Chapter 34 by John Smith

ELECTROMAGNETIC FIELDS AND WAVES

w.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a "net pole" within a surface, Q,,, = f B . d i = 0 . Since the magnetic field is uniform over each face of the box, the total magnetic flux around the box is (1 cm x 2 cm)(-2 T - 2 T + 1 T + 5 T) -t(1 cm x 1 cm)(-2 T)+ Qunknown = 0

j E-,,

.arli = -0.0002 T m2 3 B ~ , , ,COS@ , ~ = -2.0 T

The angle 8 must be 180". Because 8 i s the angle between $ and the outward normal of d i , the field into the face.

B is directed

34.2. Model: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Visualize: Please refer to Figure Ex34.2. The electric field is in the direction of the positive y-axis and the magnetic field is in the direction of the negative z-axis. Solve: Substituting into the Lorentz force law, V C ) 1.0 x 1 0 6 j-+ (-2.0 x 1073m / s) x (-0.10i T)) F,, = q( E + C x E) = (1.6 x m = (3.2; 3

F,, =

+ 1.6j) x

N = 3.58 x

8=t a d (

1.6 x

N = 26.6" 3.2 x 10-1~ N

The angle 8 is above the positive x-axis.

34.3. Model: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Solve: Substituting into the Lorentz force law,

I$,,,= q ( E + G x E) = (-1.6 x

5.0 x IO6; m / s ) x (-0.lOi T

-3) N - (8.0 x 10-l.S) N= (-3.2

= -(3.2 x 10-14)(1

- 4.8 x 10-l4j)N

34.4. Model: Assume that the electric and magnetic fields are uniform fields. Visualize: Please refer to Figure Ex34.4. The magnetic force on the negative electron by the right-hand rule is directed downward. So that the electron is undeflected, we must apply an electric field to cause an electric force directed upward. That is, the electric field must point downward. Solve: For the electron to not deflect, F;, = 4 a clC x

= eE =j E = vBsin90" = (2.0 x 10' m / s)(O.OIO T) = 2.0 x 10' V /

34.5. Model: Assume the electric and magnetic fields are uniform. Visualize: Please refer to Figure Ex34.5. Solve: The force on the proton. which is the sum of the electric and magnetic forces, is

F = F, + F,

= -Fcos30째f

+ Fsin30'3

= (-2.77; +1.603)x

34-2

Since

Chapter 34

v’ points out of the page, the magnetic force is F, = e< x B = 1.66 x lO-I33 N.Thus

- - - .

= e E = F - F , = - 2 . 7 7 ~ 1 0 - ‘ ~ ~ N ~ ~ = ~ , / e = - 1 . 7V/m 3x10~~

That is, the electric field is

= ( 1 . 7 3 ~ 1 0V/m, ~ left).

34.6. Model: Assume that the electric and magnetic fields are uniform fields. Solve:

Substituting into the Lorentz force law, = q(E+ v’ x B) 3 (9.6 x

E = -(6.0

N)(; - i )= -(1.60 x

x io5 N / c)(; - i )- (5.0 x io5 N

.)[E+

(5.0 x lo6; m / s) x (0.103 T)]

c)L= (-6.0; + i.oL ) x io5 N c

34.7. Model: Apply the Galilean transformation of velocity. Solve: (a) In the laboratory frame S, the speed of the proton is v = j(1.41 x lo6 m / s)’ +(1.41 x lo6 m / s)’ = 2.0 x lo6 m / s The angle the velocity vector makes with the positive y-axis is

e = tan-’

1 . 4 1 ~ 1 0m~/ s = 45” 1 . 4 1 ~ 1 0m~/ s

(b) In the rocket frame S’, we need to f i s t determine the vector V’. Equation 34.5 yields:

V’ = V - = (1.41 x lo6; + 1.41 x 106y)m / s -(1.0 x lo6;) m / s = (0.41 x lo6; + 1.41x 106j)m / s The speed of the proton is v’=4(O.41x1O6 m/s)’ +(1.41x106 m/s)’ =1.47x106 m / s The angle the velocity vector makes with the positive y’-axis is 8’ = tan-’

0 . 4 1 ~ 1 0m~/ s = 16.2” 1 . 4 1 ~ 1 0m~/ s

34.8. Model: Apply the Galilean transformation of fields. Visualize: Please refer to Figure Ex34.8. Solve: (a) Equation 34.15 gives the Galilean field transformation equation for magnetic fields: - - I - B’= B - - V x E C2

B is in the positive

k^ direction,

must be in the negative

= B i . For B’> B,

direction, so that

x E must be in the negative

k^ direction. Since E = Ej ,

c x E = -(Vi^)x ( E j )= -VEL. The rocket scientist will measure B’ > B,

if the rocket moves along the -x-axis. (b) For B’= B, c x E must be zero. The rocket scientist will measure B’= B if the rocket moves along either the +y-axis or the -y-axis. (c) For B‘< B, x 2 must be in the positive k^ direction. The rocket scientist will measure B’< B, if the rocket moves along the +x-axis.

34.9. Model: Use the Galilean transformation of fields. Solve: Equation 34.15 gives the Galilean transformation equations for the electric and magnetic fields in S and S’ frames:

E‘ = E+ v x B In a region of space where

B = 6, E’ = E = -1.0

1 -

B’ = B - 7V x E C-

x 106jV / m. The magnetic field is

Electromagnetic Fields and Waves

34-3

34.10. Model: Use the Galilean transformation of fields. Visualize: Please refer to Figure Ex34.10. We are given ? = 2 x lo6; m / s , B'= 1.03 T, and E' = 1.0 x 106Lv / m. Solve: Equation 34.15 gives the Galilean transformation equations for the electric and magnetic fields in S and S' frames: = 2- Qx 6'

6 = B/+-v1 - x E' C2

The electric and magnetic fields viewed from earth are

E = 1.0 x 1 0 6 iV / m E = 1.03 T +

1 -(2

(2 x lo6; m / s)x(l.Oj T) = -(1.0 x lo6 V / m)L

x lo6; m / s) x (1.0 x 1 0 6 iV / m) = 1.0; T -

3 = 0.999983 T

'OI2

(3.0 x 10' m / s) Assess: Although B < B', you need 5 significant figures of accuracy to tell the difference between them. C2

34.11. Model: Use the Galilean transformation of fields. Visualize: Please referto FigureEx34.11. We are given

9 =1 . 0 ~ 1 0m ~ ;/ s , 6 = 0.50iT, and E = (&; +&f) x

lo6 V / m . Solve: Equation 34.15 gives the Galilean transformation equation for the electric field in the S and S' frames: E' = E + X 6 . The electric field from the moving rocket is

(;

+ j)0.707 x lo6 V / m + (1.0 x lo6; m / s) x (0.50i T) = (0.707 x lo6; + 0.207 x 1063)V / m 0 = tan-'

0.207 x lo6 V / m = 16.3' above the x'-axis 0.707 x lo6 V / m

34.12. Model:

A changing magnetic field creates an electric field. Visualize: Please refer to Figure Ex34.12 in your textbook. E ( X 10-4V/m)

0 ,

(s)

34-4

Chapter 34

The line integral of E is around a closed curve that bounds the surface through which the magnetic flux is calculated. Let us travel along the closed curve clockwise and assume that 2 is positive. This is the circular line with radius r, which is concentric with the solenoid implying E and & are oriented along the same direction. To find the sign of i.d i or Om,curl your fingers around the circle of radius r in the clockwise direction. Your right thumb points to the right, in the same direction as the magnetic field. Thus, Omis positive, or

IB.d;i = BA = Bm2.Therefore,

Thus, E=-

(4n x IO-’ T m / A)(O.OIm)(400) d l

2(0.20m)

Intheintervalr=Ostor=O.l s,

-=-dl 5 A - 5 0 A I s dr 0.1 s In the interval I = 0.1 s to 0.2 s, the change in the current is zero. In the interval t = 0.2 s to 0.4 s, dI/dt = -25 A I s. Thus in the interval I = 0 s to r = 0.1 s, E = -6.28 x lo4 Vlrn. In the interval t = 0.1 s to 0.2 s , E is zero. In the interval t = 0.2 s to 0.4 s , E = +3.14 x 10“ V/m. The sign change shows that E changes direction. However, the question asks only for the field strength, which is positive. The E-versus-t graph is shown earlier, in the Visualize step.

34.13. Model: A changing magnetic field creates an electric field. Visualize:

Solenoid

Solve: (a) In Faraday’s law (Equation 34.24), the line integral of 2 is around the closed curve that bounds the surface through which the flux is calculated. To apply Faraday’s law, choose a clockwise direction around a circle of radius r as the closed curve. We assume that the electric field vector is positive when it is along the same direction as ds (everywhere tangent to the curve). The line integral in this case is

$I?-& = E2m To find the sign of d i ,curl your fingers around the circle in the clockwise direction. Your right thumb points to the right, which is the same direction as the direction of the magnetic field. Thus, 2. d;i is positive. Equation 34.24 becomes d - dB r dB fE.ds=--jB.dAd E(~IY~)=-~T‘-* E = - - dr dt 2 dt For a point on the axis, r = 0 m, so E = 0 Vlrn. (b) For a point 2.0 cm from the axis, E = - - - =dB 2 dt

(“‘i -

m ) ( 4 . 0 T I s) = 0.040 V I m

Assess: As the magnetic field due to the clockwise current in the solenoid decreases, an induced electric field is set up to oppose the decrease in the magnetic field. This is in accordance with the Lenz’s law.

Electromagnetic Fields and Waves

34-5

34.14. Model: Assume the magnetic field inside the circular region is uniform and constant. The proton accelerates due to the force of the induced electric field. Visualize: Please refer to Figure Ex34.14. Solve: Equation 34.24 is Faraday’s law:

Consider a circle of radius r centered at point b as the closed curve and choose a clockwise direction for evaluating the emf. The above integral simplifies to

using the right hand rule. Using Newton’s second law, eE er dB a = -= m 2m dt

Note that the direction of the area is determined to be along F=eE=m*

At point a the acceleration is a=

-(1.60 x

C)(l.O x lo-’ m)

2(1.67 x lo-’’ kg)

At point b, a = 0 m/s2. At point c,

(-0.10 T / s)

a = (4.8 x lo4 m / s2,up)

a = (4.8 x IO4 m / s2, down). At point d, a = (9.6 x IO4 m / s2,down).

34.15. Solve: The units of Eo(d@,/dt)are

34.16. Solve: The displacement current is defined as Idisp = .co(d@Jdt). The electric flux inside a capacitor with plate area A is = EA. The electric field inside a capacitor is E = q/q = (Q/A)/c0,and thus the electric flux is

where V, is the capacitor voltage. The capacitance Cis constant, hence the displacement current is

34.17. Model: The displacement current is numerically equal to the current in the wires leading to and from the capacitor. Solve: During the charging process, a parallel-plate capacitor develops charge as a function of time. If the charge on a capacitor plate is Q at time t, then Q = CV,, where Vc is the voltage across the capacitor plates. Taking the derivative,

34.18. Model: The electric field inside a parallel-plate capacitor is uniform. As the capacitor is charged, the changing electric field induces a magnetic field. Visualize: The induced magnetic field lines are circles concentric with the capacitor. Please refer to Figure 34.25. Solve: (a) The Ampere-Maxwell law is dt where EA is the electric flux through the circle of radius r. The magnetic field is everywhere tangent to the circle of radius r, so the integral of around the circle is simply B(27rr). The Ampere-Maxwell law becomes dE r dE 2m-B = ~ ~ p ~ ( m ’ ) -B = &,/lo-dt 2 dt On the axis, r = 0 m, so B = 0 T.

Chapter 34

34-6

(5)At r = 3.0 cm, B=

(3.0 x 10' m / s)'

( F ) ( l . 0 x 1 0 6 V / m s ) = 1 . 6 7 ~ 1 0 - 'T~

(c) For r > 5.0 cm, the electric flux @e is the flux through a 10cm-diameter circle because E = 0 V/m outside the capacitor plates. The Ampere-Maxwell law is

B ( 2 ~ r= ) ~,p,d?'

R2 dE a B = â&#x201A;Ź,p0--2 r dt

(0.05 m)'

( 3 . 0 ~ 1 0m~/ s ) ' 2(0.07m)

dE dt

(1.0 x lo6 V / m s) = 1.98 x

34.19. Model: The displacement current is numerically equal to the current in the wires leading to and from the capacitor. Solve: The process of charging increases the charge on the plates of a parallel-plate capacitor. The charge Q on a capacitor plate at time t is Q = CV,, where V, is the voltage across the capacitor plates. Taking the derivative,

I disp

= I = -d= eC - - - dVc - - - â&#x201A;Ź,A dVc - (8.85 X lo-'' C2 / N m2)(0.05 m)' (500,000 V / s) = 22.1 pA dt dt d dt 0.50 x m

34.20. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related. Solve: Using Equation 34.40, Bo = -Eo _ c

10V/m = 3.3 x lo4 Vs / m2 = 3.3 x lo-' T 3x108m/s

34.21. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related. Solve: Using Equation 34.40, E, = cBo = (3.0 x 10' m / s)(2.0 x

T)= 6.0 x 10' V / m

34.22. Model: Electromagnetic waves are sinusoidal. Solve: (a) The magnetic field is B, = Bosin(& - OX),where Bo= 3.0 pT and k = 1.00 x lo7m-I. The wavelength is

A=-=27r k

27r = 6.28 x 1.00 x 10' m-'

m = 628 nm

(b) The frequency is c 3 . 0 ~ 1 0m~/ s f=-= = 4.77 x 1014HZ ?, 6.28x10-' m (c) The electric field amplitude is

E, = cB, = (3 x 10' m/s)(3.0 x lod T) = 900 V/m

34.23. Model: Electromagnetic waves are sinusoidal. Solve: (a) The electric field is E, = E, cos (kx - ut),where E, = 20 V/m and k = 6.28 x 10' m-'. The wavelength is

(b)The frequency is c

f=-=

3 . 0 ~ 1 0m~/ s = 3.0 x 10" Hz ~ . o o x ~mo - ~

vem

20V/m =6.67xlO-'T 3 . 0 ~ 1 0 'm / s

Electromagnetic Fields and Waves

34.24. Model:

and B are perpendicular to each other and

Solve: At any point, the Poynting vector .? E p o - ' i X traveling. We have 5= B i and = 4'. So, ,. 1 -Sj=-ExBi

34-7

is in the direction of Gm.

points in the direction that an electromagnetic wave is

*E=-Ei

PO The electric field points in the negative z-direction.

34.25. Solve: (a) The units of CB are m m -xT=-x

N N -(C)(m/s) C

These are the units for E. In the unit conversion, the units of tesla are taken from the equation (b) The magnitude of the Poynting vector is

F = 4; x B .

The units of S are

In the unit conversion, we have used both N/C and V/m for the units of the electric field. Since P = N for circuits, AV=W.

34.26. Model: The electric and magnetic field amplitudes of an electromagnetic wave are related to each other. Solve: (a) Using Equation 34.40, 1 100VIm = 3.33 x lo-' T Bo =-Eo = c 3.0x1O8m/s (b) From Equation 34.47, the intensity of an electromagnetic wave is

I = -C EE ~~ 2 O

(3 x 10' m / s)(8.85 x lo-'' Cz / N m 2 ) (100 V / m)2 = 13.3 W / mz

34.27. Model: A radio signal is an electromagnetic wave. Solve: From Equation 34.47, the intensity of an electromagnetic wave is (3 x lo8 m / s)(8.85x lo-'' C 2 / N m 2 ) CE I =A E 2 = (300 x 10" 2 O 2

v / m)'

= 1.2 x 10-10w / m2

34.28. Model: The laser beam is an electromagnetic plane wave. Assume that the energy is uniformly distributed over the diameter of the laser beam. Solve: (a) Using Equation 34.47, the light intensity is I = -P= - ECEO '= A 2 '

200 x lo6 W - (3 X 10' m / s)(8.85 x lo-" C' / N m') E; * E o =2.2x10" V / m n(1.0x 104 m)' 2

(b) The electric field between the proton and the electron is

E = - - - 1-

(9 x lo9 N m2 / Cz)(1.60 x

4 7 r~ 2 ~

(0.053 x

= 5 . 1 ~ 1 0 "V / m

m)'

The laser beam's electric field is approximately half the electric field that keeps the electron in its orbit.

34-8

Chapter 34

34.29. Model: A radio wave is an electromagnetic wave. Solve: (a) The energy transported per second by the radio wave is 25 kW, or 25 x lo3J/s. This energy is carried uniformly in all directions. From Equation 34.47, the light intensity is

(b) Using Equation 34.47 again,

' E,' * 2.21 x lod 2

I=-

W / m2 =

(3 x lo8 m / s)(8.85 x

C 2 / N m2)

E: a Eo = 0.041 V / m

34.30. Solve: (a) The energy transported per second by the wave is 10 W, or 10 J/s. This energy is carried uniformly in all directions. From Equation 34.47, the light intensity is

For E, = 100 V/m, r = 0.245 m. (b) For Eo = 0.010 V/m, r = 2450 m.

34.31. Model:

An object gains momentum when it absorbs electromagnetic waves. Solve: The radiation force on an object that absorbs all the light is 'Oo0 F = -P= = 3.3 x lo4 N c 3 . 0 ~ 1 0m~/ s

Assess: The force is independent of the size of the beam and the wavelength.

3432. Model: A polarized radio wave is an electromagnetic wave. Solve: (a) From Equation 34.40, Eo Bo=-c

- lOOOV/m 3.0~1m 0 ~/ s

= 3.33 x 10" T

(b) Likewise,

B = -E= c

5 0 0 V / m =1.67x1od T 3 . 0 ~ 1 0m~/ s

The direction of the wave is into the page ( -k^ direction) and ?i is down ( -j^ direction). Using the right-hand rule

and = po-'Ex E , E is to the left ( -Idirection). (c) The wavelength of the radio wave is 3 x 1 0 ~m / s - 3 x 1 0 ~m / s =300 f 1.0 x lo6 Hz The magnetic field has half its maximum value at a point where

cos - --a - = 1 . 0 5 r a d ~ x = O . l 6 6 7 A = 5 0 m (2:x)-:

34.33. Model: Use Malus's law for the polarized light. VkudbX:

Incident

Filter axis

Transmitted

Solve: From Equation 34.52, the relationship between the incident and the transmitted polarized light is Iuanrmlurd = Z,,cos'8, where 0is the angle between the electric field and the axis of the filter. Therefore,

0.25 I , = I , cos% 3 COS^ = 0.50

0= 60"

Assess: Note that 8 is the angle between the electric field and the axis of the filter.

Electromagnetic Fields and Waves

34-9

34.34. Model: Use Malus's law for the polarized light. ~ ~ a l i z e : Incident

Filter axis

polarized light

Transmitted light

-/ Solve: From Equation 34.52, the transmitted polarized light is = I, cos28= (200 mW)cos2(90"- 25") = 36 mW Assess: Note that 8 is the angle between the electric field and the axis of the filter, which is 90" - 25" = 65"

34.35. Model: Use Malus's law for polarized light. Vkualiize:

Unpolarized incident light

Filter axis

, I

Solve: For unpolarized light, the electric field vector varies randomly through all possible values of 8.Because the average value of cos2Bis the intensity transmitted by a polarizing filter is Zrransmlaed = 3Z,,. On the other hand, for polarized light Zuansmtted = I, cos%. Therefore, Zmmaed2 -Z~,,,,,cos'8=~Z,cos28=~(350W/rn2)cos230" = 131 W / m 2

Assess: Note that any particular wave has a clear polarization. It is only in a "sea" of waves that the resultant wave has no polarization.

34.36. Model: Use the Galilean transformation of velocity. Visualize: Please refer to Figure P34.36. We will define Mary's and Tom's frame to be S and Carlos' frame to be S'. S' is moving relative to S at 3 m/s along the x-axis. Solve: (a) The velocity vector of the ball in S is ; = ( 5 m / s ) ( c o s & + s i n $ ) = ( 5 m / s ) --I6 m - + -8j )m : = ( 3 . 0 ; + 4 . 0 j ) m / s 10m 10m (b) In the S frame, Carlos' xy-coordinates are (0 m, 0 m) at the instant Mary tosses the ball. In 2.0 s, Carlos has moved a distance: Ax = (v,)At = (3.0 m/s)(2.0 s) = 6.0 m. So, Carlos' coordinates when Tom catches the ball are (6.0 m, 0 m). (c) Carlos' frame, that is, frame S', is moving at 3.0 m / s relative to frame S. The ball is also moving along the x-direction at 3.0 m/s. So, Carlos will see the ball move straight out along the y'-axis until it is caught by Tom. Y' Caught by Tom

Carlos reference

Tossed by Mary &X -I

...

. .

. ....

34- 10

Chapter 34

(d) Carlos sees the ball take 2.0 s to travel 8.0 m along the y’-axis to Tom. Thus, the ball’s velocity in frame S’ is V’ = 4.0j m / s. (e) Using the Equation 34.5,

V’ = v’-

e = (3.0; + 4.01) m 1s-3.0;

m / s = 4.01111 / s

34.37. Model: Use the Galilean transformation of fields. Assume that the electric and magnetic fields are uniform inside the capacitor.

Visualize: Please refer to Figure P34.37. The laboratory frame is the S frame and the proton’s frame is the S‘ frame. Solve: (a) The electric field is directed downward, and thus the electric force on the proton is downward. The magnetic field i is oriented so that the force on the proton is directed upward. Use of the right-hand rule tells us that the magnetic field is directed into the page. The magnitude of the magnetic field is obtained from the equation evB = eE. Solving for B,

E B=-= v

l.0x105 V / m = 0.10 T 1.0~1m 0 ~/ s

n u s B = (0.10 T, into page). (b) In the S’ frame, the magnetic and electric fields are 1 B’= B - - V x E =

-0.10kT-

(1.0 x 1 0 6 jm / s) x (1.0 x io5; v / m)

(3.0 x 10’ m / s)’

E’ = E +

- v+

= 1.0 x 105i - (1.0 x lo6;

-0.10k T

V rn / s) x (-0.10i T) = 0 m

(e) There is no electric force in the proton’s frame because E’ = 0, and there is no magnetic force because the proton is at rest in the S’ frame.

34.38. Model: Use the Galilean transformation of fields. Assume that the wire is infinite. Visualize: Please refer to Figure P34.38. The laboratory frame is frame S and the circular loop’s frame is frame S’ Solve: (a) From Equation 26.15, the electric field E due to a wire at rest is

(-,

- A A= A away from wire E = -r 2 ~ , r 2m0r

The magnetic field E at a point on the loop is zero because there are no moving charges. These are the fields measured in frame S. (b) In frame S’, E’ = E + 3x = 2 because i = 0 T. The magnetic field in frame S’ is

(c) Consider a segment of the wire of length Ax with charge AQ = &. The experimenter in the loop’s frame sees a charge on this segment passing by himher with a velocity v to the left. Thus,

AQ = AA.X = Av I =At At (d) The magnetic field in the loop’s frame is due to current I: POI POAV POAV Eo B’ = -=27l7

21tr

Jkv

EOC2 21n

Since the current I is moving to the left, the right-hand rule states the direction of B’is into the page on the top. The electric field E‘ which is due to the charge on the wire would be equal to 2. (e) As we see, the results in parts (b) and (d) are the same. (f) There will not be any induced current if the loop is made of a conducting material. This is because (i) the field is perpendicular to the area of the loop and (ii) the flux is not changing.

Electromagnetic Fields and Waves

34- 1 I

34.39. Model: Use the Galilean transformation of fields. Visualize:

Current

@;page

Wire View from left end

A current of 2.5 A flows to the right through the wire, and the plastic insulation has a charge of linear density A = 2.5 n C/cm. Solve: The magnetic field B at a distance r from the wire is

clockwise seen from left On the other hand,

E is radially out along i ,that is, E- = - ? . A , 2n~,r

As the mosquito is 1.O cm from the center of the wire at the top of the wire,

(4n x lo-’ T m / A)(2.5 A )

2n(O.O1 m)

= 5.0

x 10” T

(2.5 x IO-’ C / m)(2)(9.0 x IO9 N m’ / C ’ )

0.01 m

= 4 . 5 ~ 1 0~

where the direction of is out of the page and the direction of us call it SI, we want if= 0 T.n u s ,

2is radially outward. In the mosquit

’s frame (k

Because \7 X E must be in the direction of and E is radially outward, according to the right-hand rule be along the direction of the current. The magnitude of the velocity is

? must

(3.0 X lo8 m / s)*(5.0x lo-’ T ) V = -c=2 B =1.ox1o7 m / s E 4 . 5 ~ 1 0V~/ m The mosquito must fly at 1 .07m / s parallel to the current.

34.40. Model: uniform. Visualize:

Use Faraday’s law of electric induction and assume that the magnetic field inside the solenoid is Solenoid I

, 0

Equation 34.24 for Faraday’s law is

f E . u5 = -”[, 5.&] dt

Chapter 34

34- 12

To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric field vectors, as the figure shows, are everywhere tangent to the curve. The line integral of E then is

$i?.dS = E(2m) To do the surface integral, we need to know the sign of the flux or the integral JE.&. Curl your right fingers around the circle in the clockwise direction. Your thumb points to the right, which is along the same direction as the magnetic field 5.That is, ji.d;i = B m 2 is positive. Solve: (a) Since B = 10.0 T + (2.0 T)sin[2n(lO Hz)t], Faraday’s law simplifies to dB = -m2(2.0 T)[2~(10Hz)]cos[~z(~OHz)~] E(2m)= --Itr 2 dt

The field strength is maximum when the ,cosine function is equal to -1. Hence at r = 1.5 cm,

m) = 0.94 V

E,, = (20.0 -$)n(O.O15

m (b) E is maximum when cos[27dlO Hz)t] = -1 which means when sin[2@10Hz)t] = 0. Under this condition,

B = (10.0 T) + (2.0 T) sin [27r(10 Hz)t] = 10.0 T That is, B = 10.0 T at the instant E has a maximum value of 0.94 V/m.

34.41. Model: Use Faraday’s law of electric induction. Visualize:

Solenoid

5 Equation 34.24 for Faraday’s law is

d f E . & = --[j dt

.GI

To solve this equation, choose a clockwise direction around a circle of radius r as the closed curve. The electric field vectors, as can be seen from the figure, are everywhere tangent to the curve. The line integral of E then is

$E.&

=(2m)E

To do the surface integral, we need to know the sign of the flux or the integral d i . Curl your fingers around the circle in the clockwise direction. Your thumb points to the right, which is along the same direction as the magnetic field 5.That is i.d i = BzR’ is positive. Note that B = 0 outside the solenoid. Solve: Faraday’s law now simplifies to ,dB R’ dB E(2m) = zR‘ -jE = dt 2r dt Assess: At r = R, E = % RdB/dt ,which is the same result as Equation 34.27.

Electromagnetic Fields and Waves

34- 13

34.42. Model: Assume a uniform electric field inside the capacitor. The displacement current Iaspbetween the capacitor plates is numerically equal to the current I in the wires leading to and from the capacitor. Solve: (a) From Equation 31.40, the voltage across the capacitor that develops during charging is V, = &(I - 8 9 , where T = RC and & = 25 V is the battery emf. Since Q = CV,, we have

The maximum electric flux through the capacitor plates is

-(O.K)XIO-~

S)/(Z.SXIO-''

ison)

= 5.20 V m

Likewise, from part (a) the displacement current is

IdSP

= -C&e-iir = (0.167 A)e

- ( o . ~ o x ~ os- ~ ) / ( ~ . ~ x I o -F' ~x

ison)

= 0.044 A

34.43. Model: Use Equation 34.3 1 for the definition of the displacement current. Solve: The current in a conductor arises from the electric field E in the conductor. From Equation 28.18,

where = EA is the electric flux through the wire and, by definition, Ihsp = E, d@,/dt. Thus Ihsp = (E,/o)dI/dt. (b) Using the value for the conductivity of copper wire from Table 28.2, E, d l 8.85 x IO-" C' / N m' IdlSP = -- = ( 1 . 0 IO6 ~ A1s)= 1.48~ Q dt 6 . 0 107R-'m-' ~

34.44. Model: Assume the electric field inside the capacitor is uniform and use the Ampere-Maxwell law. Visualize:

r = 2 m

R = S m

Solve: (a) For a current carrying wire, Example 32.3 yields an equation for the magnetic field strength:

T m / A)(10 A) B - p 0 I -Ire - (471x = 1 . 0 ~ 1 0 -T~ 2m (27~)(2.0 x IO-' m) (b) Example 34.5 found that the induced magnetic field inside a charging capacitor is

where we used I,,, = dQ/dt as the actual current in the wire leading to the capacitor. At r = 2.0 mm, B=

(47r x IO-' T m / A)(2.0 x lo-' m) (10 A) = 1.6 x loJ T 771 (5.0 x 10-j m)'

34-14

Chapter 34

34.45.

Model: Assume that the electric field inside the cylinder is uniform. Use the Ampere-Maxwell law to obtain the induced magnetic field. Visualize:

di.For a closed curve of radius R, assuming a Solve: (a) The electric flux through the entire cylinder clockwise direction around the ring, the right-hand rule places the thumb into the page, which is the same direction as that of the electric field E. Thus, we will take the flux CP to have a positive sign: CP = E ( @ * ) = (1.0 x 1OXt2 v ) [ x ( 3 . 0 x

m y ] = (2.83 x lo3?) Vm

(b) From the Ampere-Maxwell law, with Z ~ u g=h0,

Because CPe is positive and the field strength increases with time, d@,/dt is positive. The line integral is therefore positive, which means that B is in the clockwise direction. This reasoning applies to both the r < R and r > R regions. Thus the field lines are clockwise and they are shown in the figure. (e) We can now make use of the Ampere-Maxwell law to find the magnitude of B for i- < R. Based upon parts (a) and (b), B ( 2 m ) = E&,

(i)? +[(

c ( m 2* ) B = â&#x201A;Ź,p0 dt

1.0 x 10s

321

= (1.1 1x 10-9 rt) T

where r is in m and t is in s. At Y = 2 mm and t=2.0 s,

B = (1.11 x 104)(2 x 10-3)(2.0)T = 4.44 x lo-'' T (d) For Y > R, the flux is confined to a circle of radius R. Thus B ( 2 m ) = go,u0( $)d2 B = 2c2r[(l.OxlOx ~)t]=(L.OOxlO-"

where r is in m and t is in s. At Y = 4 mm and t = 2.0 S,

34.46. Model: Light is an electromagnetic wave. A quick measurement of a light bulb shows that its radius is r -2.8 cm. Solve: 100 W is the energy transported per second by the electromagnetic light wave. This energy is carried in all directions. The light intensity is given by Equation 34.47:

(3 x 10' m / s)(8.85 x lo-'' -

IOOW

4n(2.8 x lo-' m)'

-Bo='.=( c

C' I N m')

E,'*

2800Vlm = 9.3 x l o d T 3 . 0 ~ 1 0 'm l s

E, = 2800

V m

Electromagnetic Fields and Waves

34-15

34.47. Model: Sunlight is an electromagnetic wave. Solve: (a) The sun's energy is transported by the electromagnetic waves in all directions. From Equation 34.47, the light intensity is

P I =-

P = IA = (1360 W / m2)(4zR2.,,,)

= (1360 W / m2)4z(1.50x 10" m)' = 3.85 x 10" W

(b) The intensity of sunlight at Mars is I=-.= p A

(1360 W / m*)(47rR:aZ,,) - (1360 W / mi)( 1.50 x 10" m 4&& 2.28 x 10" m

= 589 W / m2

34.48. Model: Radio waves are electromagnetic waves. Solve: (a) The radio transmitter is radiating energy in all directions at the rate of 21 J per second. The signal intensity received at the earth is given by Equation 34.47: 21 w I = - -P- --= P = 8.252 x W / m2 A 4 m 2 4 n ( 4 . 5 ~ 1 0 'm) ~ (b) Using Equation 34.47 again,

2(8.252 x W / m') = 7.89 x lo-'' i ( 3 . 0 x 10' m / s)(8.85 x IO-" C 2 / N m 2 )

I = ~ E ? + E = -= 2 O

V m

34.49.

Model: Radio waves are electromagnetic waves. Assume that the transmitter unit radiates in all directions. Solve: The transmitting unit radiates energy in all directions at the rate of 250 mJ per second. From Equation 34.47, the signal intensity at a distance of 42 m is

Using Equation 34.47 again, ~

I=OE**E= 2 O

E -=

2(1.13 x 10" W / m 2 ) V = 0.092 m j ( 3 . 0 x 10' m /s)(8.85 x lo-'' C2 / N m')

A few steps before 42 m, the field strength was 0.100 V/m and the door opened. The manufacturer's claims are correct.

34.50. Model: The earth is a complete absorber of sunlight. An object gains momentum when it absorbs electromagnetic waves. Solve: The radiation pressure on an object that absorbs all the light is =:=

1360 m 2 = 4.533 x lo4 Pa 3 . 0 ~ 1 0m~/ s

Seen from the sun, the earth is a circle of radius Re,

F , = p,A

and area A = M L . The pressure exerts a force on this area

prad(&L) = (4.533 x 10"

Pa)n(6.37 x lo6 rn)2= 5.78 x lo8 N

The sun's gravitational force on the earth is F

,ray

G'SUMeaih

Rs2un.b

= (6.67 x IO-" N m' / kg2)(1.99 x IO3' kg)(5.98 x (1.50 x 10" m)'

Fmd - 5 . 7 8 ~ 1 0 'N *-= 1.64 x 10-l4 Fg,, 3.53 x 10'' N That is, Fmd is 1.64 X lo-'' % of F,,,.

kg) = 3.53 x 10" N

34-16

Chapter 34

34.51. Model: Assume that the black foil absorbs the laser light completely. Use the particle model for the foil. visualize:

FIad k foil

For the foil to levitate, the radiation-pressure force must be equal to the weight of the foil. Solve: Using 34.49,

&, = p d A

= - A = - * P = CFrad = cFWeigh, = (3.0 x 10’ ml~)(25x

34.52. Model: Assume Visualize:

that

IOd kg)(9.8 mls’) = 73.5 W

the black paper absorbs the light completely. Use the particle model for the paper.

Light

For the black paper to be suspended, the radiation-pressure force must be equal to the weight of the paper. Solve: From 34.49, Fnd= p n d A= ZA/c.Hence,

1 z -CF A

= - CF A

(3.0 x 10’ m / s)(1.0 x IO-’ kg)(9.8 m / s’)

= 4.87 x 10’

(8.5 inch x 11 inchI(2.54 x lo-’ m/inch)-

rn2

34.53. Model: Assume that the block absorbs the laser light completely. Use the particle model for the block. Solve: From Equation 34.49,

Applying Newton’s second law, Fmd= ma = (100 kg)a

* a = 100kg = 0.0833 100kg

= 8.33 x lo4 m 1 s2

From kinematics,

+ 2a(s, - s,) = 0 rn’ / s’ + 2(8.33 x lo4

v: = v , ~

m / s2)(100m) J V , = 0.408 m/s

Assess: This does not seem like a promising method for launching satellites.

Model: Use the particle model for the astronaut. Solve: According to Newton’s third law, the force of the radiation on the astronaut is equal to the momentum delivered by the radiation. For this force we have ’OoO F = p radA = -P= = 3.333 x lod N c 3.0~1m 0 ~/ s

34.54.

Using Newton’s second law, the acceleration of the astronaut is

Using v1 = 1’) + a(t,- 1,) and a time equal to the lifetime of the batteries,

v,= 0 d~ (4.167 x 10“ m/s2)(36O0S ) = 1.500 x lO-‘m/s

The distance traveled in the first hour is calculated as follows: v;

* (1.500 x KO-‘ d s ) ’

- (0 d s ) ’

- v;

= 2a(4s)nrslhow

= 2 ( 4 . 1 6 7 ~10“ ds’)(aS),,

hour

(&)S)firsl

hour

= 0.270 m

Electromagnetic Fields and Waves

34-17

This means the astronaut must cover a distance of 5.0 m - 0.27 m = 4.73 m in a time of 9 hours. The acceleration is zero during this time. The time it will take the astronaut to reach the space capsule is At =

4.73 m =31,533s = 8.76hours 1 . 5 0 0 ~ 1 0m~l s

Because this time is less than 9 hours, the astronaut is able to make it safely to the space capsule.

34.55. Model: Use Malus's law for the polarized light. Visualize: Incident

unpolarized light

Filter axis

i& _-

- Io 1 - 1

/ Q

I? = c. I2c cos? + Bz

, '/II

I* = I ,

COS*

I I

Filter axis

Solve: For unpolarized light, the electric field vector varies randomly through all possible values of 8. Because the average value of cos% is %, the intensity transmitted by a polarizing filter when the incident light is unpolarize is I, = 41,. For polarized light, Itranamtled = I , cos'@. Therefore, I, = 1,COS' 45"

I3 = 12cos245"

34.56. Model: Assume that the electric field inside the capacitor is uniform. Use the Ampere-Maxwell law to find the magnetic field. Visualize:

2.0 cm

(PS)

Switch

Solve: (a) This is an RC circuit with capacitance C = .@Id = qrR2/d= 1.11 x lo-'' F = 11.1 pF. We know from Chapter 31 that the current through the wire decays exponentially as I,,,

A& - I / T (5000 A)e-1'7 = -e R

where the time constant is Z = RC = 2.22 x lo-'' s = 2.22 ps. Example 34.5 found that the induced magnetic field strength at radius r inside a charging or discharging capacitor is

= dQ1dt as the actual current in the wire leading to the capacitor. Thus where we used I*,m

(b) The graph is shown above.

___

34- 18

34.57.

Chapter 34

Solve: (a) The electric field energy density is uE= 2E,E' 1 and the magnetic field energy density is uB =

(1/2p,,)B2. In an electromagnetic wave, the fields are related by E = cB. Using this and the fact that e' = l/(%), we find = -B2 C2E, =1 "0 uE = LE E 2 = -(e-)' Eo ~2 = - ~ 2 = uB 2 2 2 2EoPo 2/42 (b) Since the energy density is equally divided between the electric field and the magnetic field, the total energy

density in an electromagnetic wave is u,, = uE + uB= 2u,. We also know that the wave intensity is I = T1 Thus

~ E,'.

34.58. Model: Dust particles absorb sunlight completely. Solve: Let R be the radius of a dust grain and p its density. The area of cross section of the dust particles is 4&, its mass is m = p ( 9 R'). Letting d be the distance of the dust grain from the sun, the gravitational force on the dust grain is

F gra"

=-G M P d2

On the other hand, the pressure on the grain due to sun's electromagnetic radiation is

For the dust particles to remain in the solar system over long periods of time, Frad< FpY. Hence,

-R'e,,,, <s= G M m GMsp4nR3 ~ d'

3d2

3, 41rcGMsp

3(3.9 x lo26w)

* R > 4rr(3.0 x 10' m / s)(6.67 x lo-''

N m2 / kg')(1.99 x lo3' kg)(2000 kg / m3)

=1 . 1 7 ~ 1 0 m~

For R > 1.17 pm, radiation pressure force is less than the sun's gravitational force, so the particle can orbit the sun. However, for R < 1.17 pm, radiation pressure force is greater than the sun's gravitational force, so the particle is gradually pushed out of the solar system. Thus, the diameter of the dust particle that can remain in the solar system for a very long period of time is 2.34 pm.

34.59. Model: Use Ampere's law and assume that the current through the resistor is uniform. Visualize:

Solve: (a) The electric field E in the resistor is

Electromagnetic Fields and Waves

34-19

For the magnetic field, we use the Ampere-Maxwell law:

POI

$ B . & = / L , I ~=~p ,O, ~ I aB ( ~ R I - ) = / . L , I -B = - -

(b) The Poynting vector

&‘EX

Bsurface

5 is directed into the curved surface. That is, into the wire. The magnitude is 1 S = -EB /.Lo

1 IRp0I = ----12R

po L 2 m

2mL

.?.d;i+

js.dA= faces

I’ R s.d;i=O+S&,, =-(2mL)= Wall 2mL

12R

34.60. Model: Use Malus’s law for the polarized light. Solve: For unpolarized light, the electric field vector varies randomly through all possible values of 8. Because the average value of cos% is 5, the intensity transmitted by the first polarkzing filter is I, = 3I,. For polarized light, I, = Io cos%. For the second filter the transmitted intensity is I? = I , cosze= +I, cos% Similarly, I3 = I? cos% = 3I, (cos2@?,and so on. Thus,

The Poynting vector is the power per unit area carried by electric and magnetic fields. Thus the Poynting flux (SAwa,,)is electromagnetic power directed into the resistor. It matches the power dissipated by the resistor (I’R), which is what we expect from energy conservation.