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Question set C C(1) (a)Write down the Coulomb 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 to the squared distance between them r1,2 . This is attraction force, if charges have different signs, and repulsion, if charges have similar sign. Mathematically it is: q1 q2 (C.1.1) F = ke 2 , r1,2 1 where ke = 4πε ≈ 9 × 109 N· m2 /C2 is so called Coulomb constant.

0 (b) Two electorns of charge −1.6 × 10−19 C are 3 × 10−5 m apart. What is the force on each?

Accordingly to formula (C.1.1), the force will be equal

F = 9 × 109

(−1.6 × 10−19 )2 = 2.56 × 10−19 N. (3 × 10−5 )2

(c)Write down Gauss’ flux theorem as it would be applied to the D field.

The total of the electric displacement flux out of a closed surface is equal to the charge enclosed divided by the permittivity. Mathematically this can be written as I D · dA = Q, (C.1.2) S

where D is the electric field, dA is a vector representing an infinitesimal element of area. (d) Using Gauss’ flux theorem, derive an expression for the D field at a distance from a large plane charged to σ C/m2 . The plane is immersed in a fluid of relative permittivity εr .

To find the D-field will build a cylinder with bases symmetrical to the plane.

If area of base is S, electrical displacement flux through one base equals DS, through both – 2DS. The flux through later surface equals 0, as D and n are perpendicular. Thus, the flux equal N = 2DS From other side, according to the Gauss’ theorem, it equals: N = q = σS After equating those two expressions, will get: D=

σ 2

As we can see, electric displacement doesn’t depend on the distance to the plane. The direction is out from the plane. (e) Find the E field at a distance of a large plane charged to σ C/m2 . The plane is immersed in a fluid of relative permittivity 6.

The electric field is connected to the electric displacement by expression: E=

D . ε0 ε

E=

σ 2ε0 εr

Using the expression we got above, will get

2 (f ) Calculate the E field at a distance of 2×10−3 m from a large plane charged to 5×10−5 C/m2 .

The plane is immersed in a fluid of relative permittivity 6.

Using the expression we got above, will get E=

5 × 10−5 = 4.7 × 106 V/m.

2 × 8.85 × 10−12 × 6

(g) What is the magnitude and direction of the force on a small particle charged to 8 × 10−6 C at a distance of 2 × 10−3 m from the charged plane of question 1(e)?

The force equals F = qE = 8 × 10−6 × 4.7 × 106 = 37.6 N.

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C(2) The figure shows a cross section of a very long coaxial cable. The inner conductor if of radius r = a. The space a < r < b is filled with dielectric of relative permittivity ε1 and the space b < r < c is filled with dielectric of relative permittivity ε2 . The outer conductor has inner radius c. The cable is grounded as shown and the inner conductor is charged with λ C/m.

(a) Draw on a suitably labled sketch the E field in and around the cable.

(b) Draw on a suitably labled sketch the D field in and around the cable. (c) Using Gauss’ Flux Theorem or some other suitible method, derive an expression for E field in the three regions: a < r < b, b < r < c, c < r < ∞.

It’s convenient to find at first electric displacement flux in the regions, as the desplacement and the ~ = ε0 εE, ~ where ε is a relative permittivity. field connected by the expression D Thus, is we will take a cylindric surface, coaxial to our cable, the flux through it will be equal (flux will be only through lateral surface of cylinder): I ~ s = D2πrl N= Dd~ S

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From Gauss’ theorem, it will be equal: N = λl After equating those expression, we will get: λ 2πr The electric displacement will be the same throughout the thickness of cable, and will be equal 0 outside.

Thus, λ λ , E2 = , E3 = 0 E1 = 2ε0 ε1 πr 2ε0 ε2 πr (d) Derive an expression for the capacitance of the cable per unit length.

By default, the capacitance is equal q C= U , where U is potential difference between capacitor plates, and q – the charge, in our case it equals λ (per unit length). The potential difference Zr2 U = E dr D=

r1

where r1 – radius of inner conductor, r2 – radius of outer conductor. In our case: Zb U=

λ dr + 2ε0 ε1 πr

Zc

a

λ λ b c dr = (ε1 ln + ε2 ln ) 2ε0 ε2 πr 2πε0 a b

b

Thus, the capacitance per unit length equals: C=

q 2πε0 = b U ε1 ln a + ε2 ln cb

(e) If a=0.5 mm, b=1.5 mm, c=2 mm, ε1 = 2, ε2 = 4, calculate the capacitance of the cable for a 5 m length.

Using the expression recieved in previous point: 2 × 3.14 × 8.8 × 10−12 × 5 C= ≈ 83pF 2 × ln 3 + 4 × ln 1.33

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C(3) The figure shows a cross section of a very long coaxial cable. The inner conductor is of a radius 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 labled sketch the B field in and around the cable.

(b) State the 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

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

B2πr = µ0 I B=

µ0 I 2πR

Outside the cable there’s no magnetic field. (d) Find an expression for the inductance of 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) If a=0.5 mm, b=2.0 mm, calculate the inductance of 5 m cable.

L ==

L=5×

1.25 × 10−6 ln 4 = 1.4 × 10−6 H. 2 × 3.14

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C(4) A long power line carries about 500 A from a wind turbine at 50 Hz. In order to develop a charging algorithm, it is decided to monitor the current by measuring the voltage induced in a 200 turn square coil of side 300 mm. (a) Assuming that no part of coil can be closer to the line than 1 m, show by means of a sketch the optimal position for the coil in order to sense the maximum voltage.

(b) Using the Ampere’s Law, find an expression for magnetic field B at a point P distance R from the line.

It follows from Ampere’s law the element of length carrying the current i induces the magnetic field at point µ0 idl sin θ , dB = 4π r2 where r – the distance between the current element and the point, θ –the angle between element and the point.

The sketch shows

dl sin θ dl sin α ds = = = dα; r r r

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R cos α

Thus, dB =

µ0 idl sin θ iµ0 cos αdα = 4π r2 4πR π

B=

Z2

µ0 i 4πR

cos αdα =

i µ0 I sin 2πνt = 2πR 2πR

− π2

(c) Find an expression for the total magnetic flux Φ linked by square coil.

Magnetic flux through little area dS equals dΦm = BdS . As magnetic field changes only in direction from the wire in a plane of coil: dΦm = N Badx R+a Z

R+a Z

R

a

µ0 I sin 2πνt Badx = N a 2π

Φm = N

µ0 I sin 2πνt R+a 1 dx = N h ln x 2π a

(d) Find an expression for the emf induced in the square coil.

According to Faradey’s law, the emf equals ε = −N µ0

dΦm R+a = −N µ0 Iν cos 2πνt ln dt a

(e) Calculate the emf induced in the squre coil when the line current is 500 A. ε = 200 × 1.25 × 10−6 × 500 cos 100πt ln 34 = 0.44 cos 100πt (f ) It is later found that the frequency of the current in fact varies between ± 5 Hz of the required 50 Hz. What are the implications for the charging algorithm?

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C(5) The figure shows a complete thoroidal iron core, uniformly wound with 200 turns of wire.

The relevant magnetisation for the magnet steel is given by the expression: B=

0.98H T 1460 + H

(a) On a suitably labelled sketch show the direction of the B-field in the ring.

(b) From the Ampere’s Law, derive an expression for the magnetic flux Φ in the ring.

The magnetic flux equals: ZZ ΦB = B · dS S

In this case: ΦB = B × πr2 To find find B-field, at first it’s needed to find H-field ditribution. For thoroid it’s equal: H=

NI 2πr

10 Using the relevant magnetisation formula given above: NI 0.98 2πr 0.98N I πr2 ΦB = πr2 = NI 1460 × 2πr + N I 1460 + 2πr

(c) Find the current in the coil for a flux density of 0.5 T. From relevant magnetisation formula 1460B H= 0.98 − B From other side NI 200I 105 H= = = I 2πr 2π0.01 π Equating those two expression gives I=

1460 × 0.5 π ≈ 0.05 A. 0.98 − 0.5 105

(d) If an air gap is made by removing a 2 mm thick radial slice of iron as shown by the dashed lines A-C on the figure, what current is required to maintain the same 0.5 T in the iron ring?

From the Ampere’s circuital law N I = Hc lc + Hg lg From the Gauss’ theoreme for magnetic field: I

~ s=0 Bd~

S

Φc = Φg Bc Sc = Bg Sg As the cross sections of gap and core are equal Bc = Bg = B 1460B 0.98 − B B Hg = µ0 1460B B NI = (2πr − lg ) + lg 0.98 − B µ0 Hc =

1 I= 200

1460 × 0.5 0.5 0.002 = 6.35 A. (2 × 3.14 × 0.05 − 0.002) + 0.48 1.25 × 10−6

(e) If the core material is now replaced by a plastic ring of relative permittivity 1.0, what current is required to maintain the same 0.5 T in the plastic ring?

In this case NI B = µµ0 H = µµ0 2πr B2πr 2 × 0.5π 0.01 I= = ≈ 125 A. µµ0 N 1 × 1.25 × 10−6 × 200 (f ) Explain why the results of 5(c)-5(e) are different.

Because of different B-H relationship in core material.