2nd Year Electromagnetism 2012:.Exam Practice

2nd Year Electromagnetism 2012:.Exam Practice These are sample questions of the type of question that will be set in the exam. They haven t been checked the way exam questions are checked so there may be typos answers in red Question 1. Spherical Charge. Consider a spherical distribution of charge: and (i) Show that the charge contained inside r is given by: ρ(r) = Q 0 r2 (3 5 ) for r < a (1) 4πa3 a2 ρ = 0 for r > a (2) and Q(r) = 0 for r > a. Q(r) = Q 0 ( r3 a 3 r5 ) a5 for r < a (3) [3 marks] Integrate using the volume element when you are spherically symmetric d 3 r = 4πr 2 dr Q(r) = r 0 ρ(r )4πr 2 dr = r 0 Q 0 (3 5r 2 4πa3 a 2 )4πr 2 dr = Q 0 ( r3 a 3 r5 a 5 ) (4) [2 marks] note Q(a) = 0 and outside r = a obviously there is no more charge so the enclosed charge outside r = a is zero.[1 mark] (ii) Derive an expression for the Electric Field. [2 mark] Use Gauss s law (for a closed surface S S E ds = 1 ϵ 0 (charge enclosed)).[1 mark] Use a spherical surface for S and that by symmetry E must be in radial direction and only a function of r so constant on S to conclude [1 mark]: 4πr 2 E r = Q(r) E r = Q(r) ϵ 0 4πϵ 0 r 2 (iii) Calculate the electrostatic potential. [3 mark] Use : E = V E r = V r E r dr = V constant [2 marks] we adjust the constant so that V = 0 for r > a. This yields after integration [1 mark]: V (r) = Q ( ) 0 1 4πϵ 0 a 4 r2 2a 2 r4 4a 4 for r < a (iv) Give an expression for the electrostatic energy and show that the total electrostatic energy is [2 marks] Q 2 0 4πa ( 8 315 )

. The electrostatic energy E is E = ρv d 3 r = a 0 ρ(r)v (r)4πr 2 dr [1 mark] and apon integration using above expressions yields the given result [1 mark]. Question 2. The Pinch. In a simple fusion device called the pinch the cylindrically symmetric field is B = B θ (r)ê θ B z (r)ẑ, where ê θ and ẑ are unit vectors in the z and θ directions. A simple model field is given by: where B 0 is a constant. B z (r) = B 0 e r2 and B θ (r) = B 0 re r2 Note For these symmetric fields B = db z dr êθ 1 r d(rb θ ) ẑ dr (i) Show that the current density is: J(r) = B 0e r2 µ 0 [2rê θ (2 2r 2 )ẑ] (5) what is the total current in the z direction?. [2 marks] Use the differential form of Ampere s law: B = µ 0 J differentiating db z dr = B 02re r 2 and 1 d(rb θ ) = (2 2r 2 )B 0 e r2 r dr the answer is then clearly given by Eq. (1). [1 mark]. Ampere s law in integral form integrating around a circle of radius r is B dr = B θ 2πr = 2πB 0 r 2 e r2 = µ 0 I where I is the total z current enclosed by the circle. Thus as r the total z current goes to zero. [1 mark] (ii) Derive an expression for the magnetic energy for this pinch when it is one metre long (in z). [2 mark] Magnetic energy is B Bz 2 Bθ 2 2πrdr = 2πrdr = πb2 0 0 2µ 0 0 2µ 0 µ 0 [1 mark for formula, 1 for doing the integral] 0 (1 r 2 )e 2r2 rdr = 3πB2 0 8µ 0 (iii) Calculate the magnetic flux through a large (r 1 with z = 0) circular loop about the pinch. [2 marks] r Φ = B da = B z (r)2πrdr B 0 e r2 2πrdr = πb 0 0 0 (iv) Suppose that the field dies away so that B 0 = B 00 e t but the form of the field (i.e. B z (r) = B 0 e r2 and B θ (r) = B 0 re r2 ) stays the same. Show that the emf for the large circular loop about the pinch is given by [3 marks] E dr = B 00 e t (6)

use Faradays law E dr = dφ dt [3 marks] and Φ = πb 0 [2 marks] (v) Suppose we make the circular loop out of wire, hook it up to a light bulb and make some light. Where does the energy in the lights come from? [1 mark] Some of the magnetic energy in the pinch is converted into electron motion via the emf. The electron motion excites atoms in the light bulb to emit light. Question 3.Current Sheet. Consider one dimensional electromagnetic waves that are polarised in the x direction. Maxwell s equations for these waves simplify to: and µ 0 ϵ 0 E x B y = E x z = B y z µ 0J x (z, t). (8) We take the current density to be: J x (z, t) = I 0 cos (ωt)δ(z) where δ(z) is the Dirac delta function. The electric field for z > 0 is: E x = 1 µ0 I 0 cos [k(z ct)]. (9) 2 ϵ 0 where ω = kc. (7) (i) Find the magnetic field, B y, for z > 0 [2 marks] Easiest is to guess and verify. Guess B y = B 0 cos [k(z ct)] then substitute into Eqs (7) and (8) where J x (z, t) = 0 for z > 0 and discover B y = 1 µ0 I 0 cos [k(z ct)] = 1 2c ϵ 0 2 µ 0I 0 cos [k(z ct)]. where I have used c = 1 ϵ 0 µ 0 (ii) Write down the form of the electric field for z < 0. Use the jump condition that E x is continuous across z = 0 to obtain an expression for E x when z < 0. [3 marks] These waves must have the same frequency as the z > 0 waves but must be propagating towards z = so they must have the form E x = E 0 cos [k(z ct)] [1 mark.. accept sine too]. For E x to be continuous the z < 0 wave and the z > 0 wave must have the same amplitude and phase at z = 0. Hence [2 marks] E x = 1 µ0 I 0 cos [k(z ct)] 2 ϵ 0 (iii) Calculate the magnetic field for z < 0 and show that the jump in the magnetic field across z = 0 is equal to µ 0 I 0 cos (ωt). [3 marks] Repeat part (ii) changing the sign of ct changes the sign of B y so B y = 1 2 µ 0I 0 cos [k(z ct)]

setting z = 0 in this expression and the expression in part (ii) we get B y = 1 2 µ 0I 0 cos (ωt) when z < 0 and B y = 1 2 µ 0I 0 cos (ωt) when z > 0 hence the jump across z = 0 is µ 0 I 0 cos (ωt) [3 marks]. (iv) Suppose we have two current sheets, the one given above and one at z = z 0 which has a radiated solution: E x = 1 2 µ0 ϵ 0 I 0 sin [k(z z 0 ct)] for z > z 0. (10) and E x = 1 2 µ0 ϵ 0 I 0 sin [k(z z 0 ct)] for z < z 0. (11) The full solution is the superposition of the two solutions. Find a value for k such that the two sheet problem has no wave going towards z = for z < 0 < z 0. [2 marks] expand sin [k(z z 0 ct)] = sin [k(z ct)] cos [kz 0 ] cos [k(z ct)] sin [kz 0 ] then for z < 0 < z 0 we get: E x = 1 2 µ0 ϵ 0 I 0 (cos [k(z ct)] sin [k(z ct)] cos [kz 0 ] cos [k(z ct)] sin [kz 0 ]) then if we set kz 0 = π/2, cos [kz 0 ] = 0 and sin [kz 0 ] = 1 and E x vanishes.

Revision Lecture

Maxwell s equations in differential form: E = ρ Gauss Law E = B Faraday s Law B =0 No monopoles B = µj µ E AmpèreMaxwell

Lecture 1 In vacuum: 0 E =0 E = B Leads to solutions; E = E 0 e i(k.r ωt)ˆk B =(E 0 /c)e i(k.r ωt) (ˆk Ê) B =0 B = µ 0 0 E Figure 1: Electromagnetic wave

Lecture 2 c!t r=ct P Et!1!2 vt O O!1 E t 2πr sin θ cδt = Q(cos θ 2 cos θ 1 )/2 0 (θ 1 θ 2 ) sin θ(v/c)

Lecture 2 E t = Q[a]sinθ 4π 0 rc B t = Q[a]sinθ 4π 0 rc 2 Figure: Accelerating Charges Radiate P = q2 z 0 2 ω 4 12π 0 c 3 Figure: Angular variation of dipole radiation

In terms of potentials V and A; B = A and E = A V Using Lorentz gauge A 1 c 2 V = 0, Maxwell s equations become; 2 V 1 2 V c 2 2 2 A 1 2 A c 2 2 = ρ/ and = µj

The timedependent solutions to Maxwell s equations are; V = 1 4π 0 [ρ]dτ r [j]dτ and A = µ 0 4π r (raθ ) A = 1 r ˆφ r A r θ A = 1 r ˆφ µ0 4π [I ]dl sin θ c µ 0 4π [I]dl sin θ r

P " r! Lecture 3 E = [ I]dl sin θ 4π 0 rc 2 ˆθ Q! r B = [ I]dl sin θ 4π 0 rc 3 ˆφ Q O Figure: Hertzian (dipole) antenna P = 2π 3 0 c dl λ 2 I 2 rms = R rad I 2 Figure: Angular variation of dipole radiation

Lecture 4 Figure: Synchrotron emission A(r, t) = µ 0 q [v] 4πr(1 ˆr [v]/c) = v V (r, t) c2 Figure: Radiation beaming

Lecture 5 Across a boundary; E is discontinuous by σ c E is continuous B is continuous B /µ is discontinuous across the boundary by j c (E) =ρ f E = B B =0 (B/µ) =j c E

Lecture 6 c ( r 0 µ 0 ) 1/2 = c/η B = ηe/c N = 1 µ E B = r 0 E 2 c t = 2η 1 η 1 η 2 r = η 1 η 2 η 2 η 1

Lecture 7 t = 2η 1 cos θ i η 1 cos θ i η 2 cos θ t r = η 1 cos θ i η 2 cos θ t η 1 cos θ i η 2 cos θ t tan θ B = η 2 /η 1 η 1 sin θ i = η 2 sin θ t sin θ cr = η 2 /η 1

(1) 0 E = ρ (2) B =0 (3) E = B (4) B = µ 0 j µ 0 0 E j = m ne 2 E 1 τ c iω In a (collisionless) plasma, j = ine2 E mω

Lecture 8 c 2 k 2 = ω 2 ω p 2 v ph = c/η; v g = cη η =(1 (ω p 2 /ω 2 ) 1/2

(1) E = ρ (2) B =0 (3) E = B (4) B = µj µ E j = m ne 2 E 1 τ c iω In a conductor, j = ne2 τ c E m = σe (Ohm s Law) So 2 E = µσ E E µ 2 2

Lecture 9 k 2 = µ ω 2 iµσω dispersion relation in metal Good conductor: k =(1i)/δ E = E 0 e (iz/δ ωt) e z/δˆx B follows by π/4. Poor conductor: k =(k 0 iα) E = E 0 e (k0z/δ ωt) e αzˆx B almost in synch.

Lecture 10 Reflectivites of metals: r 1 1R 80 ω R 1 σ N z j c E =0 Electric energy 0 Magnetic energy = ω rµ r σ 1

Lecture 10 constant current high frequency!"#"!$%&' " J!"#"!$"%&'( ( J!"# ħ! ħ" A df dτ = env B = j B (1 R)N P rad = c

Lecture 11 E =0soE to walls B =0soB surface and j TEM modes cannot exist either; B transverse TM modes E transverse TE modes For TM modes: E z = E z0 (x, y) e i(k gz ωt) with, E 0z = E 0 sin mπx a sin nπy b