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Question set ’A’ A(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 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 (A.1.1) F = ke 2 , r1,2 1 ≈ 9 × 109 N· m2 /C2 is so called Coulomb constant. where ke = 4πε 0 (b) Two electrons of charge −1.6 × 10−19 C are 1 × 10−6 m apart. What is the force on each? Accordingly to formula (A.1.1), the force will be equal

F = 9 × 109

(−1.6 × 10−19 )2 = 2.3 × 10−16 N. (1 × 10−6 )2

(c) Three electrons are arranged at each corner of an equilateral triangle of side 3 × 10−9 m. Show on a suitable sketch the magnitude and direction of the force on each electron.

Each electon acts on other two with the force (accordingly to (A.1.1)): F = 9 × 109

(−1.6 × 10−19 )2 = 2.6 × 10−11 N. (3 × 10−9 )2

Accordingly to the law of cosines, the superposition of forces is equal √ FT otal = 3F = 4.5 × 10−11 N.

2 (d) 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 (A.1.2) E · dA = , ε0 S where E is the electric field, dA is a vector representing an infinitesimal element of area. (e) Find the E-field at a distance of 1 × 10−3 m from a large plane charged to 5 × 10−6 C/m2 . To find the E-field will build a cylinder with bases symmetrical to the plane as shown on a figure.

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

σS q = ε0 ε0

After equating those two expressions, will get: E=

σ 5 × 10−6 ≈ 2.8 × 105 V/m = −12 2ε0 2 × 8.85 × 10

As we can see, electric field doesn’t depend on the distance to the plane. The direction is out from the plane. (f ) What is the magnitude and direction of the force on a small particle charged to 2×10−6 C at a distance of 1 × 10−3 from the charged plane of question 1(e)? Accordingly to the expression, recieved in previous point, force equals F = qE = 2 × 10−6 × 2.8 × 105 = 0.56 N As the plane and partical have similar charge sign, that would be a repulsion force.


A(2) The figure shows a cross section of a very long coaxial cable. The inner conductor is of radius r = a. The space a < r < b is filled with dielectric of relative permittivity ε1 , the space b < r < c is filled with dielectric of relative permittivity ε2 and the space c < r < d is filled with dielectric of relative permittivity ε3 . The outer conductor has inner radius d. The cable is grounded as shown and the inner conductor is charged with λ C/m.

(a) Using Gauss’ Flux Theorem or some other suitable method, derive an expression for the E-field in the four regions: a < r < b, b < r < c, c < r < d, d < 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 Dd~ N= S

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 λ E1 = , E2 = , E3 = , E4 = 0 2ε0 ε1 πr 2ε0 ε2 πr 2ε0 ε3 πr (b) 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=


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

λ dr + 2ε0 ε1 πr


λ dr + 2ε0 ε2 πr


λ λ b c c dr = (ε1 ln + ε2 ln + ε3 ln ) 2ε0 ε3 πr 2πε0 a b d



Thus, the capacitance per unit length equals: C=

q 2πε0 = U ε1 ln ab + ε2 ln cb + ε3 ln dc

(c) If ε1 = 4, ε2 = 1 and ε3 = 2 find the capacitance of 5 m cable for which a = 1 mm b = 2 mm, c = 2.5 mm and d = 3 mm. Using the expression recieved in previous point: C=

2 × 3.14 × 8.8 × 10−12 × 5 = 82.5pF 4 × ln 2 + 1 × ln 1.25 + 2 × ln 1.2


A(3) (a) Derive an expression for the capacitance of a parallel plate capacitor. By definition, the capacitance equals C = Uq . If will assume, that one plate of capacitor is charged with σ C per area unit. The parallel plate will be induced a charge −σ. As we could see in point (A.1.e), the electric field of charged plate σ E= 2ε0 . Inside the capacitor those fields will add, so the electric field inside the capacitor equals E= Potential difference

Zd U=

σ . ε0

σ Edx = ε0

Zd dx =

σd , ε0



where d – the distance between the plates. If the area of capacitor is S, so the capacitance will be equal C=

q ε0 σS S = = ε0 U σd d

If the capacitor is filled with dielectric, we should multiply this expression by its relative permittivity ε. (b) State any assumptions which you have made. The one plate is charged with σ C per area unit, d – distance between the plate, the space between the plates is filled by air. (c) The figure shows a cross section of a tank partiall filled with liquid ro a depth of x m. The dimensions of the tank are as shown, the depth into the page is w m. The relative permittivity of the liquid is εr and the space above the liquid is filled with air. The supply voltage V is constant. Show how the capacitance can be used to measure the depth of liquid in the tank. If the voltage is stady, the full capacitance of the tank will be equal C=

C1 U + C2 U q = = C1 + C2 , U U

where C1 = ε0 ε xw is the capacitance of part of the tank with liquid, and C2 = ε0 (L−x)w – the capacitance d d of the rest. From here we can derive an expression for x: x=

Cd ε0 w



Thus, we can find the depth of liquid, if we will know the capacitance of the tank. (d) Could this method be used to measure the depth of any liquid? No, it’s not suitable for electrolytes like the solution of salt in water, as the current will flow through it, and there won’t be any capacitance.


A(4) The figure shows a very long wire carrying a current i = I sin ωt. A small rectangular coil of 15 turns is placed in the same plane as the wire as shown.

(a) 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 (A.4.1) C

(b) Find an expression for the magnetic field B at a distance R from a very long straight wire carrying a current i.

7 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 R r= cos α

Thus, dB =

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


µ0 i 4πR

Z2 cos αdα =

i µ0 I sin ωt = 2πR 2πR

− π2

(c) Show the B field on a suitable sketch. (d) Find an expression for the total magnetic flux Φ linked by the rectangular 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 = Bhdx Za+b Za+b µ0 I sin ωt 1 µ0 I sin ωt a+b Φm = Bhdx = h dx = h ln 2π x 2π a a



(e) Find an expression for the emf induced in the rectangular coil. According to Faradey’s law, the emf equals ε=−

µ0 Iω cos ωt a+b dΦm =− 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 µ0 h a + b Φm = ln i 2π a (g) If the figure represented a circut board and we wished to minimise the mutual inductance what could be done? As the mutual inductance related to magnetic flux, it’s needed to minimize it. The easiest way is to turn the coil for 90 degrees, so magnetic flux through it would be 0. (h) If the coil was moved with a steady speed of v m/s in a direction parallel to the axis of the wire how would the emf be affected? As the emf related to the time change of magnetic flux L=


µ0 I a+b dΦm =− h ln (ω cos ωt ± v), dt 2π a

the sign is related to the direction of move.


A(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) What direction does the B-field go around the ring? (clockwise or anticlockwise) According to right-hand rule, the direction of current is clockwise. (b) Find the current in the coils for an air gap flux density of 0.5 T. From the Ampere’s circuital law N I = Hc lc + Hg lg From the Gauss’ theoreme for magnetic field: I

~ s=0 Bd~


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 (2 × 3.14 × 0.05 − 0.001) + 0.001 = 4.37 A. 0.48 1.25 × 10−6

10 (c)If an air gap flux density of 0.75 T is now required, what is the new current in the coils?   1460 × 0.75 0.75 1 (2 × 3.14 × 0.05 − 0.001) + I= 0.001 = 10.4 A. 200 0.48 1.25 × 10−6 (d) If the core material had a linear B-H relationship, what would be the ratio of the currents found in answers 5(b) and 5(c)? If the core material has a linear B-H relationship Hg = B I= µN

B µ0 µ

 2πr − lg + lg . µ

(e) Explain why your results are not as in 5(d)

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