B

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Question set ’B’ B(1) (a) Write down the Coulomb’s law of force.

Coulomb’s law of force claims, that interaction force between two point charges in vacuum directed along the line, connecting those particles, proportional to their chagres q1 and q2 and inversely proportional 2 . This is attraction force, if charges have different signs, and to the squared distance between them r1,2 repulsion, if charges have similar sign. Mathematically it is: F = ke

q1 q2 , 2 r1,2

(B.1.1)

1 ≈ 9 × 109 N · m2 /C 2 is so called Coulomb constant. where ke = 4πε 0 (b) Three point charges are located as shown on the figure. Q1 = 3 × 10−6 C, Q2 = 2 × 10−6 C and Q3 = 5 × 10−6 C. Find the magnitude and direction of the force on Q3

√ The distance from the first to the third is l13 = 0.042 + 0.032 = 0.05 m. The magnitude of force from the first charge: Q1 Q3 3 × 10−6 × 5 × 10−6 = 9 × 109 = 54N. F13 = ke l13 0.052 The magnitude of force from the second charge: F23 = ke

Q2 Q3 2 × 10−6 × 5 × 10−6 = 9 × 109 = 36N. l13 0.052

Due to law of cosins, the resulting force will be equal F ≈ 83 N. The direction see on the figure. (c) Write down Gauss’ Flux Theorem The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Mathematically this can be written as I Q E · dA = , (B.1.2) ε0 S where E is the electric field, dA is a vector representing an infinitesimal element of area. (d) Find the E field at a distance of 3 mm from a very long wire charged to 5 × 10−6 C/m.

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To find the E-field will build a cylinder with bases perpendicular and radius r to the wire and with height l.

In this case the flux trough the lateral surface of it will be equal Φ = E × 2πrl From other side, according to the Gauss’ theorem, it equals: Φ=

q λl = ε0 ε0

After equating those two expressions, will get: E=

λ 5 × 10−6 = ≈ 3 × 107 V/m 2πε0 r 2 × 8.85 × 10−12 × π × 3 × 10−3

(e) What is the magnitude and direction of the force on a small particle chared to 2 × 10−6 C at a distance from the charged wire of question 1 (d)?

The force equals F = qE = 3 × 107 × 2 × 10−6 = 60 N, it’s directed out of the wire.

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B(2) The figure shows a cross section of a very long coaxial cable. The inner conductor is of radius of r=a. The outer copper mesh has inner radius b as shown. The current in the inner conductor has direction shown.

(a) Draw on a suitably labelled sketch the B-field in and around the cable.

(b) State Ampere’s Law. The magnetic field in space around an electric current is proportional to the electric current which serves as its source, just as the electric field in space is proportional to the charge which serves as its source. Ampere’s Law states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop. Mathematically this can be written as: I B · d` = µ0 Ienc (B.2.1) C

(c) Using Ampere’s Law, or some other suitable method, derive an expression for the H field in the regions: a < r < b, outside the cable.

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H2πr = I I H= 2πR Outside the cable there’s no magnetic field. (d) Calculate the inductance of the cable per unit length. Φ I To find flux, will define a surface, through wich it flows like shown on a figure and set l to 1.

L=

Z Φ=

I BdS = µ0 2π

Zb

µ0 I b 1 dR = ln R 2π a

a

µ0 b ln 2π a (e) Why is the inductance of cable important?

Cables mostly carry AC, and the inductance acts as a resitance for it.

L ==

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B(3) The figure shows a cross section of a parallel plate capacitor. The plates are of area A and the distance between them is a+b as shown. The space 0 < y < b is filled with dielectric of relative permittivity ε2 and the space b < y < a + b is filled with dielectirc of relative permittivity ε1 .

(a) Draw on a suitably labelled sketch the E field between the plates.

(b) Draw on a suitably labelled sketch the D field between the plates.

(c) Using Gauss’ Flux Theortm, or some other suitable method, derive an expression for the E field in the two regions: 0 < y < b, b < y < a + b To find the E-field (in region 0 < y < b) will build a cylinder with bases symmetrical to upper plane as shown on a figure.

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If area of base is S, electrical flux through one base equals ES, through both – 2ES. The flux through lateral surface equals 0, as E and n are perpendicular. Thus, the flux equal Φ = 2ES From the other side, according to the Gauss’ theorem, it equals: Φ=

q AQf = ε0 ε0 ε2

After equating those two expressions, will get: Eup =

AQf 2ε0 ε2

From the same considerations we will get the field created by the lower plane: Edown = −

AQf 2ε0 ε2

E1 = Eup − Edown =

AQf ε0 ε2

Similary, E2 =

AQf ε0 ε1

(d) Calculate the capacitance of the device.

C=

QF U

Za+b Zb Za+b AQf a b U= Edy = E1 dy + E2 dy = E1 a + E2 b = ( + ) ε0 ε2 ε1 0

0

b

C=

ε0 a b ( + )−1 A ε2 ε1

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B(4) A long power line carries 1000 A at 50 Hz. (a) What is approximately the maximum voltage which could be induced in a 30 turn coil of area 50 mm2 which is situated 5 m away from the line?

The B-field at a distance R from wire see ((A.4.b)) B = µ0

i I sin 2πνt = µ0 2πR 2πR

Due to Faradey’s law ε = N µ0

I2πν cos 2πνt Iν cos 2πνt dΦm = N µ0 S = N µ0 S dt 2πR R

Emf would maximum, when cosine equals 1 εmax = N µ0

Iν 1000 × 50 S = 30 × 1.25 × 10−6 5 × 10−5 = 2 × 10−5 V. R 5

(b) State what assumptions you have made about the orientation of the coil for maximum coupling.

For maximum coupling the coil plane should be parallel to the wire. (c) If this induced voltage is a problem what orientation of coil minimizes the effect, assume that you cannot move the coil further away from the line.

Perpendicular to the wire, that would minimize the magnetic flux.

The figure shows a very long wire carrying a current i = I sin ωt. A small rectangular coil of 60 turns is placed in the same plane as the wire as shown.

(d) Find an expression for the total magnetic flux Φ linked by the rectangular coil.

Magnetic flux through little area dS equals dΦm = N BdS . As magnetic field changes only in direction from the wire in a plane of coil: dΦm = N Bhdx Za+b Za+b µ0 I sin ωt 1 µ0 I sin ωt a+b Φm = N Bhdx = N h dx = N h ln 2π x 2π a a

a

8 (e) Find an expression for the emf induced in the rectangular coil.

According to Faradey’s law, the emf equals ε = −N

µ0 Iω cos ωt a+b dΦm = −N h ln dt 2π a

(f ) Find an expression for the mutual inductance between the rectangular coil and the long wire.

Mutual inductance is derived from expression: Φm = Li L=

µ0 h a + b Φm =N ln i 2π a

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B(5) The figure shows a small magnet. The two coils each have 500 turns, are in series and produce magnetic flux in the same direction. The relevant magnetisation curve for the magnet steel is given by the expression: 0.98H T B= 1460 + H

Dimensions of the magnet are as follows: l1 = 270 mm, l2 = 60 mm, lg = 4 mm.

Area A1 = 1600mm2 , area A2 = 4000mm2 . (a) Write down an equivalent circut for finding the magnetic flux in the magnet.

Here F is magnetomotive force, R â&#x20AC;&#x201C; reluctance. Those are analogs to emf and resistance in electric circuits (b) Find the current in the coils for an air gap flux density of 0.5 T. From the Ampereâ&#x20AC;&#x2122;s circuital law 2N I = H1 l1 + H2 l2 + Hg lg

10 From the Gauss’ theoreme for magnetic field: I

~ s=0 Bd~

S

Φg = Φ2 = 2Φ1 2B1 A1 = Bg A2 = B2 = A2 B2 = Bg = B B1 = Bg

A1 B = 2A2 5

As the cross sections of gap and core are equal H2 =

1460B 0.98 − B

Hg = H1 = B I= 2N

B µ0

292B 0.98 − B/5

lg 1460 292 + + µo 0.98 − B 0.98 − B/5

(c) State any assumptions you have made.

Used Ampere’s law and Gauss Theorem.

≈ 3.3A.