot

The lines of any vector field are a family of space curves such that, at each point of space, the vector is tangent to the curve through that point. Thus the differential equation of the lines of E in a cross-sectional plane is dy/dx = Ey/ E~, in cartesian coordaintes (x. y), or

1 dp

--=pd<fJ E<J>

in polar coordinates (p, <P ).

(1)

Substitution in ( 1) of the components of E from Problem 16.21 gives

dp J.(l.841p) d<P = K 1 J;(l.S4 1p) tan <P

(2)

It is seen that the TE 11 mode of a cylindrical waveguide has the special property that the field pattern

does not change with time or with distance Calong the guide. Normally, the field lines are found by a numerical integration of the differential equation; but in this case an analytic solution is simply obtained: In J.(LS4lp) = K 2 In Isec<fJ I J 1(1.841p0 )

(3)

This is a one-parameter family of curves, where the parameter p 0 gives the radius at which a curve cuts the horizontal axis sin <P = 0. Note that the right side of ( J) does not change when <P is replaced by -<P or by <P + .1r; hence the field pattern is symmetric about both the horizontal and vertical axes, and only the quadrant 0 s, <P s, .1r /2 need be considered. As one moves along a field line through increasingly positive <P-values, the right side of (J) increases through positive values. Consequently [see Fig. 8-J(a)], PI Po increases through values greater than 1. This, together with the constraint that the field line hit the boundary p = 1 orthogonally, shows that the field line must bend away from the origin, as shown in Fig. 16-5. The line Po= 1 degenerates into a single point.

Pu"' 0

---E

----H

p,."' I -d>=UC

Fig. 16-S

CHAP. 16)

289

WAVEGUIDES

The lines of H are plotted as the orthogonal trajectories of the E lines; see Problem 16.8. By Problem 16.14 both HP and Hâ&#x20AC;˘ vanish at the points p = 1, q, = 0, .1r; hence the direction of H is indeterminate there.

16.23. A lossless air-dielectric waveguide for an S-band radar system has the dimensions a= 7.214 em and b = 3.404 em. The dominant mode propagates in the +z direction at 3 GHz. Find the average power transmitted if the excitation level of the E field is 10 kV /m. The cutoff frequency for TE 111 is

3 X lOS

/c 10 = 2a = 2(0_07214) = 2.08 GHz and (19bis) yields

- = (10')2 (7.214)(3.404)10- ~1- (2.08) 4

P.w

4(377)

= 117.4 w

16.24. In a lossless air-dielectric cylindrical waveguide with a 1 em radius the transmitted power in the dominant mode at 15 GHz is 2 W. Find the level of excitation for the magnetic field. The cutoff frequency for TE 11 is (see Table 16-2): Uo

kTEu = -x;. 2.1ra

3 X Hf ( . ) (1.841) =8.79GHz 2Jr 1 x 10 2

so that (4) of Problem 16.15 becomes (see also Problem 16.16): 2= Solving,

~ 1H11 1 2(Jrl0- 4 )(1S/8. 7W v'1- (8. 79/15)

3 7

[0.239)

IHul = 0. 11 A/em.

16.25. A section of X-band waveguiude with dimensions a = 2.286 em and b = 1.016 em has perfectly conducting walls and is filled with a lossy dielectric (od = 367.5/lS/m, ÂŁ, = 2.1, llr = 1). Find the attenuation factor, in dB/m, for the dominant mode of propagation at a frequency of 9 GHz. The cutoff frequency of TE10 is f }ciO

= u0 = (3 X 10S)tV2l = 4.SJ GHz 2a 2(0.02286)

and (22) gives (second form):

a (dB/m)=(377/~)(367.5x w-6)x8.69=0.48 d 2v'1- (4.53/9) 2 The reader should verify that the underlying approximation,

wpdod <81 f:J~o,

holds for the data.

16.26. An X-band air-dielectric rectangular waveguide has brass walls (llw = llo , Ow = .,...I+ 16 MS/m) with a = 2. 286 em and b = 1.016 em. Find the dB/m of attenuation due to ~ wall loss when the dominant mode is propagating at 9.6 GHz.

:ca.

At the cutoff frequency of the dominant mode, Uo

3 X lOS _ = 6.56 GHz 10 2

k10 = 2a = 4 _572 x the surface resistance of the brass is

Jr(6.56 X 109 )(4Jr X 10- 7 ) 16 x 1~

40.24 m!J

and, by

16.27

(CHAP. 16

WAVEGUIDES

290

0.04024 { ~.6) 0.02286 + 2(0.01016)(6.56/9.6)2 dB 1m = - X869=0214 a ...( ) 377 6.56 (0.02286)(0.01016)V1 - (6.56/9.6) 2 • •

< ),

An air-dielectric cylindrical waveguide (a= 5 mm) operates in the TMo1 mode at frequency f = 1.3/cTMOt. Find the dB/m of attenuation due to wall loss in a short section of copper (ow= 58 MS/m). First derive an expression for P.ou(O), following Section 16.8. By (1) of Problem 16.10, £,01 (r, ¢') = £ 01 J0(x01 r/a). Then (3) of Problem 16.5 gives the tangential magnetic field at the wall as [J~(v) = -J1 (v)): "") _jw~oXmEmJI(Xm) _iEmJ1(Xm) ( f ) H .,.o1 (a, "f' 2 -kcTMOia 'lo hTMOI and, since H.,. 01 is constant, (24) gives P.._(O) =

lR.[ 1Eml2'1o~(xm) (-l-)2](2.7ra) /cTMOl

Next find P.TMm by the method of Problem 16.15.

(1)

By Problem 16.15,

(_1 )( V_J1 _ (/cTM01) )l1(Xmr) I o 2

EriJI "=jkTMmEm 11 (Xo1') = j£01 kcTMCll a /cTMol

_iEm(f /fcTMm) J (Xmr) H4>0lI 'lo a while Hri)1= £.,. 01 = 0.

Thus the time-average Poynting vector is 2 2 S _ !£ H• _ IEml (I flcTM01tV1- UTMm/!) P(Xm') - z riJI .,.01 2'1o I a

Integrating over a cross section,

') rdrd¢1=-T 2A L"Ol J:(v)vdv=Arfi(xm) [o Lo:z,. 1:(X~ 0 Xm o Combining these results,

a __ P..__(O_) _ ---;==R==·==:==.. w-

2~TMOI- 1]oO\ft- (lcTMOl//) 2

For the data,

u0 3xHf /cTMo1 =-Xm = ( 21to 21t 5 x 10

(2.405) = 22.99 GHz

f= (1.3)(22.99) = 29.89 GHz R, =

{;j;:., =

r-------~------~ 7 9

1t(29.89 X 10 )(41t X 10- ) = 0.0451 Q 58 X l<f 0.0451 a = 0.0374 Np/m = 0.325 dB/m ... (377)(5 X 10 3 )V1- (1/1.3)2

-v-;:-

Supplementary Problems 16.18. Determine the condition(s) under which a magnetic field with H.(x, y, z, t) = Kcos87.3x cos92.4y cos(21t/t -109.1z)

can exist in free space.

Ans.

I= 8.0 GHz

(2)

CHAP. 16)

291

WAVEGUIDES

16.19. Obtain the critical wave number for a 4-GHz wave propagating in a medium with p, = 1 and 2.2, if the phase shift constant (wave number) is 54° per em. Ans. 81.1 rad/m

£,

16.30. If Hz(x, y, z, t) in Problem 16.28 represents the axial field of a TE21 wave in a rectangular waveguide, find (a) the guide size, (b) the critical wave number, (c) the guide wavelength. Ans. (a) 7.2cm by 3.4cm; (b) 127.1 rad/m; (c) 5.76cm 16.31. The S-band waveguide of Problem 16.18 is used in the X-band at 9 GHz. Identify the modes that could propagate in the guide. Ans. TE01 , TEm , TE10, TE 11 , TEm, TE21 , TEJO , TE11 , TE40; TMu , TM21 , TM11 16.31. In Problem 16.19, what other modes could propagate at the given frequency? Ans. TE01 , TEu, TE21, TE11; TMot 16.33. A C-band waveguide for use between 3.95 and 5.85 GHz measures 4.755 em by 2.215 em. For air dielectric, calculate the dominant mode cutoff frequency and the guide wavelength when the operating frequency is 4.2 GHz. Ans. 3.155 GHz, 10.82 em 16.34. The WC-50 cylindrical waveguide with air dielectric is used in the frequency range 15.9-21.8 GHz for dominant-mode propagation. Calculate the cutoff frequency for an inside diameter of 1.270 em. Also obtain the cutoff frequency for the TMo1 mode. Ans. 13.84 GHz, 18.08 GHz 16.35. An air-dielectric L-band rectangular waveguide has a/b = 2 and a dominant-mode cutoff frequency of 0. 908 GHz. If the measured guide wavelength is 40 em, find the operating frequency, the guide Ans. 1.18 Ghz, 16.52 em by 8.26 em, 15.7 rad/m dimensions, and the wave number. 16.36. For the waveguide in Problem 16.35 find the lowest frequency at which a TE21 mode would propagate. Ans. I > 2.569 GHz 16.37. A V-band waveguide for use between 26.5 and 400Hz has inside dimensions 0.711 em by 0.356em. (a) Calculate the dominant-mode critical wave number for air dielectric. (b) If the measured guide wavelength is 1.41 em, what is the operating frequency? Ans. (a) 441.86rad/m; (b) 29.98GHz 16.38. The WC-19 air-dielectric cylindrical waveguide is used for dominant-mode operation in the 42.458.10 GHz range. Find the inside diameter for the specified cutoff frequency of 36.776GHz. Ans. 0.478cm 16.39. A Ku-band air-dielectric guide with alb= 2 is used in the 12.4-18.8 GHz range for dominant-mode operation with a cutoff frequency of 9.49 GHz. What are the inside dimensions? Ans. 1.58 em by 0. 79 em 16.40. Find the radius and guide wavelength in an air-dielectric cylindrical waveguide for the dominant mode at I = 30 GHz = 1.5fcTEu . Will the TM 11 mode propagate under these conditions? Ans. 0.44 em, 1.34 em; No 16.41. Solve Problem 16.40 for the guide with a lossless dielectric of Ans. 0.296 em, 0.903 em; No

£,

= 2.2.

16.42. A K-band rectangular waveguide with dimensions 1.067 em and 0.432 em operates in the dominant mode at 18 GHz. Find the cutoff frequency, guide wavelength, phase velocity, and wave impedance, if the dielectric is air. Ans. 14.06 GHz, 2.67 em, 4.81 x lOS m/s, 604.2 Q 16.43. Solve Problem 16.42 if the guide is filled with a lossless dielectric of Ans. 9.93 GHz, 1.44 em, 2.54 x lOS m/s, 319.6 Q

£, =

2.0.

292

WAVEGUIDES

(CHAP. 16

16.44. Calculate the radius and guide wavelength for a TM 11 mode at f = 30 GHz = l.SfcTMu air-dielectric cylindrical waveguide. [Compare Problem 16.40.) Ans. 0.915 em, 1.342 em

in an

16.45. For an (m, n) mode operated below its cutoff frequency, the cutoff attenuation factor is defined as ll'c, = -jkm, . Calculate acTEn , in dB/em when a lossless air-dielectric guide, 2.286 em by 1.016 em is operated at 9.4 GHz. Ans. 23.9 16.46. In a certain cross section of a rectangular waveguide the instantaneous components of E are

E. =0

Sketch this E field and identify the mode of operation.

Ans.

See Fig. 16-6; TE 11

Fig. 16-6

16.47. The air-dielectric waveguide of Problem 16.23 transports 200 W of average power at 2.6 GHz. excitation level of the field. Ans. 143 V/cm

Find the

16.48. If a lossless dielectric having ÂŁ, = 1.8 is inserted in the waveguide of Problem 16.47, calculate the Ans. 106.8 V /em excitation level for the transport of 200 W. 16.49. The air-dielectric waveguide of Problem 16.24 is filled with a lossless dielectric having ÂŁ, = 2.1. Find the power transported in the dominant mode, if the excitation level and frequency are unchanged. Ans. 0.09 A/em 16.50.

Show that result (2) of Problem 16.27 can be rewritten as (frequency dependent) skin depth.

a .. =

-" , 0..,0 Uw'ITMOI

where 6.., is the

Chapter 17 Antennas (by Kai-Fong Lee)

17.1 INfRODUCilON Maxwell's equations as examined in Chapter 14 predict propagating plane waves in an unbounded source-free region. In this chapter the propagating waves produced by current sources or antennas are examined; in general these waves have spherical wavefronts and direction-dependent amplitudes. Because free-space conditions are exclusively assumed throughout the chapter, the notation for the permittivity, permeability, propagation speed, and characteristic impedance of the medium can omit the subscript 0; likewise the wave number (phase shift constant) of the radiation will be written {J = ro .....r,ii = ro I u.

17.2 CURRENf SOURCE AND mE E AND H FIELDS The vector magnetic potential A defined in Section 9. 7 gives the phasor fields in the region outside of the current source as 1 u H=-VXA=-VXA ll 'I 1 1 u E=--VXH=---VXVXA=-VXVXA jro£ iroll£ ifJ

in which u = 3 x 108 m/s and fJ The phasor A is itself given by

(1)

= 120Jt Q. (2)

In (2), r is the distance between the observation point and the source current element Is dv. significance of the factor e-iPr becomes clear when A is transformed to the time domain:

The

Ills cos ro(t- r/u) dv

4Jtr

vol

Thus A at the observation point properly reflects conditions at the source at earlier times--the lag for any given source element being precisely the time rI u needed for the condition to propagate to the observation point.

17.3 ELECTRIC (HERTZIAN) DIPOLE ANI'ENNA The vector potential set up by the infinitesimal current element of Fig. 17-1 is. by (2). -jfjr

A(P) =!!!..__(I dt)az 4Jtr In spherical coordinates, az =cos 8a, - sin 8a8 ;

Idt{J 2 • Hq. = 4 Jt sm 8e

relations ( 1) yield

-·p,[i{Jr + {J1] r 1

2 2

293

294

ANTENNAS

[CHAP. 17

/II I I I I

1::---~~---y

-..

'-..J

_ 2/ df

E, - 'I 4Jt fJ cos 8e E8

_ -

._1_]

-j{Jr[_1_ _ {J2r2 I {JJrJ

_1__ ._1_]

/ df 2 • -j{Jr[. _!_ 'I 4Jt {J sm 8e 1 {Jr + {J2r 2 I {J3r3

Attention will be restricted to the far field, in which terms in 1/r2

All other components are zero. or 1/r3 are neglected.

j/ dt{J . 8 _,..., H =--Sin e ... 4> 4Jtr

far field

jl dt{J .

(3)

= 1 ] - - SID 8e-JfJr = f]H 4Jtr

It is clear that (3) represents a diverging spherical wave which at any point is traveling in the +a, direction with an amplitude that falls off as 1/r. The power radiated by of the Hertzian dipole is obtained by integrating the time-averaged Poynting vector,

!Jt•.,1 = ~ Re (EX H*) (Section 14.12), of the far field over the surface of a (large) sphere. Prad =

i2"i" 0

!Jt•.,1 • r 2 sin 8 d8 dcpa,

L L" U 2Jf

Re (£8 H;)]r2 sin 8d8dcp

= 'l(fJI dtf =

f1.1tP

(dt)

12Jt

(4)

17.4 ANfENNA PARAMETERS The radiation resistance Rrad is defined as the value of a hypothetical resistor that would dissipate a power equal to the power radiated by the antenna when fed by the same current, thus, Prad = VliR.ad or Rrad = 2P.ad//~ where / 0 is the peak value of the feed point current. For the Hertzian dipole, from ( 4), 2

2Jtf} (dt) (df\ Rrad=J A =790 TJ

(Q)

The pattern function F( 8, cp) gives the variation of the far-zone electric or magnetic field magnitude with direction. For the Hertizian dipole this reduces to F(8) =sin 8, since lEI and IHI are independent of c/J.

CHAP. 17)

295

ANTENNAS

The radiation intensity U( 6, cp) is another measure of antenna performance; it is defined as the time-averaged radiated power per unit solid angle. From Fig. 17-2, U((J ""') = dPrad = I!Jfavgl dS' ' 'I' dQ dS' /r2

2I!Jf I avg

/~ava /

_.._-:,.

·~-...:::::r-

dS'

Fig. 17-2

Because U is independent of r (by energy conservation), the far field may be used in its evaluation. For the Hertzian dipole, U(B) =

df\ STJ (IT)

2 2

(5)

sin (J

Polar plots of the pattern function and radiation intensity distribution for the Hertzian dipole are given in Fig. 17-3. :.

(b)

(a) o F{fl)

Half

U(6)

Fig. 17-3

In Fig. 17-3(b), the half-power points are at (J =45° and (J = 135° and the half-power beam width is therefore 90°. In general, the smaller the beam width (about the direction of Umu), the more directive the antenna. Directive gain D( 6, cp) of an antenna is defined as the ratio of the radiation intensity U( 6, cp) to that of a hypothetical isotropic radiator that radiates the same total power U0 • For the isotropic radiator, u.o_-Prad 4n

Then

D(B, cp)

U(B, cp)

4nU(6, cp)

The directivity of an antenna is the maximum value of its directive gain: D = 4nUmax max Prad

296

ANTENNAS

[CHAP. 17

For the Hertzian dipole, ( 4) and (5) give

I df\ 2 2 (4Jr) ~ ( T) sin

v(B.rp>=

(~Jf)e:02

=1.5sin

and

Dmax = 1.5

(6)

The radiation efficiency of an antenna is Erad = P.ad/ P;n , where P;n is the time-averaged power that the antenna accepts from the feed. The (power) gain G(B, rp) is defined as the efficiency times the directive gain:

where PL is the ohmic loss of the antenna. A lossless isotropic radiator has a power gain At times the power gain of an antenna is expressed in decibles, where

G0 = 1.

GdB

= 10 log10- = 10 log10 G Go

17.5 SMALL CIRCULAR-LOOP ANTENNA Also known as the magnetic dipole, a small loop in the z = 0 plane, carrying a phasor current produces radiating E and H fields with characteristics similar to those of the Hertzian dipole, but with the directions of E and H interchanged. In the far zone,

Ia~,

H6 = -

(f32rca2)Je-if3r . sm

4rcr

/ (J /

/"'

I I I I I I I

Fig. 17-4

The radiation resistance of the small loop antenna is found as part of Problem 17.6: (20 fJ)(/3 2rca 2) 2â&#x20AC;˘

Rrad =

17.6 FINITE-LENGTH DIPOLE

The expression (4) for the radiated power of the Hertzian dipole contains the term (dt/}.y which suggests that the length should be comparable to the wavelength. The open-circuited two-wire transmission line shown in Fig. 17-5(a) has currents in the conductors that are out of phase, so that the far field nearly cancels out. An efficient antenna results when the line is opened out as

297

ANTENNAS

CHAP. 17) U2

1-l

-----

-: ----

l1(z')

>~~:~:-........

Al2

IL/2

~..,

t--

...

/2(z')

---1

I /

(a)

ILI2

(h)

Fig. 17-S

shown in Fig. 17-5(b), producing current phasors /1

(z') = lm sin P(~- z')

and

(0< z' < L/2) (-L/2<z' <0)

The two currents are exactly in phase at mirror-image points in the y axis, and they vanish at the endpoints z' = ÂąL/2. The two legs form a single dipole antenna of finite length L. Note that the current at the feed point (z' = 0) is related to the maximum current by fo = lm sin PL. 2

The far field is calculated by means of (2) and (1), under the assumption

and

where the pattern function is given by cos (P~cos cos (P~) F(O)=--------s-in_e________

0)-

The antenna can also be assigned an effective length (write h,(O)

sine IL/2 =lo

/(z') = lm sin JJ(L/2 -lz'l)]:

l(z')eilh' cos 6 dz'

-vz

21 F( 0) Plo

=--.!!!

which has the units of length and contains all the pattern information. For L up to about 1.2A the antenna patterns resemble the figure eight, becoming sharper as L approaches 1.2A. In the other limit, as L ~A, the pattern is that of the Hertzian dipole shown in Fig. 17-3(a). As L becomes greater than l.2A, the patterns become multilobed. See Fig. 17-6. The radiation resistance of a finite dipole of length (2n -l)A/2 (n = 1, 2, 3, ... ) can be shown to be Rrad = (30 fJ) Cin [(4n- 2).7r], where . ( ) LX 1 - cosy d C10 X "" y 0 y is a tabulated function. For n = 1 (half-wave dipole), 1.64 (see Problem 17.8).

Rrad

= 30(2.438) = 73 Q

and

Dmax

ANTENNAS

298

(CHAP. 17

(al

=L =

~ ~ (hi= L =

~12

lei c L =

3~12

Fig. 17-6

17.7 MONOPOLE ANTENNA A conductor of length L/2 normal to an infinite conducting plane [Fig. 17-7(a)] forms a monopole antenna. When fed at the base the resulting E and H fields are identical to the dipole's. This is evident when the image of the monopole is positioned below the conducting plane as shown in Fig. 17-7(b).

...

T Lf2

\ \

/( :l

I /

I (:l

( \

,---:

Ia I

lm<lge current

1 1

,,.,"

(hl

Fig. 17-7

As the monopole radiates power only in the region above the conducting plane, the total radiated power is one-half that of the corresponding dipole. From Rrad = 2Prad/il; it follows that the radiation resistance is one-half the value for the dipole. Thus, for L/2 = A/4 (quarter-wave monopole), Rrad = 36.5 fJ. 17.8 SELF- AND MUTUAL IMPEDANCES With respect to its feed, an antenna is equivalent to a load impedance Za = Ra + jXa , where Ra = Rrad + RL, and RL is ohmic resistance. The reactance Xa is not easily calculated; it is a function of the radius p of the conductors for dipoles and monopoles. Figure 17-8 illustrates the variation of both Ra and Xa for monopoles of length L/2; the figure also applies to dipoles of length L if vertical scale values are doubled. Thus the half-wave dipole has Ra = 73 fJ and, roughly independent of p, Xa =40 fJ. (It can be shown that as p--+ 0, Xa--+ 42.5 fJ.) When a second antenna is placed adjacent to a first antenna, a current in one will induce a voltage in the other. Consequently, a mutual impedance Z 21 = V21 / 11 = R 21 + jX 21 exists in the system. For two side-by-side half-wave dipoles with very small conductor size, R 21 and X 21 vary with the separation d as shown in Fig. 17-9.

ANTENNAS

CHAP. 17]

299

Length of monopole. L/2'1<.

Fig. 17-8. (Source: Edward C. Jordan/Keith G. Blllmain, Electromagnetic Waves tmd Rtullating Systems, 2nd eel.,ÂŠ l9Ci8, p. 548. Reprinted by permission of Preutice-Hall, IDe., Euglewood Cliffs, N.J.)

Fig. 17-9. (Source: Weeks (1968), Antm~~~~ En&I~~Urlft&. Reproduced by permissiou of McGnw-Hill, IDe.)

17-9 mE RECEMNG ANTENNA An antenna in the far field of a transmitter extracts energy from what is essentially a plane wave and delivers it to a load impedance Z 1 â&#x20AC;˘ In Fig. 17-lO(a) the dipole antenna lies along the z axis and the incident wave has a Poynting vector ~. The open-circuit voltage is equal to the product of the effective length h.,( 6) and the magnitude E of the projection of E onto the plane of incidence. [For the coordinate system of Fig. 17-lO(a), E= VE';+ E~.) Voc =h.,( 6)E

ANTENNAS

300

[CHAP. 17

(h)

(ll)

Fig. 17-10

The pattern for the receiving antenna is identical to that of a similar transmitting antenna. The available power Pais the maximum power which the receiving antenna can deliver to a load, which occurs when Z 1 = From the equivalent circuit of Fig. 17-10(b),

z:.

= h.,( 6)2 ÂŁ2 BRa

The effective area A.,( 0) for an antenna is a hypothetical area such that when multiplied by the power density of the incident wave, E 2 /2TJ, it results in the available power. 2 2 2 ÂŁ ) _ _ h.,(6) E ) 2( A.,( 6 ) ( '1 - Pa BRa or A.,( 6) =h.,( 6) ~a 2

It can be shown that the effective area is related to the directive gain by

A.,(O, cp) 0( 6, cp)

2 --=....:....._....:....:.=-

4.7r

When both a transmitting and a receiving antenna are considered, the power Prad 1 radiated by antenna 1 and the available power Pa 2 at the receiving atenna 2, are related by the Friss transmission formula,

Pa2 Prad 1

--=

D1(fJ1, cfJ1)A.,2(fJz, cfJ2) --=-' ---=--::....::..::---::=--=-....:....:::o. 2

4.7rr

Here, r is the separation of the two antennas. Angles 6 1 and cp 1 specify the direction of the receiving antenna as seen from the coordinate system of antenna 1. Similarly 6 2 and cp 2 specify the direction of the transmitting antenna as viewed from the coordinate system of antenna 2.

i1i

r!!!tt7.10 LINEAR ARRAYS A far-field pattern with a narrow beam width and high gain can be achieved by forming an array of identical antenna elements, each with the same orientation as shown in Fig. 17-11. The pattern function of the array is equal to the pattern function of an individual element multiplied by an array factor f(x). In Problem 17.15 it is shown that, for a uniformly spaced array of N elements where dis

301

ANTENNAS

CHAP. 17]

p (1, 6, 4>)

,. Fig. 17·11

the spacing N-1

f(X)

l,ei/Jndcosx

n=O

The angle sin (J cos t/J.

is the angle between the array axis and the line OP; by geometry, cos x = /, = a,ei""' (n = 0, 1, ... , N- 1),

If the elements are progressively phased so that N-1

J<x> = L

a,ei"<"'+/Jdco•x>

n=O

or, defining u = a

+ f3d cos x, N-1

Ji(u) =

a,einu

(7)

n=O

The overall pattern function will be a maximum when lfi(u)l is a maximum, which occurs for u = 0. If a= 0 (the individual antennas are all in phase), then u = 0 implies x = ±90°; i.e., peak radiation occurs at right angles to the line of antennas. This is called a broadside array. On the other hand, if the phasing a= -{M is imposed, u = 0 implies x = 00; this is an endjlre array. A uniform array has all antenna currents equal in magnitude. For a 0 = a 1 = · · · = aN_ 1 = 1, ( 7) becomes _sin (Nu/2) i(N- 1)ut2 e sin (u/2)

1 u)Ji(

(8)

Thus, the main peak or lobe of the radiation pattern, centered on u = 0, has "height" lfi(O)I = N. The two jlrst nulls of the pattern [zeros of lfi(u)l], occur at u = ±2rr/N. The separation of the two first nulls can be used to define the beamwidth. Concentrating on the plane (J = 90°, one finds: broadside unifonn endfire unifonn

. 1 2rr 2A l:irp = 2 sm--=fjNd

l:irp=4sin- 1

0 = [iii '/fiNd 'Jt:M

where the approximations are for the case Nd »A. The sidelobes occur approximately midway between the nulls. The ratio of the main lobe to the first sidelobe is N sin (3rr /2N) which approaches the value 3rr /2 for large N.

302

ANTENNAS

[CHAP. 17

17.11 REFLECTORS The gain of an antenna element can be enhanced by means of a reflector.

Gains of from 6 to

12 dB can be obtained by using a half-wave dipole and a corner reflector such as that shown in Fig. 17-12(a ). (A flat sheet reflector results when 1J1 = 180°.)

( orncr reflector

H .. rw H· dlpl<.

Fig. 17-U

The effect of a reflector with 1J1 = 180°/ N (N = 1, 2, 3, ... ) can be calculated by the method of images. The actual reflector is replaced by 2N - 1 image dipoles, which together with the actual driven dipole, constitute an evenly spaced circular array, alternating in polarity [Fig. 17-12(b)]. Superposition of the far fields yields . E=J1Jloe

-jflrcos(~cose)2N-• . L (-l)"eJflSsinllcos(rnJ!-)ae

2:nr

sm B

(9)

n~o

For high gain applications, the parabolic reflector driven by a source located at its focus, as shown in Fig. 17-13, is widely used. The directivity of the parabolic reflector is proportional to the aperture radius a and the aperture efficiency ~:

The aperture efficiency depends on a variety of design factors; a reasonable value half-power beam width can be estimated from the formula HPBW = ll7°{A./2a ).

ILE

Fig. 17-13

55%. The

303

ANlENNAS

CHAP. 17]

Solved Problems 17.1.

A center-fed dipole antenna with a z-directed current has electrical length L/}, ~ /o. (a) Show that the current distribution may be assumed to be triangular in form. (b) Find the components of the vector magnetic potential A. (a)

Since

L 1 fJ ( -L2 -lzl ) < {J-L2 = .rr-~). 10 we have I(z') = Im sin fJ{~ -lz'l)

=ImfJ{~-Iz'l) = 2~;, (~-lz'l) where I;,= Im fJ2L (b)

PIL/2 l(z')a. (e-ifl') pi' (L) . - - dz'= 2pl' e-iflr IL/2 (L --lz'l )dz'a.=___!!!e-'fl'a.

A=4.rr

-L/2

4.rrLr

-LI2

4.rrr 2

from which

(L)

. 8 = -pi:, . A 9 =-A.sm - - e-'.fl'sm8 4.rrr 2

pi:, .fl A =A cos8=-e-''cos8 ' â&#x20AC;˘ 4.rrr 2

and A.,. =0.

17.2.

i!i.

I.!!

(a) Find the current required to radiate a power of 100 W at 100 MHz from a 0.01-m Hertzian dipole. (b) Find the magnitudes of E and Hat (100m, 90Â°, 00). (a)

(a)

df\2

3xllf ).=1()8=3m

Rrad

=790( A} = 8.78 X 10-J Q

~8.78x 200 w-

- I 51

3 -

(This extremely high current illustrates that an antenna with a length much less than a wavelength is not an efficient radiator.) 7]{JI dt .

lEI = -4-- sm 9()0 = 0.95 v /m .rrr

(b)

17.3.

IHI = 2.52 X 10- 3 A/m

Two z-directed Hertzian dipoles are in phase and a distance d apart, as shown in Fig. 17-14. Obtain the radiation intensity in the direction (6, rp). Since cos a

= sin 8 sin 1/J, d r1 = r - - cos a

The far electric field is then

=r -

- sin 8 sin tp

a8 ,

j{J7](I dt) -

Z:rrr

d sm . (} sm . ""' r2 = r + 2 'Y'

and

with

( d .

e 'fl' sm 8 cos ~ sm (} sm 1/J

)

304

ANTENNAS

[CHAP. 17 p

Fig. 17-14 Consequently,

U For d <!!:A

= r 2 ( : ; ) = 7](~~~t)

sin 2 (}cos

(tJ ~sin (}sin tJ>)

the cosine term is nearly 1 and

17.4.

7](/JI df)

• 2 (}

S.n-2

The far electric field of two Hertzian dipoles at right angles to each other (Fig. 17-15), fed by equal-amplitude currents with a 90° phase difference, is E =jf3~l df) e-ill1(sin nr

(J-

j cos (J cos rp)a8

+ (j sin rp)a~]

Find the far-zone magnetic field, the radiation intensity, the power radiated, the directive gain, and the directivity. E9 71

H• = - =

jfJ(I dt) 111 . . e- '(sm (}- j cos(} cos tJ>) 4.n-r

E. jfJ(I dt) -Ill• . H 9= - - = e smtJ> 7] 4.n-r 2

_ r E · E• _ 7](/JI df) (l . 2 • 2 fJ) U+smtJ>sm 322 .7r 2 7] Prad

I 0

I" .

Usm fJd(}dtf> =

7](/JI df)2 .7r 6

D( (}, tf>) = P..n-U = i(1 + sin2 tJ> sin2 fJ) rad

Dmu = D(90°, 90°) = ~

ld'Oa ~?tj\

Fig. 17-IS

jldl

.. y

CHAP. 17)

17.5.

305

ANTENNAS

A Hertzian dipole of length L = 2 m operates at 1 MHz. Find the radiation efficiency if the copper conductor has ac =57 MS/m, ~-'â&#x20AC;˘ = 1, and radius a= 1 mm. As defined in Section 17.4,

where R,.d is the radiation resistance and RL is the ohmic resistance. than the skin depth

The radius a is much greater

1 1 6=---=-mm VIifpa~ 15 so that the current may be assumed to be confined to a cylindrical shell of thickness lJ.

RL=---=0.0840 ac (2.1ra )lJ

o>(i) = (790 o>( ~0 2

R,od = (790 E,od

17.6.

Mlllhclld

= o.035 o

0.035 9

= O.l1 = 29.4%

Find the radiation efficiency of a circular-loop antenna, of radius a= .n-â&#x20AC;˘ m, operating at 1 MHz. The loop is made of A WG 20 wire, with parameters aw = 0.406 mm, a= 57 MS/m, and 1-'r = 1. At 1 MHz the skin depth is lJ = 0.667 pm. thickness lJ, the ohmic resistance is

Assuming the current is in a surface layer of

2 RL =.!. ( .1ra ) =0.206 Q a 2.1ra,.lJ

Taking the far-zone magnetic field from Section 17.5,

from which

and

17.7.

Find the radiation resistance of dipole antennas of lengths (a) L

= A/2

and (b) L

(2n - 1) 2, n = 1, 2 ...

(a)

1 Prad = -2

L2"'L"' IE 12 8

- 2 r sin() d() d<fJ = 30/;, 'lo

60/~

2Prad "' {cos ( p Rrad =~=---a10

L o

"' {cos (P~cos ())-cos (P~) 2

. sm ()

~cos ()) - cos ( ~)

. () sm

d()

ANTENNAS

306

For the half-wavelength dipole L == J../2 and /.., ==

[CHAP. 17

sin (P~)

= /0

Letx =cos B

1 ~) ttt 1 1

R- = 60

cos ( -d1-x)

= 30 { 1 + cos 2 _1 1-x

JU + 1 + cos JU} ttt 1+x

Since the two terms within the brackets are equal 1

R,ad = 30J

JU) ttt

+cos 1 +x

(

-1

letting y = n(1 + x) 2

1 "

R,ad == 30

{1- ycosy) dy = 30 Cin (2n)

R,ad == 30(2.48) = 73 Q (b) For L = (2n- 1)

a similar approach yields R,.d = 30 Cin [(4n- 2)n] Q

17.8.

Find the directivity Dmax of a half-wave dipole. From Section 17.6, for

/JL/2 == n/2, lo

IHI=.,. 2nr the maximum being attained at

cos (~cos 2 sin B

B == 90". U

m..

e) whence

It follows that 2 '11 IH 1 J1l~ == r 2 .,. 2max == 811"2

From Problem 17.7,

Thus

17.9.

_ 4nUmax m .. -

P,ad

Cin (2n)- 1. 64

A 1.5-l dipole radiates a time-averaged power of 200 W in free space at a frequency of 500 MHz. Find the electric and magnetic field magnitudes at r = 100m, () = 90". From Problem 17.7,

R,ad=(30Q)Cin(6n)==l05.3Q, 10 ==

For a 1.5J.. dipole,

llol = 11... 1.

and so

== /2fiOO) = 1. 95 A 'If2P: R:; '1105:3

307

ANTENNAS

CHAP. 17]

From Section 17.6,

= llmll

IH.,.(100 m, 90")1

21tr

IF(90")1 â&#x20AC;˘-lOOm

1Eo(100 m, 90")1 = (120n)(3.1

10- 3 )

=2 1

) (1) = 3.1 mA/m

1t 100

= 1.17 V/m

17.10. Obtain the image currents for a dipole above a perfectly conducting plane, for normal and parallel orientations. The basic principle of imaging in a perfect conductor is that a positive charge is mirrored by a negative charge, and vice versa. By convention, electric currents are attributed to the motion of positive charges. Hence, for the two orientations, the image dipoles are constructed as in Fig. 17-16.

A(TUAL

lliPOLF A(TUAL

DIPOLE

77777777777777777777777777777.

!7 -,

i ~/ I

I I I I

-------J I

IMAGE

DIPOLE

-1

L-------

0--

Ill

--0/,. ~

-------1

l ____ - - -

IMAGE

DIPOI

(h) Parallel

(a) Normal

Fig. 17-16

17.11. Calculate the input impedances for two side-by-side, half-wave dipoles with a separation d = )./2. Assume equal-magnitude, opposite-phase feed-point currents. The two feed-point voltages are given by

"'=

where

l,Zn + /2Zo2

Z, 2 = Z 2,; consequently,

Z2== It; =Z22+ (/â&#x20AC;˘) /;_ Z,2 12

For half-wave dipoles Fig. 17-8 gives j28 Q. Then, with I,= -/2 , Zt

Z 11

= Z 22 = 73 + j42.5

and Fig. 17-9 gives

= Z2 = 73 + j42.5- ( -12.5- j28) = 85.5 + i10.5

Z 12 = -12.5-

17.12. Three identical dipole antennas with their axes perpendicular to the horizontal plane, spaced )./4 apart, form a linear array. The feed currents are each 5 A in magnitude with a phase lag of n/2 radians between adjacent elements. Given Z 11 = 70 Q, Z 12 = -(10 + j20) Q, and Zn = (5 + jlO) Q, calculate the power radiated by each antenna and the total radiated power.

ANTENNAS

308

Z 1 =~ - = Zu 11 Similarly,

Z 2 = 70 Q

[CHAP. 17

(h) 1

+ (~) - Z 12 + - Z 13 = 70 + e-'"12( -10- j20) + e-1"(5 + jlO) = 45 Q 11

and Z 3 = (85 - j20) Q.

It follows that

P,.d, = ~ llrl Re (Z,) = !{25)(45) = 562.5 W 2

P,.d2 = 875 W

P,ad) = 1{)65.2 W

for a total of 2500 W.

17.13 . Two half-wave dipoles are arranged as shown in Fig. 17-17, with #1 transmitting 300W at 300 MHz. Find the open-circuit voltage induced at the terminals of the receiving #2 antenna and its effective area.

•

Fig. 17-17 For a half-wave dipole

(l,_,=lmax),

Section 17.6 gives

2 h,(8) = fJand, at 300 MHz,

fJ = 2:rr.

cos (icos e) . "' smu

For # 1, !,_" =

f2P::. = v-n /2(300) = 2.87 A VIC::

and the far field at angle 8 1 is of magnitude

8,)

cos (~cos 1E(6,)1 = 'lfJim h,(6 1) = 'llm _ _~--4:rrr 2:rrr sm 8, Consequently.

Substituting the numerical values gives The effective area of antenna #2

•

IVod = 0.449 V.

A,(90°) =

;ad

lh.(WOW = 0.131 m

2 •

17.14. For the antenna arrangement of Problem 17.13 find the available power at antenna #2 .

309

ANTENNAS

CHAP. 17) From Section 17.9,

17.15. Derive the array factor for the linear array of Fig. 17-11 (redrawn as Fig. 17-18).

Fig. 17-18

The far electric field of the nth dipole

= 0, 1, ... , N- 1)

is, by Section 17.6,

By superposition, the field at Pis N-l

Âˇ

-jtJr

E(P)=}: E.,= 11Je [F(8)f(x))afl 2Jrr .. -o where the array factor N-l

/(X)= }: I,eiflndcosx

acts as the modulation envelope of the individual pattern functions F( 8).

17.16. Suppose that Fig. 17-11 depicts a uniform array of N = 10 half-wave dipoles with d = }../2 and a= -n/4. In the xy plane let fl> 1 be the angle measured from the x axis to the primary maximum of the pattern and 4>2 the angle to the first secondary maximum. Find

tPt - fP2 . For 8 = 1r12.

x = tP

and the condition u 1f

0 = - - + 1f cos t/J, 4

The first two nulls occur at u

= 21r IN

for the primary maximum yields

and u = 41f IN.

The first secondary maximum is approxi-

310

ANTENNAS mately midway between, at u

= 3:rr/N;

3.n 10

(CHAP. 17

hence,

:rr

= -4+:rrcostP2

¢'2 = 56.63°

Then ¢'•- ¢'2 = 18.89".

17.17. A z-directed half-wave dipole with feed-point current /0 is placed at a distance s from a perfectly conducting yz plane, as shown in Fig. 17-19. Obtain the far-zone electric field for points in the xy plane.

Fig. 17-19

The effect of the reflector can be simulated by an image dipole with feed-point current -/0 • We then have a linear array of N = 2 dipoles, to which Problem 17.15 applies. Making the substitutions N-+2

x-tP r-+r +scos tP

we obtain (to the same order of approximation) E( P) = aa

i'le -iP!•+•""" •> 2.nr "~"0 e-illr

=--

:rrr

• I • 2jloeiP.""" "' sin ({Js cos tP) '-----"' '----.....------' F(90") /(

sin ({Js cos tP )aa

17.18. For the antenna and reflector of Problem 17.17 the radiated power is 1 W and s = O.U. (a) Neglecting ohmic losses, compare the feed-point currents with and without the reflector. (b) Compare the electric field strengths in the direction (6 = 90", t/J = 0°) with and without the reflector. (a)

With the reflector in place, the input impedance at the feed point is

z, = z .. - z.2 = (73 + j42.5)- Z12

CHAP. 17]

311

ANTENNAS But Fig. 17-9 gives, for d: 2s: 0.2A, so l . O..lth

With the reflector removed,

Z 12 =(51- j21) Q.

Thus Z, = (22 + j63.5) Q

and

= '.JfiiG = /2(ij = 0.302 A ~ '.J 22

Z 1 = (73 + j42.5) Q

fo..ithout =

and

~ = 0.166 A

At P(r, 90", 0"), one has, from Problem 17.17,

(b)

'l/Owith . 1t nr 5

IE withl=--smand, from Section 17.6 IE .

wtthout

Hence

I = 'l/Owithout 21rr

ll:w;,hi/IEwithoutl = 2(0.302/0.166) sin 36° = 2.14.

17.19. A half-wave dipole is placed at a distance S = )./2 from the apex of a 90" corner reflector. Find the radiation intensity in the direction ( 6 = 90", 4J = 0°), given a feedpoint current of 1.0 A. For

tjJ=90" and {JS=n,

(9)ofSection17.11yields

£11(90", 0") =i'lloe-illr (1)[-1-1 + (-1) -1] = -j2'l(l.O)e-illr (V/m) 2nr 1ft

Then

U(90", 0") = r21Efl(90",

2'1

2'1 = 76.4 W/sr 1t2

17.20. A parabolic reflector antenna is designed to have a directivity of 30 dB at 300 MHz. (a) Assuming an aperture efficiency of 55%, find the diameter and estimate the half-power beam width. (b) Find the directivity and HPBW if the reflector is used at 150MHz. (a)

A directivity of 30 dB corresponds to Drna. = 1000,

(T2na) ~ 2

Dma•=

and

}, = 1 m at 300 MHz.

Jn:: 2a=; '.J~=13.58m },

and HPBW "= (l17°)(M2a) = 8.62°. (b) Halving the frequency doubles the wavelength; hence, from (a), Dma.= 1? =250=24dB

and

HPBW = 2(8.62°) = 17.24°

Supplementary Problems 17.21. The vector magnetic potential A(r, t) due to an arbitrary time-varying current density distribution J(r', t) throughout a volume v· may be written as

A(r, t) =_l!_JJJJ(r', t -lr- r'l/u) dv' 4n lr- r'l v·

312

ANTENNAS

[CHAP. 17

u = 3 x I<f m/s. Obtain A(r,t) for a Hertzian dipole at the origin carrying current I(t) =

where 10 e- •a, ,I

(-r > 0).

Ans.

IJ(/o d£} e -(r 4n lrl

lrl/u)lr

17.22. For the Hertzian dipole of Problem 17.21, determine H(r, 6, ¢')under the assumption

Ans.

lrl ~ UT.

. 8 -(r -lrllu)l• - tJ(lo df) sme a.,. 4nu-r lrl

17.23. Consider a Hertzian dipole at the origin with angular frequency w. Find the phases of E, and £ 9 relative to the phase of H.,. at points corresponding to (a) {Jr = 1, (b) {Jr = 10. Assume 0 < 6 < 90". Ans. (a) E, lags H.,. by 90", E 6 lags H.,. by 45°; (b) £,lags H.,. by 90", £ 6 and H.,. are almost in phase

17.24. A z-directed Hertzian dipole /, d£ and a second that is x-directed have the same angular frequency w. If /, leads lx by 90", show that on the y axis in the far zone the field is right-hand, circularly polarized. 17

.zs.

Find the radiated power of the two Hertzian dipoles of Problem 17 .3, if d ~ J... Ans .

4n'l (I d£) 2 3 ).-,

17.Z6. A short dipole antenna of length 10 em current distribution. Find (a) the 10- 7 H/m; (b) the maximum power 1.0. Ans. (a) 42%; (b) 0.63; (c)

and radius 400 tJm operates at 30 MHz. Assume a uniform radiation efficiency, using a= 57 MS/m and IJ = 4n x gain; (c) the angle fJ at which the directive gain is 54.71°

17.1:7. Consider the combination of a z-directed Hertzian dipole of length 6.f and a circular loop in the xy plane of radius a, shown in Fig. 17-20. (a) If/, and I.,. are in phase obtain a relationship among/,, I.,., and a such that the polarization is circular in all directions. (b) Is linear polarization possible? If so, what is the phase relationship?

I.,.

Ans.

J.. 6.f

(a)

f."' Zna 2

(b)

Yes.

The currents must be out of phase by 90"

Fig. 17-2.0 17.28. A 1-cm-radius circular-loop antenna has N turns and operates at 100 MHz. Ans. 515 resistance of 10.0 Q.

Find N for a radiation

17.29. A half-wave dipole operates at 200 MHz. The copper conductor is 4061Jm in radius. Find the radiation efficiency and maximum power gain, if a= 57 MS/m and IJ = 4n x 10- 7 H/m. Ans. 99.26%, 1.63 17.30. Obtam the ratio of the maximum current to the feed-point current for dipoles of length (a) 3J../4, (b) 3J../2. Ans. (a) 1.414; (b) -1

17.31. A short monopole antenna of length 10 em and conductor radius 400 ~Jm is placed above a perfectly

313

ANTENNAS

CHAP. 17]

conducting plane and operates at 30 MHz. Assuming a uniform current distribution, find the radiation Ans. 73.36% efficiency. Use a= 57 MS/m and IJ = 4n x 10 7 H/m. 17.3Z. Two half-wave dipoles are placed side-by-side with separation 0.4J.. If / 1 = 2/2 and #1 is connected to a 75-Q transmission line, find the standing-wave ratio on the line. [Recall that the reflection coefficient r is (Z 1 - Z 0 )/(Z 1 + Z 0 ) and the standing-wave ratio is (1 + lfl)/(1 -If!).] Ans. 1.63 17.33. A driven dipole antenna has two identical dipoles as parasitic elements; both spacings are 0.15J.. Given that Z 12 = (64 + jO) Q and Z 13 = (33- j33) Q, find the driving-point impedance at the active dipole. Ans. (29.36 + j65.93) Q 17.34. In Fig. 17-21(a) a half-wave dipole operates as a recetvmg antenna and the incoming field is E = 4.0e+'Znxay (mV/m). Let the available power be Pa 1 • In Fig. 17-21(b) a 3A/2 dipole lies in the xy plane at an angle of 45° with the y axis. The same incoming field is assumed, and the available A ns. 0. 748 power is Pa 2 • Find the ratio Pa 1/ Pa 2 •

,.l h/2

1----+---•

dipole

(h)

Ia)

Fig. 17-Z1 17.35. Find the effective area and the directive gain of a 3A/2 dipole that is used to receive an incoming wave of 300 MHZ arriving at an angle of 45° with respect to the antenna axis. Ans. 0.173 m2 • 2.18

17.36. Consider a uniform array of ten z-directed half-wave dipoles with spacing d = )./2 and with a= 00. With the array axis along x, find the ratio of the magnitudes of theE fields at ~(100m, 900, 0°) and P2 (100 m, 90", 30°). Ans. 11.36 17.37. Eleven z-directed half-wave dipoles lie along the x axis, at x = 0, ±A/2, ±)., ±3)./2, ±2)., ±5A/2. Let the feed-point current of the nth element be /, = / 0 ei"a. A half-wave dipole receiving antenna is placed with its center at (100m, 900, 30°). (a) Determine a and the orientation of the receiving dipole such that the received signal is a maximum. (b) Find the open-circuit voltage at the terminals of the receiving antenna when /0 = LOA. Ans. (a) a= -0.866.1r; (b) 2.1 V 17.38. A half-wave dipole is placed at a distance S = A/2 from the apex of a&' comer reflector; the feed current is 1.0 A. Find the radiation intensity in the direction (B = 90", tP = 00). Ans. 76.4 W/sr 17.39. Two parabolic reflector antennas, operating at 100 MHz and 200 MHz, have the same directivity, 30 dB. Assuming that the aperture efficiency is 55% for both reflectors, find the ratios of the diameters and the half-power beamwidths. Ans. 1.414, 0.707

Appendix A Sl Unit Prefixes Factor

Factor

Prelb

Symbol

1018 101s 1012 109

ex a peta tera giga mega kilo hecto deka

10~1

p T G

10~2

M k h da

10~"

l(f l(f 10

10~3 w~6

10~12

~1s

10~18

Prelb

Symbol

deci centi milli micro nano pico fern to atto

d c m 1-'

n p f a

Divergence, Curl, Gradient, and Laplacian Cartesian Coordinates.

aAx + aAy + aAz ax ay az V X A= (aAz _ aAy)a + (aAx _ aAz)a + (aAy _ aAx)a ay az az ax y ax ay z av +-a av +-a av VV=-a ax ay y az z V2V = iJlV + iJlV + iJlV ax 2 ay 2 az 2 V. A=

Cy6ndrical Coordinates 1a 1 aA.,. aAz V·A=--(rA ) + - - + -

r ar r aq, az (aA,- aAz) 1 [ a ) -aA,] 1 aAz VXA= ( - -aA.,.) -a r aq, az- a+ az ara.,. +-r -(rA ar aq, z av 1av av VV=-a +--a +-a ar r aq, az z 1 a ( aY\ 1 iJlv iJlv v v =; ar r a;J+ r 2 aq, 2 + az 2 r

Spherical Coordinates 1 a 2 1 a . 1 aA.,. V·A=--(rA ) + - - - ( A 9 sm0)+---r2 ar r rsin () alJ r sin () aq,

a (A.p sm. O)- aAa] alJ aq, a,+;1 [ sin1 () aA, aq, - ara (rA.p) ]8a +;1 [ ara (rAa)- aA,] alJ a<P av 1 av 1 av vv = ar a, + ; alJ 8a + rsin () aq, a.,. 2 a~ a ( . a~ +-~1 iJlv V2 v =1- a - (r - + 2 1 r2 ar ar r sin () alJ smlJalJ r 2 sin2 () aq, 2 1

V X A= rsin ()

[

315

Appendix B SAMPLE Screens From The Companion Interactive Outline As described on the back cover, this book has a companion Interactive Schaum's Outline using MathcadÂŽ which is designed to help you learn the subject matter more quickly and effectively. The Interactive Outline uses the LIVE-MATH environment of Mathcad technical calculation software to give you on-screen access to approximately 100 representative solved problems from this book, along with summaries of key theoretical points and electronic cross-referencing and hyperlinking. The following pages reproduce a representative sample of screens from the Interactive Outline and will help you understand the powerful capabilities of this electronic learning tool. Compare these screens with the associated solved problems from this book (the corresponding page numbers are listed at the start of each problem) to see how one complements the other. In the Interactive Schaum's Outline, you'll find all related text, diagrams. and equations for a particular solved problem together on your computer screen. As you can see on the following pages, all the math appears in familiar notation, including units. The format differences you may notice between the printed Schaum's Outline and the Interactive Outline are designed to encourage your interaction with the material or show you alternate ways to solve challenging problems. As you view the following pages, keep in mind that every number. formula, and graph shown is

completely interactive when viewed on the computer screen. You can change the starting parameters of a problem and watch as new output graphs are calculated before your eyes; you can change any equation and immediately see the effect of the numerical calculations on the solution. Every equation, graph, and number you see is available for experimentation. Each adapted solved problem becomes a worksheet you can modify to solve dozens of related problems. The companion Interactive Outline thus will help you to learn and retain the material taught in this book more effectively and can also serve as a working problem-solving tool. The Mathcad icon shown on the right is printed throughout this Schaum's Outline, indicating which problems are included in the Interactive Outline.

a ..-J+

Mathcad

For more information about system requirements and the availability of titles in Schaum's Interactive Outline Series, please see the back cover. Mathcad is a registered trademark of MathSoft, Inc.

317

318

MATHCADSAMPLES

Electric Field Due to a Charge Distribution Over an Infinite Plane (Schaum's Electromagnetics Solved Problems 2.12 and 2.24, pp. 23 and 29)

Statement

System Parameters

A charge of uniform density p covers a plane in Cartesian space. Find E on the side of the plane containing the origin.

c =1

B =-3

A =2

The equation of the plane is

R =6

Ax-r B·y+ C·z=R

Ps :=0.3·m

Permittivity of free space:

Solution

0 := 8.854·10- 12. farad m

The electric field for a uniform, infinite sheet of charge is given by the expression 2·1t

p 8 ·r·z

E(r,z) -

4·1t·E

o· ( r2 + z2)

dr dql·a

which you can derive using a cylindrical coordinate system, finding the distance vector from any point in the plane (use z = 0 plane for convenience) to any point above or below the plane. Load the Symbolic Processor from the Symbolic menu, select the expression above, and choose Simplify. The symbolic processor reduces this expression to

1 Izl E(r,z) ·=-·p 8·--·az

(z·Eo)

Emag = - -

2·Eo

319

MATHCAD SAMPLES The z-dependence cancels, but the sign of z determines the direction. If the field is taken above the plane, the sign is positive; below, it is negative. (Notice that this isn't true if the charge distribution has an r or+ dependence.) For the given parameters, E mag= 16.94â&#x20AC;˘ volt m All that remains to solve this problem is to find the unit normal vector to our given plane. The unit normal vectors for a plane are plus or minus the following expression:

vector

a n :=-:----:-

Ivector!

0.5) (0.3

= -o.8

To determine which side of the plane contains the origin, draw the plane. The intercepts on the three axes are given by: 0 PI:=

P2 =

10.-----------r---------~

P3 :=

0 0

perspective scale factor s=4

-s

If you change the equation of the plane, you may need to adjust the scale factor to get a better view of the plane. The equations which generate this picture may be found to the right of this screen.

320

MATHCAD SAMPLES It is evident from this sketch that the unit vector on the side of the plane containing the origin is produced by a negative sign. The electric field anywhere on this side of the plane is

EmagÂˇ(-an)

--9.1 ) 13.6 .v~t (

-4.5

Editor's Note: The equations which generate the live picture of the plane are not shown here, for simplicity. They would. of course, be accessible in the Electronic Book companion.

321

MATHCAD SAMPLES

Flux Density (Schaum's Electromagnetics Solved Problem 3.11, p. 40)

Statement

Determine the flux crossing a 1 mm by 1 mm area on the surface of a cylindrical shell at a point P given a flux density D.

2·x

System Parameters

1coul

D(x,y,z) ·c 2·( 1- y) · m

[

4·z

This is the given flux density in cartesian coordinates.

At point P, which is given in cylindrical coordinates:

53.2·deg

area =(l·mm)

J!C

Solution

=0.929•rad

=10- 6 -coul

First convert the point P on the shell to Cartesian coordinates.

P0 ·cos(P1) p• = P0 ·sin(P 1)

p2 Then, at P,

322

MATHCAD SAMPLES If the radius r of the cylinder is large, the 1 mm2 area is essentially planar, with a directed area vector normal to its surface (no z component) beginning at P. P'

0.6)

S := P'

- s as lsi

dS :=area·a S

=(~-8

·mm

The flux is given by the dot product of the flux density and the area vector:

d't' =D p·dS

d't' =-4.1 "J.lC

The negative sign indicates that flux crosses this differential surface in a direction toward the z axis, rather than outward in the direction ofdS. This is a good example of the type of simplifying assumptions used by engineers. Given the capabilities of Mathcad, the problem could have been solved exactly, rather than assuming the small surface to be planar. If you'd like to see this sort of problem solved exactly, examine Chapter 3.

Editor's Note: The boldface, underlined text in the paragraph above indicates a hyperlinked piece of text. If you were working on a computer, double-clicking on the bold text with the mouse would take you to the file indicated by the text. Also, the slight difference between the answer here and on page 40 is due to the difference in numerical accuracy used in the two calculations.

323

MATHCAD SAMPLES

Method of Images (Schaum's Electromagnetics Solved Problem 7.22 and Supplementary Problem 7 .34, pp. 109 and 111)

Statement

System Parameters

(a) Find the capacitance per unit length between a cylindrical conductor of radius a and a ground plane parallel to the conductor axis, at a distance h from it. The conductor carries a charge distribution p. (b) Double the conductor diameter to see the effect on the capacitance.

a ·=2.5·cm

h =6.0·m

w-6. coul m

- 12 .,

pF = 10

·•ilrad

Permittivity of free space:

-12

= 8.854· 10

farad

·-~

Solution

(a)

A useful technique in problems of this kind is the method of images. Take the mirror image of the conductor in the ground plane, and let this image conductor carry the negative of the charge distribution on the actual conductor. Now suppose the ground plane is removed. It is clear that the electric field of the two conductors obeys the correct boundary condition at the actual conductor, and by symmetry, has an equipotential surface where the ground plane was. Thus, the field found in this way is the field in the region between the actual conductor and the ground plane. See the figure below for clarification. a

+ h

' , J ,r \,, , '

\ \

y.' ,

' .,.~ -c .,.-' ' ...... ---

, /I - ---

--..-~-Pa

Approximating the actual and image charge distributions by line charges at their centers, Potential at radius a due to the actual conductor:

p ( a) V real = - - - · I n 2·1t·E em 0

(See Chapter 5 for the potential from a line charge.)

324

MATHCAD SAMPLES To find the potential due to the image conductor, find the distance from the imaginary line charge to any point on the circumference of the real conductor .

~ r . =h+-Jh~a· Potential at point P due to the image conductor:

Vimage ·= __ -_P_.Jn(~) 2·n·to em V a =I . Il·Hr ·volt

V a =V real + V image

Similarly, the potential of the image conductor is

-v•.

Thus, the potential

difference between the two conductors is 2V8 , and the potential difference at the ground plane is 2V,/2

8•

The capacitance per unit length is then given by C = Q/(LV):

c -

C =9.0l·pF

Observe that the capacitance per unit length for the whole source-image system (and more generally, for any pair of parallel cylindrical conductors with center-to-center separation 2h) is one-half the value found above, since it contains the same charge, but twice the voltage drop.

(b)

If the radius is doubled in the original problem, a .=2·a the resulting capacitance, C2x, is

C 2X = 10.15·pF m

Recall that

C =9.0I·pF

for comparison.

Editor's Note: As in the previous example, differences in numerical accuracy will affect how closely these answers match those on pages II 0 and 1 I I.

325

MATHCAD SAMPLES

Magnetic Potential (Schaum's Electromagnetics Solved Problem 9.12, p. 147)

Statement

System Parameters

Given the general vector field A, in cylindrical coordinates, find the curl of A at point P.

_ 5-e-r·cos(q,) A(r,q>) .0

Coordinates for P:

3 q, =-·n·rad

r := 2

Solution

z =0

-5·cos(ell)

The curl of A is given by:

curl

·=[.!..(~A(r.ell))!!_A(r.ell) ]·a r dell dz r 2

~A(r,ell) )-a,..

curl,.. = (!!...A(r,ell)0 "' dz dr

curl

=L[~(r·A(r.ell))~A(r,ell)o]·a z r dr dell

1:1

curl r + curl ell + curl z

-2 5 ) curl= 0 ( :-0.34

Try creating different functional dependencies for A to see the effect on the curl of the vector field. Remember when changing the variables in the vector to also change the variables in the left-hand side of the definition. If A were a magnetic field, does the resulting current density agree with what you'd expect to find?

326

MATHCAD SAMPLES

Magnetic Work and Power (Schaum's Electromagnetics Solved Problems 10.13 and 10.14, pp. 163 and 164)

Statement

System Parameters

Find the work and power required to move the conductor shown in the figure one full tum in the positive direction at a rotational frequency of N revolutions per minute, if B is a radial field. Find the work to move the conductor over some portion of the cylinder at a constant speed. If the current direction is reversed over half the cylinder, what is the total work required for one revolution?

1 rpm=min

B 0 =- 3.5·I0- ·tesla

N =35·rpm

r .= 25·mm L := IOO·mm

ell 1 := O·rad y

Solution

ell 2 =1t·rad

The force on the conductor is

F =I·LX Br

The applied force to move the conductor is equal in magnitude and opposite in direction, and so it is in the positive ell direction. The work required to move the conductor through one full revolution is given by the line integral over the force.

327

MATHCAD SAMPLES

-4

W =-2.749-10

•joule

Multiply by N revolutions per minute to arrive at the total power. P:=N·W

To find the work done for only part of a revolution, and for a magnetic field that varies with angle, find the applied force over part of a rotation.

w = 87.5•jlJ

If the current direction changes on the side of the conductor between and 2K. the work will be the same, because the sign of the integral changes. K

w =87.5•jlJ 2·W =175•jlJ

The total work for a single revolution is and the power is

-4

N·2·W =1.021•10

•watt

328

MATHCAD SAMPLES

Underconstrained Magnetic Circuits (Schaum's Electromagnetics Solved Problem 11.16, p. 185)

Statement

The magnetic circuit shown below consists of nickel-iron alloy in part 1 and cast steel for part 2. In part 1, the mean length is d 1 and the area 5 1. In part 2, the mean length is d2 and area is 5 2. Find the flux densities 8 1 and 8 2.

System Parameters d 1 =IO·cm

F cc40·amp

Solution

=2.25·cm 2

=8·cm

=3·cm

Given the information, it is possible to tell what the sum of the magnetomotive forces (mmfs) are in the structure, but not the individual magnetic fields. Since there isn't an explicit function to solve simultaneously for the nonlinear flux density, this problem will require an iterative solution. The following expressions must hold true:

H l"f..li·S I

H2=

--f..l2·S2

Making a guess that the greater percentage of the mmf drops in the material of lower permeability (not a bad guess given the results of the analysis in Chapter 11) let's suppose:

H1-

percent·

(:I )

percent = 0.2

The variable percent is defined globally below with the final result.

329

MATHCAD SAMPLES Read in the data points:

data

=READPRN(bhiow)

- ( data<O> , data<4>) vI = Isphne 4 NiFe(H) cinterp( vi,data<O> ,data< > ,H)·tesla

Now use the function NIFe to find the flux density. B I = I.079•tesla

From the flux density, find the magnetic field in part 2:

B 1·S I B2-=-S2

data2 = READPRN( bhhigh)

. ( data2<2> , data2<O>) v2 := Jsphne

steel( B)

=interp(v2, data2<2 >, data2<0>, B). amp m

Percentage drop In part 1:

B 1 = 1.079•tesla

=464.958-amp m

High field data used - always check what range Is In effect!

percent=20·%

B 2 = 0.809 •tesla

Adjust the value of percent until the mmfs around the circuit and from the coil match (they aren't matched now!)

330

MATHCAD SAMPLES It is also possible to let Mathcad do the iterating for you automatically. The until instruction causes values of the percentage to be calculated in a vector until the mmf matching condition is met. First element in the vector:

perc ·= 100·%

Number of iterations required:

Correct percentage:

n .=0 .. 1000

m · = last( perc)

percent ·c perc m- 1

m=853

percent= 14.8•%

Flux densities at this percentage:

B 1 = 1.028 •tesla

B 2 =0.77l•tesla

Editor's Note: The two data files used in this example, bhhigh and bhlow, are supplied with the Electronic Book companion. The data are taken from the graphs shown on page 175 of this Schaum's Outline. Differences in interpolation and numerical accuracy will affect how closely this answer matches that in the original solved problem on page 185.

331

MATHCAD SAMPLES

Traveling Waves (Schaum's Electromagnetics Solved Problem 14.1, p. 226)

Statement

System Parameters

A traveling wave is described by the function y(z, t, m, p). Sketch the wave at t 0, and at t t.. when it has advanced one eighth of a wavelength in free space. Repeat for a wave with twice the frequency and the samet1.

·--1Hz . sec

Yo:= 10

y( z, t, m, P> := y 0 -sin( P·z-

Solution

m·t)

Permittivity of free space:

12 0 = 8.854·10- • farad m

Permeability of free space:

J.1.

7 0 ·c4 ·1t·l o· · henry m

First calculate the necessary wave parameters for both plotted frequencies.

P1 :=m·JJJ.o·Eo

p2 ·=2·m·JJJ.o·Eo

p 1 = 3.336•10- 3 p2 =6.671·10-3

. m

The wave number is twice as large for twice the frequency.

The wavelength is half its former value.

332

MATHCAD SAMPLES The waves will be plotted at times

·=O·sec

Plot the wave ever a single cycle of p1 . Jt

2·Jt

z =O·m,0.05·- .. p1 p1

)., I

>{z.l,lll,~ 1) >{z.l ,,ro, ~ 1)

0~~~--~----~~~----4-----~~

500

1000

1500

2000

y(z,l,2·1ll,~ 2) y(z,lt,2·1ll,~ 2)

0~--~~~~--~~----~~~--~

500

1000

1500

2000

At time 11, the first wave has advanced

=235.46•m

1..2

=235.46•m

The second wave has advanced by the same distance, but this distance is now a quarter of a wavelength.

Editor's Note: The eJlipsis used to define the variable z indicates that it is a Mathcad range variable. The variable z assumes all the values in the given range, in steps of O.SJtJp1, so that the traveling wave can be plotted over this range. As in previous examples, differences in numerical accuracy used will affect how closely this answer matches that in the original solved problem on page 226.

INDEX AC resistance, of transmission lines, 237 Air-gap line. negative, 183, 184 Air gaps, cores with, 177 Ampere (unit), 76, 174 Ampere turns (unit), 174 Am~re's law, 135-153, 137 for magnetic circuits, 174, 176-177 Antenna parameters, 294-296 Antennas, 293-313 available power of, 300 directivity of. 295-296 effective area for, 300 effective length of, 297 electric dipole, 293-294 linear arrays of, 300-301 monopole, 298 ohmic loss of, 296 power gain of, 296 radiation efficiency of, 296 receiving, 299-300 self-impedance of, 298-299 small circular-loop, 296 Array factor, 300 Arrays: endfire, 301 linear, of antennas, 300-301 uniform, 301 Associative law, I Attenuation, per-meter, 255 Attenuation factor, 280-281 total, 281 Available power of antennas, 300 Avogadro's number, 86 Axial components, transverse components from, 275 Axial fields, 274-275 determination of, 276-2n

Canesian coordinate system (Cont.): del operator in, 49-50 differential displacement vector in, 59 divergence, curl, gradient, and Laplacian in, 314 divergence in, 47-49 electric flux density in, 39 field vector in, 274 gradient in, 63 Laplace's equation in 114-115 in one variable, 116-117 product solution of, 117-118 Laplacian of vector in, 216 Maxwell's equations in, 217-218 position vectors in, 6 Characteristic impedance, 240 good, Maxwell's equation solutions for, 220-221 in motion: through time-dependent fields, 196-197 through time-independent fields, 195-196 parallel, inductance of, 171 perfecl, imaging in, 307 Conservative fields, 60 Conservative propeny of electrostatic field, 60 Constant currents, 76 Continuity of current, 82-83 equation of, 82 Contour, closed, 60 Conveclion current, n Convection current density (J), n Coordinate system, divergence, curl, gradient, and Laplacian in, 314 Coordinate systems, 3-4, (See also Canesian coordinate system; Circular cylindrical coordinate system; Spherical coordinate system) Coordinates, 3 Core lengths, 174 Cores, with air gaps, 177 Coulomb (unit), 13 Coulomb forces, 13-31 Coulomb's law, 13-14, 101 Critical wave number, 282 Cross product of two vectors, 2 Curl, 47 in coordinate systems, 314 divergence of, as zero scalar, 139 of gradient as zero vector, 139 of vector field, 137-139 Current(s) (I). 79 conslant, 76 continuity of (see Continuity of current) displacement (see Displacement current) time-variable, 76 Current density, 76 conduction, 192 convection, n displacement, 192 magnetic field strength and, 139 total, 193 Current elements, magnetic force on, 155-156 Current filament, vector magnetic potential for, 141 Current law, Kirchhoff's, 82 Current sheet, 81 at boundary, 106-107

B (see Magnetic flux density) B-H curve, 174, 175 Back-voltage in inductor, 172 Beam width, half-power, 295 Biot-Savan law, 135-137 Boundary, current sheet at, 106-017 Boundary conditions: across interface of two dielectrics, 207 at interface of two dielectrics, 100-101 conductor-dielectric, 83-85 Boundary reflection coefficient, 240 Boundary relalions, for magnetic fields, 205-206 Capacitance, 95-113 definition of, 96-97 equivalent, 97 of transmission lines, 237 Capacitors: energy stored in, 98 multiple-dielectric, 97-98 parallel-plate, fringing of, 43-44 Canesian coordinate system, 3-4 curl in, 138-139

333

334 Current sheet density, 81-82 Current source, phasor fields outside, 293 Cutoff frequency, 2TI-278 Cutoff wavelength, 286 Cylindrical conductors, inductance of, 171 Cylindrical coordinate system (see Circular cylindrical coordinate system) Cylindrical guides, 284 D (see Electric flux density) D'Arsonval meter movement, 161 DC resistance, of transmission lines, 237 Decibel (unit), 229 Del operator, 49-50 Delay time, 248 Density: charge (see Charge density) current (see Current density) energy, 71 flux (see Flux density) Depth of penetration, 220 Determinants, 2 Dielectric-conductor boundary conditions, 83-85 Dielectric constant, 13 Dielectric free-space interface, 102-103 Dielectric losses, 280-281 Dielectrics: boundary conditions across interface of two, '1ff1 boundary conditions at interface of two 100-101 perfect, Maxwell's equation solutions for, 219-220 polarization of (see Polarization of dielectric materials) two, in multiple-dielectric capacitors, 97 Differential line element, 5 Differential surface element, 5 Differential volume, 4-5 Diffusion, 278 Dipole: finite-length, 296-298 magnetic, 296 Dipole antennas, electric, 293-294 Dipole moment, electric, 95 Directivity, of antennas, 295-296 Dispersive medium, 219 Displacement current, 192-204 definition of, 192-193 Displacement current density, 192 Displacement flux, 32 Displacement vectors, 13 Distributive law, I Divergence, 47-58 in cartesian coordinates, 47-49 in coordinate systems, 314 of curl as zero scalar, 139 definition of, 47 of electric flux density, 49 of gradient of potential function, 114-115 negative, 47 of zero, 52 Divergence theorem, 50-51 Dominant mode of waveguides, 278-279 Dot product of two vectors, 1-2 Double-stub matching, 245-247 Drift velocity, 76 E (see Electric field intensity) Effective area for antennas, 300 Effective length of antennas, 297 Electric component of force, 155

INDEX Electric current (see Current entries) Electric dipole antennas, 293-294 Electric dipole moment, 95 Electric field intensity (E), 13-31, 114 definition of, 14-15 due to point charges, 21 fixed-charge, 99 fixed-voltage, 98-99 flux density and, 34-35 motional, 195 potential function and, 63-64 tangential component of, 100 units of, 15 Electric fields: magnetic fields combined with, 155 point charges causing, 21 static, energy in, 64-65 work done against, 59 work done by, 59 Electric flux, 32-46 definition of, 32-33 Electric flux density (D), 33-34 antisymmetrical, 127-128 divergence of, 49 electric field intensity and, 34-35 fixed-charge, 99 fixed-voltage, 98-99 normal component of, 100 Electric potential: of charge distributions, 61-62 definition of, 60 of point charges, 61 between two points, 60-61 Electric susceptibility, 95 Electromagnetic waves, 216-236 Electromotive force, 172 Electron-gas theory, 76 Electron-hole pairs, 78 Electron mobility, 85 Electrostatic field, 59-75 conservative property of, 60 Endfire arrays, 301 Energy: instantaneous rate of, leaving volume, 226 in static electric fields, 64-65 stored in capacitors, 98 Energy density, 71 Energy differences, 72 Equation of continuity for current, 82 Equipotential surfaces, 63 Equivalent capacitance, 97 Farad (unit), 96 Faraday homopolar generator, 201 Faraday's law, 171-172 integral form of, 194 two-term form of, 196 Ferromagnetic materials, 174 Fidd lines, 288-289 Field vector, 274 Fields: axial (see Axial fields) conservative, 60 electric (see Electric fields) electrostatic, 59-75 magnetic (see Magnetic fields) radial, 148 time-dependent, conductors in motion through, 196-197

INDEX Fields (Cont.): time-independent, conducton in motion through, 195-196 transverse, 274-275 vector (see Vector fields) Finite-length dipole, 296-298 Fint nulls, 301 Fixed-charge electric field intensity and electric ftux density, 99 Fixed-voltage electric field intensity and electric ftux density, 98-99 Flux: displacement, 32 electric (see Electric ftux) magnetic, 140 Flux density: electric (see Electric ftux density) magnetic (see Magnetic ftux density) Flux Jines, 32-33 Flux linkage, 169 Forces: Coulomb, 13-31 electromotive, 172 in magnetic fields, 154-168 magnetomotive, 174 moment of, 157 Fourier sine series, 130 Free charge, 83 Free space, Maxwell's equation solutions in, 220 Free-space interface, dielectric, 102-103 Free-space permeability, 140 Frequency of harmonic wave, 217 Fringing of parallel-plate capaciton, 43-44 Friss transmission formula, 300 Gauss' divergence theorem, 50 Gauss' law, 34 Gaussian surfaces, special, 35-36 Generator, Faraday homopolar, 201 Geometrical factor, 237 Gradient, 62-63 in coordinate systems, 314 curl of, as zero vector, 139 divergence of, of potential function, 114-115 Guide wavelength, 277 H (see Magnetic field strength) Half-power beam width, 295 Half-power points, 295 Helical motion, 155 Henry (unit), 140, 169 Hertzian dipole antennas, 293-294 High-frequency lines, 241 Homopolar generator, Faraday, 201 I (see Current) Imaging in perfect conductor, 307 Impedance: characteristic, 240 intrinsic, 218, 222 mutual, of antennas, 298-299 self-impedance, of antennas, 298-299 wave,276 Impedance matching, 243-244 Impendance measurement, transmission line, 247-248 Incidence: angle of, 222 normal, interface conditions at, 221-222

Incidence (Cont.): oblique, 222 plane of, 222 Induced voltage, 172 Inductance, 169-191 definition of, 169-170 internal, 172-173 mutual, 173 self-inductance, 172 of transmission lines, 237, 238 Inductor, back-voltage in, 172 Infinite line charge, 17 Infinite plane charge, 18 Infinity, zero reference at, 61 Instantaneous power, 231 Interface conditions at normal incidence, 221-222 Internal inductance, 172-173 Intrinsic concentration, 87 Intrinsic impedance, 218, 222 Intrinsic semiconducton, 78 Inverse-square law of point charge, 17 Iron-core magnetics, 174 Isotropic radiator, 295

J (see Conduction current density; Convection current density) Kirchhoff's law, 82, 176 Laplace's equation, 114-134 in cartesian coordinate system (see Cartesian coordinate system, Laplace's equation in) definition of, 114 explicit forms of, 114-115 Laplacian, in coordinate systems, 314 Legendre polynomial: higher-order, 130 of order n, 119 Lenz's law, 194 Lever arm, 157 Line charge, 16 infinite, 17 Line charge density, 16 Line element, differential, 5 Linear arrays of antennas, 300-301 Lorentz force, 155 Lossless lines, 241 transients in, 248-250 Lossless waveguide, power transmitted in, 279-280 Lossy waveguide, power dissipation in, 280-281 Magnetic circuits, 173-174 Am~re's law for, 174, 176-177 parallel, 178 Magnetic component of force, 155 Magnetic dipole, 296 Magnetic field strength (H), 135, 206 current density and, 139 tangential component of, 206 Magnetic fields: boundary relations for, 205-206 electric fields combined with, 155 forces and torques in, 154-168 static, 135 time-variable, 192 Magnetic ftux, 140 Magnetic ftux density (B), 140, 206 normal component of, 206

335

336 Magnetic force: on current elements, 155-156 on panicles, 154 Magnetic moment: of planar coil, 157-158 of planar current loop, 158 Magnetic potential, vector, 141-142 Magnetization curves, 174 Magnetomotive force. 174 Mass-action law, 87 Matching: double-stub, 245-247 impedance, 243-244 single-stub, 244-245 Maximum value theorem, 116 Maxwell's equations. 114, 205-215 free-space set, 208 general set. 208 interface conditions at normal incidence, 221-222 solutions for good conductors, 220-221 solutions for panially conducting media, 218-219 solutions for perfect dielectrics, 219-220 solutions in free space, 220 Mean value theorem, 116 in special case, 120-121 Method of images, 109-110 Mho (unit), 80 Mil, circular. 87 Mobility, 76 electron, 85 Mode cutoff frequencies, 277-278 Moment: electric dipole, 95 of force, 157 magnetic (see Magnetic moment) Monopole, quarter-wave, 298 Monopole antennas, 298 Motion: charges in, 76 conductors in (see Conductors in motion) helical, 155 Motional electric field intensity, 195 Multiple coils, 177 Multiple-dielectric capacitors, 97-98 Mutual impedance, of antennas, 298-299 Mutual inductance. 173 n-type semiconductor materials, 78 Negative air-gap line, 183, 184 Neper (unit), 229 Net charge. 82 in region. 32 Net charge density, 82-83 Newton (unit), 13 Nl rise and Nl drop, 176 Nulls, first, 301 Oblique incidence, 222 Ohmic loss of antennas, 296 Ohm's law, 76 point form of, 77 Operating wavelength, 277 Onhogonal surfaces, 3-4 p-type semiconductor materials, 78 Parallel conductors, inductance of, 171 Parallel magnetic circuits, 178

INDEX Parallel-plate capacitors, fringing of, 43-44 Parallel plate geometrical factors. 238 Parallel polarization, 224 Parallel wire geometrical factors, 238 Panicles, magnetic force on, 154 Pattern function, 294 Penetration, depth of, 220 Per-meter attenuation, 255 Period of harmonic wave, 217 Permeability, 140 free-space, 140 relative, 140 Permittivity, 13 relative, 13, 95-96 Perpendicular polarization. 223-224 Phasor fields, 293 Phasors, 239 Planar coil. magnetic moment of, 157-158 Plane, of incidence, 222 Plane charge, infinite, 18 Plane waves, 218 Point charges: causing electric fields, 21 electric field intensity due to, 21 electric potential of, 61 inverse-square law of, 17 in spherical coordinate system, 38 work done in moving, 59 Point form of Ohm's law, 77 Points, 3 electric potential between two, 60-61 Poisson's equation, 114, 127 Polar form, 239 Polarity, 194 Polarization of dielectric materials, 95-96 parallel, 224 perpendicular, 223-224 Position vectors, 5 Potential: electric (see Electric potential) vector magnetic, 141-142 Potential difference, 61 Potential function (V): divergence of gradient of, 114-115 electric field intensity and, 63-64 Power: available, of antennas, 300 complex, 226 dissipated in lossy waveguide, 280-281 instantaneous, 231 Poynting vector and, 225-226 Jransmitted in 1ossless waveguide. 279-280 work and, 156-157 Power gain of antennas, 296 Poynting vector, 226 powerand,225-226 Propagation constant, 216 Quarter-wave monopole, 298 Quaner-wave transformer. 260

R (see Resistance) Radial fields, 148 Radiation efficiency of antennas, 296 Radiation intensity, 295 Radiation resistance, 294 Radiator, isotropic, 295 Receiving antennas, 299-300

INDEX Rectangular-guide formulas, 284 Reflection: angle of, 222 Snell's law of, 222 Reflectors, 302 Refraction, Snell's law of, 222 Relative permeability, 140 Relative permittivity, 13,95-96 Relaxation time, 83 Reluctance, 176 Resistance (R), 80 radiation, 294 surface, 281 Resistivity, 76 conductivity as reciprocal of. 87 Right-hand rule, 136

Scalar function, gradient of, 62-63 Scalar triple product, 7-8 Scalars, I zero, divergence of curl as, 139 Self-impedance, of antennas, 298-299 Self-inductance, 172 Semiconductors, 78 Sheet charge, 16 Sheet current, vector magnetic potential for, 141 Sl unit prefixes, 314 Sl units, rationalized, 13 Sidelobes, 301 Siemens (unit), 77, 80 Single-stub matching, 244-245 Sinks, 47 Sinusoidal steady-state transmission-line excitation, 239-241 Skin depth, 220 Skin effect, 172-173 Slotted lines, 247 Small circular-loop antennas, 296 Smith Chart, 241-243 Snell's law: of reflection, 222 of refraction, 222 Solenoids, inductance of, 171 Sources, 47 Space, free (see Free space) Spherical coordinate system. 3-4 curl in, 139 differential displacement vector in, 59 divergence, curl, gradient, and Laplacian in, 314 divergence in, 48 gradient in, 63 Laplace's equation in, 115 product solution of, 119 point charge in, 38 potential in, 64 Spherical shells, concentric, 67 Standing-wave ratio, voltage, 241 Standing waves, 224-225 Static electric fields, energy in, 64-65 Static magnetic field, 135 Stokes' theorem, 142 Surface charge density, 16 Surface element, differential, 5 Surface resistance, 281 Surfaces: equipotential, 63 orthogonal, 3-4 Susceptibility, electric, 95

TE (transverse electric) waves, 276 Time constant, 83 Time-dependent fields, conductors in motion through. 196-197 Time-distance plots, 248-250 Time-independent fields, conductors in motion through, 195-196 Time-variable currents, 76 Time-variable magnetic field, 192 TM (transverse magnetic) waves, 276 Toroids, inductance of, 171 Torque: definition of, 157 in magnetic fields, 154-168 Transformer, quarter-wave, 260 Transients in lossless lines, 248-250 Transmission, angle of, 222 Transmission formula, Friss, 300 Transmission lines, 237-273 distributed parameters, 237-238 double-stub matching, 245-247 impedance mathcing, 243-244 impedance measurement, 247-248 incremental model, 239-240 per-meter attenuation, 255 single-stub matching, 244-245 sinusoidal steady-state excitation, 239-241 slotted, 247 uniform, 237 Transverse components from axial components, 275 Transverse electric (TE) waves, 276 Transverse fields, 274-275 Transverse length, unit, charge transport per, 81 Transverse magnetic (TM) waves, 276 Traveling waves, 226-227 Triple product: scalar, 7-8 vector, 7 Tuner circle, 245 Uniform arrays, 301 Uniform transmission lines, 237 Uniqueness theorem, 115-116 Unit vectors, I, 3

V (see Potential function) Vector(s), I absolute value of, I component form of, I cross product of two. 2 displacement, 13 dot product of two, 1-2 position, 5 Poynting (see Poynting vector) projection of one, on second, 7 unit, I, 3 zero, curl of gradient as, 139 Vector algebra, 1-3 Vector analysis, 1-12 Vector fields, 47 curl of, 137-139 Vector integral, 14 Vector magnetic potential, 141-142 Vector notation, I Vector sum, 14 Vector triple product, 7 Vector wave equations, 216 Velocity, drift, 76

337

338 Voltage: around closed contour, 171 induced, 172 of self-inductance, 172 Voltage drop, 80 Voltage standing-wave ratio, 241 Volume: differential, 4-5 instantaneous rate of energy leaving, 226 Volume charge, IS Volume current, vector magnetic potential for, 141 Wall losses, 280, 281 Wave equations, 216 Waveimpedance,276 Wave number, 275 critical, 282 of radiation, 293 Waveguides, 274-292 dominant mode of, 278-279 lossless, power transmitted in, 279-280 lossy, power dissipation in, 280-281

INDEX Wavelength: cutoff, 286 guide,2n of harmonic wave, 217 operating, 277 Waves: electromagnetic (see Electromagnetic waves) plane, 218 standing, 224-225 traveling, 226-227 Weber (unit), 140 Work: definition of, S9 done against electric field, 59 done by electric field, 59 done in moving point charges, 59 power and, 156-157 Zero, divergence of, 52 Zero reference at infinity, 61 Zero scalar, divergence of curl as, 139 Zero vector, curl of gradient as, 139