2nd Year Electromagnetism 2012: Practice Problems on VectorCalculus.

Lecturer: Steve Cowley

These problems are optional and should be entirely revision – hopefully they will help you remember vectorcalculus from last year.

(i) Let ψ = ψ0e−r2 where r2 = x2 + y2 + z2 is the radius in spherical polar coordinates and ψ0 is a constant.

Show that

∇ψ = −2rψ0e−r2 (1)

where r = xi+ yj+ zk

(ii) Let ψ = ψ0 sin(kx)e−ky where k and ψ0 are constants. Show that

∇2ψ = 0 (2)

(iii) Let ψ = ψ01r sin(kr − ωt) where: r is the radius in spherical polar coordinates, ψ0 is a constant, the

constants k and ω are related by ω = kc and, c is the velosity of light . Show that ψ satisfies the waveequation:

c2∇2ψ =∂2ψ

∂t2(3)

(iv) Find the curl of the vector field A = yi − xj. Draw the ”field lines” of A. Show that we can writeA = ∇ψ × k and find ψ.

(v) Consider a vector field A(x, y, z) = ∇a ×∇b where a = a(x, y, z) and b = b(x, y, z) are arbitrary scalarfields. Show that ∇ ·A = 0. Also find C such that ∇×C = A. Is C unique?

(vi) Let P = f(r · d)(r× d) where r = xi+ yj+ zk and f is an arbitrary function. Show that ∇ ·P = 0.

(vii) Let ψ = ψ0eik·r where ψ0 is a constant and k is a constant (wave) vector. Show that ∇ψ = ikψ.

(viii) Let A = A0eik·r where A0 is a constant vector. Show that ∇ ·A = ik ·A and ∇×A = ik×A.

2nd Year Electromagnetism 2012: Homework 1.

Lecturer: Steve Cowley

Question 1. Dipoles. This should be entirely revision – but hopefully it will get you warmed up.Consider a (model) polar diatomic molecule with charge +q at +∆r and −q at −∆r.

(i) Give an exact expression for the electric field.For this molecule we define the dipole moment as p = 2q∆r.(ii) Give a simple expression for the electric field when ∆r ≪ r – write your answer in terms of p.(iii) Describe qualitatively what would happen to two isolated dipoles placed near each other – how wouldthey move?

Question 2. Spherical Charge Distributions.

Let the charge density ρ be given by: ρ = ρ0 for r < a and, ρ = 0 for r > a where ρ0 = constant, a=constant and r =

√x2 + y2 + z2. Use Gauss’s law to find the electric field E for r < a and r > a.

Question 3. Cylindrical Charge Distributions.

Let the charge density ρ be given by: ρ = ρ0ra for r < a and, ρ = 0 for r > a where ρ0 = constant, a=

constant and r =√x2 + y2. Use Gauss’s law to find the electric field E for r < a and r > a.

Question 4. Electric Field of Arbitrary Stationary Charge.

In this question we derive the formula for the electric field from a stationary distribution of charge.(i) Consider a spherical distribution of charge; i.e. ρ = ρ(r) with r =

√x2 + y2 + z2. Show that in this case,

∇ ·E =1

r2∂

∂r(r2Er) =

1

ϵ0ρ(r). (1)

where Er is the radial component of E. What about Eθ and Eϕ?(ii) Integrate to find a formal expression for Er.(iii) Suppose ρ = 0 for r > a. Express Er for r > a in terms of the total charge.(iv) Let

ρ = ρ0 =2q

4πrr20e(− r2

r20

). (2)

Find an explicit expression for Er. Find Er in the limit r0 → 0.

In the limit r0 → 0, ρ0 is essentially zero everywhere except close to the origin where it tends to infinity.Thus in this limit ∫

V

ρ0A(r)dV → A(0)

∫V

ρ0dV = A(0)q. (3)

if the volume V contains the origin. We can define the three dimensional Dirac delta function as:

δ(r) = limr0→0

(2

4πrr20e(− r2

r20

)

). (4)

We can therefore write the charge density of a point charge as ρ = qδ(r).

(v) Argue that for a point charge q at r = r′ we may write ρ = qδ(r− r′).HARDER(vi) Find E = G(r, r′) for ρ = ϵ0δ(r− r′) i.e. find the solution to:

∇ ·G = δ(r− r′). (5)

We often refer to functions like this as Green’s functions.HARD(vii) Hence show that,

E(r) =

∫ρ(r′)

(r− r′)

4πϵ0|r− r′|3dV ′ (6)

where the integration is over all r′ and |r− r′| is the distance between r and r′.

Question 5. Magnetic Field from a Steady Current.

(i) Calculate the magnetic field inside and outside a long straight conductor with circular cross section ofradius R that carries a uniform current density with total current I.

(ii) Now suppose that inside the conductor, there is a cylindrical hole of radius a whose axis is parallel tothe axis of the conductor and a distance b from it (a + b < R). Assume again that the current density isconstant and that the total current is I. Show that the magnetic field inside the hole is uniform and equal

toµ0bI

2π(R2 − a2). (7)

Draw a picture showing the orientation of the fields and the hole.

Question 6. Coulomb Explosions.

In this question you are expected to make estimates – the more accurate the better. In the basement ofBlackett high powered lasers were used to blast the electrons off clusters of a few thousand atoms. Supposeall the electrons are suddenly removed from a cluster of one thousand carbon atoms. The carbon nucleiithen explode outwards – why and how fast? Hint consider the energy. How does the energy of the explosionscale with the size of the cluster?

2nd Year Electromagnetism 2012: Homework 2.

Lecturer: Steve Cowley

Question 1. Solenoids. In this question the integral form of ampere’s law

(i) Show that the field inside a long cylindrical solenoid is B = µ0nI where n is the number of turns per unitlength and I is the current in each turn (see above). A thin solenoid of length 10cm with 600 turns carriesa current of 20 amperes, calculate the field inside the solenoid.

(ii) Now suppose that the solenoid is wrapped into torus – a doughnut shape – this is sometimes called atoroidal solenoid. Calculate the field inside the torus. If you want help see

http://www.youtube.com/watch?v=jdsUQs9w0uw

(iii) The fusion experiments JET and ITER (http://www.iter.org/) are based on huge toroidal solenoids.Schematically shown above. The field created by the 18 superconducting ”toroidal field coils” (like the blueones above) in ITER is 5.2 tesla at a radius of about 6 meters. What is the current in each toroidal fieldcoil?

Question 2. Charge Conservation.

(i) Deduce charge conservation from Maxwell’s equations. (Former exam question).

(ii) Suppose magnetic monopoles (magnetic charges) exist and that magnetic charge is conserved. Let ρBbe the density of magnetic charge and JB the current density of magnetic charge. We take units so that,

∇ ·B = µ0ρB . (1)

Write down the equation for the conservation of magnetic charge.

(iii) Modify Faradays law to make Maxwell’s equations conserve magnetic charge.

Question 3. Induction and Transformers.

Consider a 10cm long solenoid (solenoid A) of radii approximately 1cm with 400 turns.

(i) Let the current in the A, IA, change with time slowly (IA = I0(t)) so that you may ignore the displacementcurrent. What is the instantaneous magnetic flux in the solenoid?

(ii) A voltmeter is connected across the solenoid. The meter will measure∫E · dl integrated along the

wire. Derive an expression for the voltage measured (ignoring the resistance). Calculate the voltage whenI0 = sin [100πt]

Consider now two 10cm long solenoids of radii approximately 1cm: solenoid A with 400 turns and solenoidB with 800 turns. B is fitted snugly inside A see diagram below.

B

B

VA

VB

Double solenoid – solenoid B inside solenoid A

IB

IA

(iii) Let current IA flow in A and current IB flow in B. What is the field inside both?

Now we disconnect B and connect it to the volt meter. We put a slowly changing current IA = I0(t) throughthe coil A.

(iv) Calculate the induced voltage across B.

(iii) In the general case where both IA and IB are changing show that the voltages measured across A, VA,and across B, VB , satisfy VB/VA = 2.

Question 4. Pinch Effect.

MAGPIE in the basement of Blackett squeezes plasmas with magnetic fields and makes very dense and hot

plasmas. In this question you will estimate the forces involved.

(i) Show that if we assume the current in the (neutral) plasma is carried by many electrons and that theions are stationary then

J = −neev (2)

where ne is the electron number density (the number of electrons per unit volume) and v is the averagevelocity of the electrons.

(ii) Recall that the force on a single electron is −e(E+ v ×B). Hence show that the force per unit volumeon the electrons and ions in the plasma (and hence on the plasma) is:

J×B (3)

if we assume that the plasma is neutral so that the electron and ion charge densities acancel i.e. ene = qni

where q is the ion charge.

(iii) Magpie makes a cylindrical shell of current flowing in plasma made from vaporized wires. The currentis taken to be zero except in a shell ofradius 0.01 metres and thickness 0.001 metres where:

J = J0z J0 = constant. (4)

Calculate the magnetic field (ignoring displacement current).

(iv) Calculate or estimate the radial force per unit length on the plasma. Show that it squeezes the plasma.

(v) Suppose the initial plasma is one tenth the mass density of solid aluminium and the total current is onemega-amp. Estimate the time to halve the shell radius.

2nd Year Electromagnetism 2012: Homework No. 3.

Lecturer: Steve Cowley

These questions require lectures 5 and 6.

Question 1. Circular Polarisation. (i) Calculate the real part of the complex wave solution toMaxwell’s equations in a vacuum:

E = E0(x+ iy) exp ik(z − ct). (1)

where x and y are unit vectors in the x and y directions respectively.

(ii) Calculate the magnetic field of the wave.

(iii) Describe the field seen by an observer at z = 0. What happens if we replace (x+ iy) with (x− iy).

(iv) Calculate the motion in the wave of an electron which starts from rest at (0, 0, 0) at time t = 0. Youmay ignore the magnetic force – when is this approximation valid?

Question 2. General 1D Solution of the Wave Equation.

Consider the wave equation for linearly x polarized waves travelling in the ±z directions:

∂2Ex

∂t2= c2

∂2Ex

∂z2(2)

(i) Transform Eq. (2) to the independent variables q = z − ct ands = z + ct and show that:

∂2Ex

∂s∂q= 0 (3)

(ii) Hence show that the general solution of Eq. (2) is:

Ex = E+(q) + E−(s) = E+(z − ct) + E−(z + ct) (4)

where E+ and E− are arbitrary functions.

(iii) Calculate the general form of the magnetic field in terms of E+ and E−.

(iv) Suppose that at t = 0:

Ex = E0e−αz2

and By = 0 (5)

calculate the fields for t > 0. Describe the solution.

Question 3. Spectrum.

We define the Fourier transform of Ex by:

Ek(t) =

∫ ∞

−∞Ex(z, t)e

−ikzdz (6)

(i) Calculate Ek(t) for the pulse E+ calculated in Question 2.

(ii) Calculate the ”spectrum”, i.e. |Ek|2. How does the width of the spectrum (in k) depend on the widthof the pulse in z?

Question 4. General Plane Waves.

We may represent a general electromagnetic plane wave by (real part of the complex exponentials:

E = E0e(ik·r−iωt) and B = B0e

(ik·r−iωt) (7)

(i) Show that Faraday’s law: ∂B∂t = −∇×E becomes:

−iωB0 = −ik×E0 (8)

(ii) Hence show that Maxwell’s equations for plane waves in a vacuum become:

ω

kB0 = n×E0 and

ω

kE0 = −c2n×B0 (9)

where n is the unit vector in the direction of k and k = |k|. Hence show that ωk = ±c.

(iii) Let us define two constant unit vectors e1 and e2 by e2 = n×e1 and e1 = e2×n. Show that n = e1×e2.

(iv) Show that the real part of E = E0(e1 + ie2)E0e(ik·r−iωt) yields circularly polarized waves.

Question 5. Some Common Waves.

(i) Look up the frequency or wave length of your cell phone carrier signal. How big is the wavelengthcompared to the size of the phone?

(ii) What is the wavelength of sodium orange yellow light? How big is a sodium atom?

(iii) What is the typical wavelength of a hospital x-ray?

2nd Year Electromagnetism 2012: Homework No. 4.

Lecturer: Steve Cowley

Question 1. Energy Conservation in Waves.

This question uses results developed in Lecture 8.

(i) Derive an expression for the Poynting flux for a wave Ex = E+ cos [k(z − ct)]. What is the time averageflux?

(ii) The Rutherford lab has a plan to produce 10 petawatt (1016) laser with a wavelength of approximately910 nanometres. The laser would be pulsed 300J in 3×10−14s. This laser can be focussed onto a small spotto intensities of 1027Wm−2. Calculate/estimate the typical electric field in the focus.

(iii) Compare the electric field in the focus to the the typical electric field seen by an electron in a hydrogenatom. Hint what is the approximate distance of an electron from the nucleus in hydrogen?

(iv) Estimate the typical kinetic energy of an electron in the focus. How does it compare with the rest massenergy of an electron (mec

2)? Can you do this relativistically?

Question 2. Radiation from a Current Sheet.

This question expands on the model developed in class (Lecture 7).

(i) Suppose we take a current sheet source,

Jx(z, t) = I0 cos (ωt)δ(z) (1)

Write down the radiated waves.

(ii) Write down a formula for the energy flux in the waves. Show that the energy conservation relation(Lecture 8.) is obeyed.

(iii) In JET, the European fusion experiment in Oxfordshire, we heat the plasma with waves from an antennathat is almost a sheet. The Antenna is about 2 metres square and radiates waves at frequency ∼ 3×108s−1.Calculate the approximate wavelength of the radiated waves.

(iv) The antenna on JET puts about 30 megawatts of power into the plasma. Approximating the antennaas a sheet calculate the amplitude of the oscillating current in the antenna.

(v) [HARDER] Suppose we have two current sheets, i.e. :

Jx(z, t) = I0 sin (ωt+ ψ)δ(z − z0) + I0 sin (ωt− ψ)δ(z + z0) (2)

Where the phase, ψ, and z0 are constants. Calculate the radiated fields. What values of the phase andz0 minimize the radiation towards z → −∞? What values simultaneously maximize the radiation towardsz → ∞ and minimize the radiation towards z → −∞?

Question 3. Radiation from an Extended Source – HARD

This question is advanced – but still worth a shot.

(i) We begin by redoing the current sheet of Lecture 7. with complex sources. Let us write a current sheetat z0 as:

Jx(z, t) = I0(z0)e−iωtδ(z − z0) (3)

Where the constant I0(z0) is labelled by the position of the sheet. Show that for z > z0 with this source:

Ex(z, z0, t) = −1

2

õ0

ϵ0I0(z0)e

ik(z−z0−ct) By(z, z0, t) = −1

2µ0I0(z0)e

ik(z−z0−ct) (4)

and that for z < z0 with this source:

Ex(z, z0, t) = −1

2

õ0

ϵ0I0(z0)e

−ik(z−z0+ct) By(z, z0, t) =1

2µ0I0(z0)e

−ik(z−z0+ct) (5)

where we have labelled the fields with an argument z0 to show that they depend on z0.

(ii) We can construct an extended source by adding sheets together – integrating.

I0(z) =

∫ ∞

−∞I0(z0)δ(z − z0)dz0 (6)

(iii) By integrating the equations for Ex(z, z0, t) and By(z, z0, t) over z0 show that

Ex(z, t) =

∫ ∞

−∞Ex(z, z0, t)dz0 and By(z, t) =

∫ ∞

−∞By(z, z0, t)dz0 (7)

satisfy the 1D equations with an extended current source I0(z)e−iωt i.e.

∂By

∂t+∂Ex

∂z= 0 and µ0ϵ0

∂Ex

∂t+∂By

∂z= −µ0I0(z)e

−iωt (8)

(iv) Hence show that when z > a and I0(z) = 0 for z > a:

Ex(z, t) =

∫ ∞

−∞−1

2

õ0

ϵ0I0(z0)e

ik(z−z0−ct)dz0 = −1

2

õ0

ϵ0eik(z−ct)

∫ ∞

−∞I0(z0)e

−ikz0dz0 (9)

Describe this solution.

2nd Year Electromagnetism 2012: Homework No. 5.

Lecturer: Steve Cowley

Question 1. Spherical Waves.

Waves from a small source are spherical waves that begin to look like plane waves far from the source. Herewe look at scalar spherical waves – of course EM waves are vector waves.

(i) In three dimensions the wave equation is

∂2ϕ

∂t2= c2∇2ϕ. (1)

We assume spherically symmetric waves thus ϕ = ϕ(r, t). Using the form of ∇2 in spherical polar coordinateswrite down the partial differential equation for ϕ.

(ii) Let ϕ = A(r, t)/r. Show that A obeys the one dimensional wave equation:

∂2A

∂t2= c2

∂2A

∂r2. (2)

(iii) Write down solutions for A(r, t).

(iv) Suppose the source of the waves is localised at (or very close to) r = 0. Find the wave solutions thatradiate away from the source.

(v) Explain physically why the amplitude of the wave decreases as 1/r.

Question 2. Multipole Moments.

For a stationary charge distribution the scalar potential is given by:

V (r) =

∫ρ(r′)

4πϵ0|r− r′|dv′. (3)

See Lecturer 9.

(i) Show that for r ≫ r′

1

|r− r′|∼ 1

r+

r · r′

r3. . . , (4)

where |r| = r.

(ii) Show that the potential far from the charge distribution (i.e. when r ≫ r′) – has the form:

V (r) =Q

4πϵ0r+

r · p4πϵ0r3

, (5)

and give expressions for the charge Q and the dipole moment p.

(iii) Water molecules have a dipole moment – estimate how big it is.

Question 3. Vector Potential and Waves.

Find the vector and scalar potentials for an Electromagnetic wave traveling in the z direction and polarisedin the x direction.State which gauge you use.

Question 4. Some Numbers.

(i) Find the approximate wavelength of red light.

(ii) Find the wavelength of the photon emitted in the n = 2 to n = 1 transition in hydrogen. How does itcompare to the size of the hydrogen atom.

(iii) Find the wavelength of BBC radio 4.

(iv) Find the wavelength of the gamma ray photons emitted whenan electron and positron anihilate eachother.

27th February, 2012

Electricity and Magnetism IIProblem Sheet 6: Vacuum Waves and Radiation

1. Wave-vectors: A sinusoidal wave of wave number k and angular frequency ω may be writtenas Ψ(r, t) = Ψ0 exp i(k.r− ωt) where Ψ0 is a constant vector (e.g. one of the components of theEM field). For such waves, propagating in the directions specified below, deduce the componentsof k and hence write down expressions for the waves.

(a) z direction;

(b) −x direction;

(c) in the yz plane at 45 to y and z;

(d) in the xy plane at angle φ from x towards y .

2. Complex Phase: In complex form, an EM wave is described by E(r, t) = E0 exp i(k.r− ωt),where E0 is a constant vector with complex components. If E0 is in the y direction, show thatthe real (physical) wave electric field can be written as E = E0 cos(k.r − ωt + φ)y , wheretanφ = E0i/E0r and E0r and E0i are the real and imaginary parts of the complex field E0.

3. Elliptical polarisation:

(a) An elliptically polarised wave propagates in the +x direction in vacuum. The (real physical)electric field of the wave is,

E = E0y cos(kx− ωt) y + E0z sin(kx− ωt) z.

Express this field in the complex form.

(b) Show that the magnetic field of the wave in complex form is

B(x, t) = (1/c)(E0zeiπ/2 y + E0yz)ei(kx−ωt).

What is the real physical magnetic field?

(c) Use these results to show that the time-averaged energy density in the electric field is givenby 〈UE〉 = 1

4ε0(E0y2 + E0z

2) , and that the time-averaged energy density in the magneticfield, 〈UB〉, is equal to 〈UE〉.

(d) Calculate the time-averaged energy flux and show that 〈N〉 = c〈U〉x , where 〈U〉 is thetotal average energy density. Hence give a physical interpretation to the Poynting flux.

4. Dipole Antenna: A simple Hertzian antenna, stretching from z = +dl/2 to z = −dl/2, has atime-varying current I flowing in it. In lecture 3, we derived the electric field generated in theradiation field in such an antenna (i.e. assuming that the wavelength of the radiation is muchlarger than the antenna, λ = (2πc/ω) dl);

E =1

4πε0

[I] dl sin θ

rc2θ

(a) If the current flowing is given by I = I0 sinωt, show that the radiation field can be expressedas a plane electromagnetic wave of form;

E = E0 cos(k · r− ωt) n

where n is a unit vector normal to the direction of propagation k. Give a value for k andE0.

(b) Hence give an expression for the mean energy flux radiated by the antenna, (you may usethe expression derived in question 3 for the Poynting flux).

(c) By integrating over all angles, show that the total (mean) power radiated can be writtenin the form 〈P 〉 = RradIrms

2, and write Rrad in terms of the wavelength of emission λ. Youwill require the integral

∮sin2 θdΩ = 8π

3 , where the integral is over all solid angle.

(d) A one meter long antenna is used to send radio waves to taxis at a frequency of 50 MHz.The receivers can detect an electric field of 0.01 Vm−1. Calculate the antenna’s effectiveresistance. Estimate the power needed to cover London with one transmitter, and thecorresponding current required.

5. Synchrotron Radiation: In Lecture 2 we derived the formula for radiation from an acceleratingcharge. For acceleration a in the z−direction the fields are given by;

E =Q[a] sin θ

4πε0 rc2θ; B =

Q[a] sin θ

4πε0 rc3φ

(a) Find the instantaneous energy flux in the waves at a distance r as a function of θ and theretarded acceleration [a].

(b) Show that the total instantaneous radiated power can be written as:

W =Q2[a]2

6πε0c3(Larmor equation)

(c) Use this result to calculate the power of synchrotron radiation radiated by an electron ofspeed v ( c) traveling in a circle in a magnetic field B = Bx.

(d) Synchrotron radiation cools plasmas and can be a problem for fusion. Estimate the syn-chrotron radiation power from a plasma with 1022 electrons and temperature of 108 K in amagnetic field of 5 T. Approximately how long does it take for the plasma to lose half it’senergy/temperature. Assume that v is sufficiently below c to use the above formula.

6. Circular Accelerators: When β ⊥ β, the relativistically correct expression for power loss ofcharged particles is given by,

Pobs =q2

4πε0

2

3cγ4β2

(a) The Large Hadron collider accelerates protons up to a final design energy of 7 TeV in a ringof radius r = 4.3 km. Calculate the relativistic factor of the protons γ and hence calculatetheir β = v/c and (1− β).

(b) Hence calculate β, and the rate at which a proton irradiates energy. If the accelerator stores3× 1014 protons in the ring, calculate the total energy radiated per second, and compare itto the energy stored by the ring. You might also like to compare the energy in the protonbeam, to the energy of a TGV train (∼400 tons) travelling at 150 km/hour.

(c) The LHC used to house an electron-positron collider (LEP). If instead of protons, LHC wereused to accelerate 1014 electrons/positrons to 7 TeV, calculate the total radiated power loss.Can you see why the LHC is a proton collider?

5th March, 2012

Electricity and Magnetism IIProblem Sheet 7: Dielectrics

1. EM waves in a dielectric: An HIL medium of permittivity ε and permeability µ has zerofree charge density, and zero conductivity. The electric field of a wave propagating in the +xdirection in the medium is given by E(x, t) = E0 cos(kx− ωt) y.

(a) Use Faraday’s law to calculate the magnetic field of the wave. Check that these fields alsosatisfy the Ampere-Maxwell equation.

(b) Show that the energy density in the electric and magnetic fields of the wave are equal atevery (x, t). Show that the Poynting vector is given by N = c′U x where U is the totalenergy density.

2. Cerenkov radiation: Energetic particles with velocity with v/c = β are fired into a dielectricof refractive index η. As demonstrated in figure 1, a front of radiation is produced in an opticalshock travelling at an angle θc.

θcβ

Figure 1: Cerenkov radiation

Find an expression for θc and hence calculate the minimum energy required for an electron toemit Cerenkov radiation in:

(a) glass (η = 1.5)

(b) a helium cell at atmospheric pressure (η = 1.000036)

3. Normal incidence on a dielectric: Let the region z < 0 be filled with a dielectric withdielectric constant ε1 (or equivalently refractive index n1) and the region z > 0 be filled with adielectric with dielectric constant ε2 (refractive index n2). The incident wave in the region z < 0has the form:

Ex = Exi cos (ωt− kz)

(a) What are the appropriate boundary conditions at z = 0?

(b) Calculate the amplitudes of the transmitted and reflected waves.

(c) Show that the sum of the (moduli of) energy flux in the reflected and transmitted waves isequal to the energy flux in the incident wave.

4. Fresnel coefficients: For the case of a wave incident on a dielectric interface with the electricfield normal to the plane of incidence:

(a) Show that application of the boundary condition on the normal component of the B-field(along with the condition of continuity on the E field) leads to Snell’s law.

(b) Now using the condition for B‖ at the boundary obtain the Fresnel relations for this case:

t =2η1 cos θi

η1 cos θi + η2 cos θtr =

η1 cos θi − η2 cos θtη1 cos θi + η2 cos θt

(c) Combine the Fresnel relations with Snell’s law to show that:

r =sin(θt − θi)

sin(θt + θi)and t =

2 cos θi sin θtsin(θt + θi)

(d) Similarly show that for the polarisation of the beams in the plane of incidence, the reflectioncoefficients are given by:

t =2η1 cos θi

η2 cos θi + η1 cos θtr =

η1 cos θt − η2 cos θiη2 cos θi + η1 cos θt

5. Brewster’s Angle and Total internal reflection: State the conditions necessary for both;(i) Brewster’s Angle and (ii) Total internal reflection. Calculate both Brewster’s angle, θB andthe critical angle θc (at which total internal reflection occurs) at the boundary of two dielectricwith refractive index η1 and η2. (You may use the Fresnel coefficients derived above).

Hence calculate:

(a) θB and θc for propagation through both a glass-air interface and an air-glass interface.(η = 1.5 for glass).

(b) The maximum permissible entrance angle of a light ray in an optical fibre if the core hasη = 1.6 and the cladding has η = 1.5.

(c) θc for diamond (η = 2.52). Can you see why diamonds are cut with multiple angled faces?

12th March, 2012

Electricity and Magnetism IIProblem Sheet 8: Plasma

1. Evanescence: An EM wave is totally internally reflected at a dielectric interface lying in they-z plane at x = 0. The evanescent E-wave is described by Et = Et e

i(kt·r−ωt) y where kt =iktxx + ktzz. (ktx and ktz are real.)

(a) Use Faraday’s law to show that the evanescent B-field is given by

Bt = (−ktx + iktz)(Et/ω)ei(kt·r−ωt)

(b) Hence determine the time-averaged Poynting vector in (i) the z-direction, and (ii) thex-direction, to show that in the evanescent wave there is energy flow in the z-direction(parallel to the interface), but not in the x-direction (normal to the interface).

(Hint: remember the Ponyting vector is the product only of the real parts of the waves.)

2. EM waves in plasmas: Consider an em wave propagating in the z direction in vacuum towardsa plasma, which has an electron density n for z > 0 and a boundary at z = 0.

(a) Write out Maxwell’s equations applicable to a plasma in differential form.

(b) Give an expression for the current density jc in term of the resultant fields for a collisionlessplasma, stating any other appropriate assumptions.

(c) Assuming solutions of the form E = Exei(kz−ωt) x and B = Bye

i(kz−ωt) y, show that forz > 0:

ωBy = kEx; and ω2 = k2c2 + ω2p,

where the plasma frequency is ωp =

(ne2

ε0m

)1/2

.

(d) In the region z < 0 the incident wave is Exiei(kz−ωt). There is also a reflected wave with

the same frequency. Thus for z < 0,

Ex = Exiei(kz−ωt) + Exre

−i(kz+ωt).

Derive an expression for By in this region.

(e) Show that when ω < ωp the electric field in the region z > 0 is given by

Ex = Expe(−kpz−iωt)

and find kp. Find the magnetic field in this region in terms of Exp, ω and kp.

(f) At the boundary to the plasma, z = 0, the electric field and magnetic field must be contin-uous. Show that this implies:

Exi + Exr = Exp and kExi − kExr = ikpExp.

Hence show that the amplitude of the reflected wave is equal to the amplitude of the incidentwave.

3. Power loss in collisionless plasma Write down the expression for the power dissipated by EMfields in a conducting material of volume τ , in terms of the electric field, E and the conductioncurrent density, jf . Hence show that for a sinusoidal EM wave propagating through a collision-free plasma, there is no power dissipation. You may use your expression for jc derived in thelast question. Similarly, using the results of the last question within the plasma, show that thetime average Poynting flux into the plasma is also zero.

1

4. Wave velocities and cut-offs: A wave propagating in a collisionless plasma satisfies thedispersion relation ω2 = ω2

p + c2k2, where ωp is the plasma frequency.

(a) Show that the group velocity of the wave is given by vg = c(1 − (ωp/ω)2)1/2 and that thephase (vφ) and group velocities satisfy vgvφ = c2.

(b) By requiring k to be real for an EM wave to be able to propagate, find an expression for themaximum (critical) density to which a wave of angular frequency w can propagate, (youmay use the expression for the plasma frequency obtained above).

(c) The density in a discharge plasma is monitored using microwaves. A microwave antennaproducing an em wave of variable frequency, is monitored by a receiver placed on the otherside of the plasma from the antenna, as shown in the diagram. When the discharge isactive, the receiver detects a drop in transmitted power when the frequency falls below 50GHz. Calculate the peak density of the plasma.

Microwave source

Microwave detector

plasma

5. Dispersion of Pulsar emissions: Pulsars are magnetised neutron stars that emit pulses ofradio waves of between 1 to 50 milliseconds duration at intervals of a second or less. Therepetition rate of the emission is stable to 1 part in 108. It is believed that the radiationis emitted in a cone that sweeps past the earth as the pulsar rotates (like the light from alighthouse), producing a pulse of radio waves of multiple frequencies. The higher frequenciesarrive first due to dispersion of the em waves in the tenuous inter-stellar plasma. Within monthsof the discovery of pulsars, this information was used to determine their distance.

(a) The density of electrons between the stars is roughly 105 m−3. What is the plasma frequencyin this plasma?

(b) Show that for a pulsar emitting a frequencies f1 and f2, the time delay between arrival ofthe pulses is given by;

∆t ∼=df2p2c

[1

f22 − 1

f12

]where fp = ωp/2π and d is the distance to the pulsar.

(c) For pulsar CP 0328, three packets of central frequencies 151, 408 and 610 MHz are observed.The 610 MHz pulse arrives first followed by the 408 MHz pulse 0.365 seconds later, withthe 151 MHz a further 4.18 seconds later. Calculate the distance from earth to CP 0328 –check that both time delays give the same distance (almost).

6. Grazing-angle x-ray reflections: In the design of an imaging x-ray telescope it is necessaryto reflect 1 keV x-rays off the surfaces of metallic mirrors. Assume that the conductivity of themetal is plasma-like, with a plasma frequency corresponding to a vacuum photon energy of 6eV. What is the maximum allowed angle which the incident x-rays can make with a metallicmirror and still experience total reflection? (You may assume that the x-rays are travelling in avacuum before striking the metal.)

2

16th March, 2012

Electricity and Magnetism IIProblem Sheet 9: Conductors and waveguides

1. Reflection from a conductor: An infinite plane electromagnetic wave is incident from vacuumfor z < 0 onto a good conductor for z > 0 where j = σE.

(a) State mathematically the condition for σ so that the displacement current may be ignored.Show that in this limit for the conductor, Faraday’s law and Ampere’s law become:

∇×E = −∂B

∂t; ∇×B = µσE

(b) Show that; E = Ex0e−(z/δ)+i(z/δ)−iωt x; B = By0e

−(z/δ)+i(z/δ)−iωt y

are solutions to these equations for z > 0, with δ =

(2

µ0σω

)1/2

and |Ex0| =ωδ√

2|By0|.

(c) The reflection coefficient for normal incidence onto a surface of refractive index η fromvacuum is given by;

r =1− η1 + η

.

Given an expression for the refractive index of a metal (bearing in mind that it is nowcomplex). Show that the energy reflectivity of the surface is R = |r|2 ≈ 1−

√8ωε0/σ.

(You may assume that µ ' µ0).(d) Calculate the reflectivity of a gold mirror, σAu = 4.1 × 107 (Ωm)−1 for red (λ = 630 nm),

and blue (λ = 420 nm) light.

2. Good conductors: The table below shows the conductivities and relative permitivities εr(where ε = εrε0 , and ε0 = 8.85× 10−12 Fm−1 ) of several substances.

Substance σ (Ωm−1) εrAluminium 4× 107 1Sea Water 3 10

Pure Silicon 0.1 1Distilled Water 10−3 10

Polythene 10−15 2

(a) Calculate the limiting frequency below which each of these becomes a “good conductor”

(b) Which of these substances is a good conductor at 100 kHz, and which a weakly conductingdielectric i.e. a good insulator?

(c) Communications are attempted with a submarine at this frequency (100 kHz). Estimatethe maximum depth to which communication is still possible.

3. Skin depth: The conductivity of copper is σCu = 5.7×107 (Ωm)−1, and µ ' µ0 = 4π10−7 Hm−1.

(a) A screened room is used to prevent electromagnetic interference (“noise”) at frequencyaround 1 MHz from reaching a detector. How thick must the copper wall be to attenuatethe the intensity of the noise by a factor 10−3?

(b) Estimate the frequency at which the resistance of a copper wire of radius 0.1 mm starts torise above its DC value. What is the corresponding value for a wire of 1 mm radius?

1

4. AC Resistance:

(a) The electric and magnetic fields of a plane wave propagating in the x direction in a goodconductor of conductivity σ, permittivity ε, and permeability µ are given by:

E(x, t) = E0e−x/δei(x/δ−ωt) y and B(x, t) = (µσ/ω)1/2E0e

−x/δei(x/δ−ωt+π/4) z

where kc =√µσω/2. Show that the time-averaged Poynting vector is

〈N〉 =1

2

√σ/2µω E2

0 e−2x/δ x.

(b) Calculate the average power per unit volume, P = 〈jc ·E〉, converted from the EM field ofthe wave into heat. Show that ∂

∂x〈N〉+〈jc ·E〉 = 0. Give an interpretation of this equation.

5. Waveguides: Consider waves propagating in a vacuum between two (“perfect”) conductingplates at x = 0 and x = a. Let the waves have frequency ω. We consider two dimensional waveswith E and B varying in x and z but not y.

(a) Show that Maxwell’s equations in vacuum,

∇ ·E = 0; ∇ ·B = 0;∂B

∂t= −∇×E; µ0ε0

∂E

∂t= ∇×B

yield1

c2∂2E

∂t2= ∇2E.

(b) Show that the one dimensional wave,

Ex = Ex0 cos (kz − ωt)

is a solution if ω = kc and that it satisfies the boundary conditions (for a good conductor)at x = 0 and x = 1.

(c) Show that the two dimensional wave,

Ey = Ey0 sin(nπx

a

)cos (kz − ωt)

is a solution if ω2 = k2c2 + n2π2c2/a2 and that it satisfies the boundary conditions (for agood conductor) at x = 0 and x = 1 if n is an integer.

(d) Give an expression for the group velocity of the two dimensional waves.

(e) Show that if n = 1 and ω < πc/a no traveling wave solution of this two dimensional formexists.

(f) For ω < πc/a find a solution for Ey that decays exponentially in z.

6. TE modes: EM radiation propagates down a long, perfectly conducting metallic waveguidewith width a, and height b. The EM wave is considered to be of the transverse electric (TE)form, with an E-field only in the y direction, E = E0 sin(mπz/b)ei(kx−ωt) y, where m is aninteger.

(a) Use Faraday’s law to show that the corresponding B-field is:

B =

(E0

ω

)[imπ

bcos(mπz

b

)x + k sin

(mπzb

)z

]ei(kx−ωt)

and show that it satisfies the correct boundary conditions at z = 0 and z = b.

(b) Show that there is a time-averaged Poynting flux in the waveguide only in the positive xdirection, and that for m = 1, it is 〈N〉 = (kE2

0/2ωµ0) sin2(πz/b) x.

2

2nd Year Electromagnetism 2012: Practice Problems on VectorCalculus.

Lecturer: Steve CowleyAnswers in red

These problems are optional and should be entirely revision – hopefully they will help you remember vectorcalculus from last year.

(i) Let ψ = ψ0e−r2 where r2 = x2 + y2 + z2 is the radius in spherical polar coordinates and ψ0 is a constant.

Show that

∇ψ = −2rψ0e−r2 (1)

where r = xi+ yj+ zk

(ii) Let ψ = ψ0 sin(kx)e−ky where k and ψ0 are constants. Show that

∇2ψ = 0 (2)

(iii) Let ψ = ψ01r sin(kr − ωt) where: r is the radius in spherical polar coordinates, ψ0 is a constant, the

constants k and ω are related by ω = kc and, c is the velosity of light . Show that ψ satisfies the waveequation:

c2∇2ψ =∂2ψ

∂t2(3)

use wave equation in spherical polars.

c21

r2∂

∂r(r2

∂ψ

∂r) =

∂2ψ

∂t2. (4)

just differentiate.

(iv) Find the curl of the vector field A = yi − xj. note A = r × k. Draw the ”field lines” of A. Circlesaround zero. Show that we can write A = ∇ψ × k and find ψ. Note r = ∇r2/2 thus ψ = r2/2.

(v) Consider a vector field A(x, y, z) = ∇a ×∇b where a = a(x, y, z) and b = b(x, y, z) are arbitrary scalarfields. Show that ∇ ·A = 0. Also find C such that ∇×C = A. Is C unique?Use formula from the formula sheet ∇ · (A × B) = B · ∇ × A − A · ∇ × B don’t get confused with theA in this formula and our A thus ∇ · A = ∇ · (∇a × ∇b) = ∇b · ∇ × ∇a − ∇a · ∇ × ∇b = 0 because ofthe identity ∇ × ∇a = ∇ × ∇b = 0. Finding C is easiest by inspection or trial and error. Notice that∇b · A(x, y, z) = ∇b · ∇a × ∇b = 0 = ∇b · ∇ × C. Thus using the vector identity above we can show∇ · (∇b × C) = 0 We therefore try choice a vector parallel to ∇b so that we satisfy this condition. LetC = g∇b then is ∇ × C = A = ∇g × ∇b then clearly g = a or C = a∇b satisfies the equation. But C isnot unique the general solution is C = α∇b +∇χ where χ is any scalar field. χ does not contribute to Abecause of the identity ∇×∇χ = 0.

(vi) Let P = f(r · d)(r× d) where r = xi + yj + zk, f is an arbitrary function and d is a constant vector.Show that ∇ ·P = 0.Proof 1.: The quick way to do this is to orient the coordinates so that d points in the z direction i.e.d = dk. Then P = df(zd)(yi−xj) – note the x component has no x dependence and the y component has noy dependence. Then use the usual cartesian form of the divergence to show it trivially has zero divergence.

Proof 2.: Use identity from formulary ∇ · fA = ∇f ·A+ f∇ ·A with f = f(r · d) and A = r× d. Usingvector identity from above (part (v)) we can show ∇ · (r × d) = 0 because ∇ × r = ∇ × d = 0. Thus∇ ·P = ∇f · (r× d). Let r · d = u then ∇f = ∇u df

du = d dfdu . Then ∇ ·P = df

dud · (r× d) = 0 Q.E.D.

(vii) Let ψ = ψ0eik·r where ψ0 is a constant and k is a constant (wave) vector. Show that ∇ψ = ikψ.

(viii) Let A = A0eik·r where A0 is a constant vector. Show that ∇ ·A = ik ·A and ∇×A = ik×A.

2nd Year Electromagnetism 2or2; rromework 1.Lecturer: Steae Cowley

Question 1-' Dipoles. This should be entirely revision - but hopefuily it will get you warmed up.Consider a (model) polar diatomic molecule with charge fq at *Ar and _q at _Ar.

(i) Give an exact expression for the electric field.

Et ' \ = f i :o l )/*7re,fr-ar13

q (e+ af )h|T eol!+ aJ: I'

-q+q

For this molecule we define the dipole moment 6 p:2gAr.(ii) Give a simple expression for the electric field when A. <<

" * write your answer in terms of p.

E:N l l -a: l -3 = rrs-oryi?"= l&-ur). ts-or) i" '* ( .2- 10r, ! + !1r1-t4-;r , - tF . ,* ' ) - ' i i l , -

r# jto gcf otr..' frr*n g*rk Sg -a -b:

F/n\ - t ( :+ 'sg -L7="/ o,7tol t t 7!PfGr,

(iii) Descrihe qualitatively what would happen to two isolated dipoles placed near each other - how wouldtnev mover

L:i iq **Fcaie so t i^""F +ue) .oF"[r*.*f i (-"") tLe,^ -ot..*t"r ,^, i l .*.. t **t-; f

elect'o"*t^e

3 sr € .rr* sp,r.$ur-. k#^ oi\r"t J ' p*-.uh$*

Question 2. Spherical Charge Distributions.

Letthechargedensi typbegiyenby: p-po forr < oand. p:0 forr > owherep0: consranr) a:constant and r: \FTple. url Guurrjs law to find the electric fierd E for r < a and r > a.

6 e.auJ-

h1trTfr= fn*o*t * r>a

,tl e sp!^er ir:*i.- Su$* ce rs$i*: r

Syr*eh5 q,q.e**-b9 Sg t Fs# *. C

= l ( #tJeo

- \vFr l a lor D

4 \ '

' r tIe-*l '

l+t /?5

t=Po

*

t r tPo>f -f*^ra7, v*o

E.' ft r' 7/

" L6

fo, t<a

In the limit 16 J 0, po is essentially zero everywhere except close to the origin where it tends to infinity.Thus in this limit

[ ,oolr)0, -+A(o) [ ,oor:A(o)a.JV JV

if the volume V contains the origin. We can define the three dimensional Dirac delta function as:

, r (") : t i^ ( )2 , . ' -+ ' )ro-+tl \

47lrr[- I

We can therefore write the charge density of a point charge as p : qd(r).

(v) Argue that for a point charge Q at r: r'we may write p : q6r - r/).HARDER

l,Ja\.1 urog i-oo heA jrrf il,-[t- eciovdrr..("n, s, tu",,f Yr'e cuin'^ is J

Yt' -rka^ *s o!*" ' tst: l '

(vi) Find E - G(r, r') for p : €od(r - r') i.e. find the solution to:

V.G:d(r-r ' ) .

We ofben refer to functions like this os Green's functi,ons.

g = €" =) IJ - fs-r ,) = G/r, r , \r.

ltlf lt-s'12

HARD(vii) Hence show that,

E(r): I oe,t nffifiav, (6)where the integration is over all r/ and l" - "'l

is the distance between r and rt.

= V' ( p::', g $,! ) J,'t 't - ' 'z J 'e , t . , ,^"*L J &-u'

€. go u

Div , .J;h^ t laf t t r

f Yt 1; gl'f '

:) E = * \

tut ') e(r r') otlg' ile,4re (6 ) so*r* "f f,*u* o**og e.r/ ,"*rL.^lu$ l;3- '

Question 5. Magnetic Field from a Steady Current.

(i) Caiculate the magnetic field inside and outside a long straight conductor with circular cross section ofradius R that caries a uniform current densitv with total current I.

(3)

(4)

lc /

->et t Cuttt*l tk$*i.^.u fe*f

l t "LR

6-

8r l** .*S b ' -

5r.* &,t.=)l t r i - rB*: Fuf

>R n*-*

l r\ t rs ide ur i te i )ur:

IlIr gs

-(r, t )= ./l"f , f

alflz?

= p"f I l r<eR'

f.f = -hri) /roJ Bo2TrR7 \- -' A-

\ - i **

f<R

.u 6r,A9 = rzr [n - ! t ,At, . f i

ri** )

Hse q*[*fu

$* flareuuiL \e,rgth

r f

Question 3. Cylindrical Charge Distributions.

Letthe chargedensi ty pbegiven by: p: paf , forr <aand, P:0 forr > a,where p0: constant ' a:

constant and r : tFiF Use Gauss's law to find the electric field E for r ( a and' r > a'

"t , ' ' l i :& fo'€* b

t,td

Question 4. Electric Field of Arbitrary Stationary Charge'

In this question we derive the formuia for the electric field from a stationary distribution of charge.

(i) Consider a spherical distribution of charge; i.e. p - p(r) with r : JPWTF. Shovr that in this case,

v.E: '^ !o 'n) : !p(r) .T'Clr €0

where E, is the radial component of E. What about Es and' E4?

Ee= E(, = 0 bn s3n**?5

(ii) Integrate to find a formal expression for 8,.

(1)

(ii i) Suppose P:0 for ,,l.ur*Ot"ss.E" for r > ain terms of the total charge'

-T6TAL (rdlr f i6e : ot t ' ' f ( ' ' ) t r ' * f iJ a

'o ' '

- - ) Er= |

' r , ' pr , , \ l r ' = f ( t ' ' 7(") ' t ' " *€*€"r"J [ ' r " 6, f )" t L;4z

2q r-#tp : po: ;:::t e 'o

. Find E" in the limit rs -+ 0.

t , q Idr ' = "* l I - gT' ( i"l

rbl--* I

II

; '

(iv) Let

Find an explicit expression for 'O"

ft^) l

" / r (c ' ) aLr

(r \

rL1 -

I

a l f -a( fQ_e

*9" Jra* II

t-tsr

| .Ll l /4 ' \ lb*/ \ l

uf

\!

ff

. ":: ,f r'lf4[ Eb I

(ii) Now suppose that inside the conductor, there is a cylindrical hole of radius a whose axis is parallel tothe axis of the conductor and a distance b from it (a * b < R). Assume again that the current density isconstant and that the total current is 1. Show that the magnetic field inside the hole is uniform and equal

4Y

to

t - , , Ty= cutte,t l de"sig =

ff i-a:

t' a\F_L-*r

4 l l t -o

FobI2"G'z - a?\

E' f La, Lf = t l\Y

8= 9f ia 'p^J t -

( totr*ro. ,o= I?ooo e. ( a,6* a.h,p.s

L6. lo,^r )

=Hass = Laoo, l2-,Wprnton = ! . f ,xro'?t P,

cctrfs6.r *e*rfo ^ ?- $rr^*l : 2xro-3 Lg*-s

=) o. * l. ( 7t to-?*

* *v^ : 3, 'o-rtts _-----_

lf - ro 6*/" tJ

= lga2-

Fietct is 6lu* lo egl,wr'tcr n4'J,*s R

cgt,t^c(f,r 6p;q^ 'a-eli;"r x ceJve aF I

Cur/e*l- de"ro' ier .*tce!, /Nsrog HbLf

scale with the size of the cluster? ,

ndu,fo,*

luk hn# tf o.e 6' sp[oa oF n ci"crge '

[,^^ &,ro!,o, I ad lu'f '4ew'

cen-oo a! +,t+ - f .*,,"*f *?^*,lr ; PgfJ o^otker

- b u= o o.'*cl c,a.rre-f'c'(*'+tlil -S. \'^rirte -fap l^ole+t'€

B * -G*?)q: -r"-ui:"?JE')* +. rf ,t)Draw a picture showing the orientation of the fields and the hole.

Question 6. Coulomb Explosions.

In this question you are expectecl to make estimates the more accurate the better. In the basement ofBlackett high powered lasers were used to blast the electrons off clusters of a few thousand atoms. Supposeall the electrons are suddenly removed from a cluster of one thousand carbon atoms. The carbon nucleiithen explode outwa.rds - *hy and how fast? Hi,nt car*ider the energy. How does the energy of the explosion

%t*r"f",t!, . - - -V

f

a3

' Ao(t- t "?4ii6oa I ao)

&;?60 f I

_p4nfur

\ / -

(= yvnv l? '?,

/4,| / 1,: I r.r., r". g \ t- j ' d.i.

| lii€oa L z"= )

=te$* =t q ' = ?v,o-" it5 eo 5 +116o4

So( e:1otht E Uf AGY(-^.

^rYuTg*ta!. fne'$3

2nd Year Electromagnetism 2Ol2: Flomework 2.

Lect,.trer: Steue Cowley

Question 1, Solenoids. In this question the integral form of ampere's law

The magneticfield isconcentratedinto a nearlyunilorm iieldin the centerof a longsolenoid. Thefield outsideis weak anddivergent.

(i) Show that the field inside a long cylindrical solenoid is B : 1L,snI whete n is the number of turns per unitiength and I is the current in each turn (see above). A thin solenoid of length 10an with 600 turns caniesa current of 20 amperes, calculate the field inside the solenoid.

(ii) Now suppose that the solenoid is wrapped into torus - a doughnut shape - this is sometimes called atoroidal solenoid. Calculate the field inside the torus. If y-ou want help see

http: //www.youtube.com/rvatch?v 1j dsUQs9w0uw

lsmr Pob&l tFS .od3

Pnmary trentlo|@t craul

P€6nr a@trs curmllhcdffry t.tnildtua circril|

\k5*zl9 x 4r *v

-Pa,/ , t t 'To,*ud;. i "n: i . i

9Y sl = ' t7:28,f i : F* T j' /t'to uT t)-T*r& L

To.odel roeNt( l*ld

rdf) ' t( i . , . e t . i r . . . .e- rq , ( -J: # a"- ' t ' ! : \ r r"r ' 61 : i I 16,a i

.,;------,,"l*-'-.

,.-r.- r'-'"

*\ .p. - u, ,ArT rI r)o " I-:*-\- avT lP 'l\ . /

\ ^.--l!--"-*^---,-#

(iii) The fusion experiments JET and ITER (http:ffwww.iter.orgf) are based on huge toroidal solenoids.Schematically shown above. The field created by the 18 superconducting "toroidal field coils" (like the blueones above) in ITER is 5.2 tesla at a radius of about 6 meters. What is the current in each toroidal fieldcoll I

Lrf ir 735 T ! - / #12.6 , t | =

lvNu eJ$6mdl,

Question 2. Charge Conservation.(i) Deduce charge conservation from Maxwell's equations. (Former exam question).

ro€o.) F, vxE *r*?d+

(ii) Suppose magnetic monopoles (magnetic charges) exist and that magnetic charge is conserved. Let p6

be the density of magnetic charge and Je the current density of magnetic charge. We take units so that,

= )"1 =)dr

Y -B: [ ropo.

Write down thg equation for the conservation of magnetic charge.

)P* * v. =- =,0-6

d+

9.€ = E =) E )0 ' fr.' df

(1)

(iii) Modify Faradays law to make Maxwell's equations conserve magnetic charge.

\a/t l<

A+

P [+e1 = 'v/'r\ ( - - ) /*r [ '+ V- (F*h" fe"l.or, )- \ /

- Vr F * F*t'at [".,''^, ,, --=)xtat tcr.ra , => *5* 'n

t \ - -\

- t r \ - t rer . [o,edq [oCev lro dfs = V-lF*h'ler*\

*lr-* t / F-of*o "T.€r"r :.) l'- E-x&* "T.er"' : - F" 3g

Question B. rnduction and Tbansformers. ?# = &xE - flo Tg

Consider a l\cm long solenoid (solenoid A) of radii approximately lcm with 400 turns.

(i) Let the current in the A, 14, change with time slowly (In : 1o(r)) so that you may ignore the displacement

current. What is the instantaneous magnetic flux in the solenoid?

S= ft-eo at su[e,^oi& , 7i to-* vaL \r1 = fuig a. 4*og E = ,t'^"nl-'O'l s

4x togTa = ;u."n Io 5

).5 r ro-( 3,

= FLtJt: V n-*, f f i66:7,

s / 67r^t ,u-*f* P

("U *- i l = f l "^*" : leK'o-d. IoJi-"- & d,v

(m I

\s r r :Uo* ro*,

€.tdrrrd

4oo pr" ^ sg.

'tF

= 6Xto-+ S, = UQL'rAc,;fr645b*SS*

4I; 'dr

)o (1 ,"-+ +Y' j aO,\L

il = E,(A,e*)

&fid$drs

fJ LiU,.ef =r1s6 fr nrft

(ii) A voltmeter is connected across the solenoid. The meter will measure / E dt integrated along the wire. r _ ,1Derive an expression for the voltage measured (ignoring the resistance). Cctc-tof,c h^e .rol"r.re ,*t,e*a Io ; Sin[mn tl.

.o, I r*tf]roo ii os !mr *J

Consider now two 10crn long solenoids of radii approximately lcm: solenoid A with 400 turns and solenoidB with 800 turns. B is fitted snugly inside A see diagram below.

Double solenoid - solenoid B inside solenoid A

(iii) Let current Ia flow in A and current 16 flow in B. what is the field inside both?

t ( -4& = 4ooo

Y' lU:8ooo

Now we disconnect B and ionnect it to the volt meter.the coil A.

We put a slowly changing current IA: Io(t) through

(iv) Calculate the induced voltage across B.

UqLT&Gf

lgArundD

I

= iE'dg rJrou

hulrr \

ao" f , $oo F. ,,rnS f, = tkY to uLI"

atf f i

(iii) In the general case where both I,a and 16 are changing show that the voltages measured across A, V4,

Ne,'.F

P"P -

and across B,Va, sat isfy VBfVa:2.

-7- *' ' sr ig 4i'r-u e" i-.t8= t : ' t

Vi l 'hooFtur lsy+ VB'

velocitv of the electro"r,;U*/_--:=---7

l --Al

|AC !t" Il 3 l--'-nlo

-=>*AV

, l - l f 5"Fi- ' cL 'uge r /ose' ' r1 5 ' ' h '** f*

' 'N-u*bn'

"$*.* io*- ' 1un x*sf ou"= r 'e' S 5*ly 1

, C)rarqe, G*os: ix$ ,g -* - e e lY l6F

tut;* (ta,v**L i;#' !'"''a"

+ lsat-)

Question 4. Pinch Effect.MAGPIE in the basement of Blackett squeezes plasmas with magnetic fields and makes very dense and hotplasmas. In this question you wili esiimate the forces involved.

(i) Show that if we assume the current in the (neutral) plasma is carried by many electrons and that theions are stationary then

J : -lleev ( r \

whete n, is the electron number density (the number of electrons per unit volume) and v is the average

Cd NSt 0r i i l

f t4o :?+t:| -T.r \

x * €vt"y 5t

- ,j = -RY'!gY

(ii) Recall that the force on a single electron is *e(E * v x B). Hence show that the force per unit volumeon the electrons and ions in the plasma (and hence on the plasma) is:

JxB (3)

if we assume that the plasma is neutral so that the electron and ion charge densities acancel i,.e. en" : qn6where q is the ion charge"

Tnoe \ ,Fe te l ,n*, , e . 16r*u Fr;n(g 6tr P/&*.7tcL65 ld Alt ' - f

a-v,-:,/t

IJet "I t_

I c,^d' g ,^oo",lzn,at t :-

a*€r o7o 6*Q.

€ ?r*+q' 'E)= - e (v " F)n. d't

lx ? *l=e \

\ , /JU F.tt !t1 i5

$f.r&ut S Fp;. ,&ytz& t j

AJ f"n i** d

(iii) Magpie makes a cylindrical shell of current flowing in plasma made from vaporized wires. The currentis taken to be zero except in a shell ofradius 0.01 metres and thickness 0.001 metres where:

J : Joz Jo: constant. f =* J". Z77r 3 (4)

Calculate the magnetic field (ignoring displacement current); trox ta- sn a.Ti

(iv) Calculate or estimate the radial force per unit length on the plasma. Show that it squeezes the plasma.

T,\

A/p-1* nAb,Ac F:s$f : =

0 $i ir-rj rf 1-rr$fF

, f ,9 i : , rvr t ro ' rcs - i J i ' ' t * : ' - t* ( i?; ; -**** ln* ' [n ' ]

/ gld t -., f1 rB I rir.5el

ry- i *"

Fo-

A.Ttr

I ti rT

IB

loTA t-

A!^ct lne (fLr,l/C"*i':" t g *etq l,tutPti *-r'"ap

(v) Suppose the initial plasma is one tenth the mass density of solid aluminiumtrEstimate the time to halve

the shell radius.

t4 = qTrr .3 .

-2, t

2lTx7.7 xb- Ys

P' 'q(-l

-17L77*/0 ' rx l0- . :

s--- , . -1: !#

3Irxy 'o-"x8"&r/s-3

P' ' = 7?oo kql* t1""

D -

o.t l . lo "F*uf*t - l-,to lo

TOI e, c r At.l :'

- t IDeJ

^irr( [€K4rh

I

tr

.;

k= A.,rJc,r.h o*" -1f loIA7T'r

8,$vrg'3

\ * loto ^r ' '42

l0 - -LO, 'La N /A t ' t L

d.tslc.*slL *r 12J- a-t-7

' ' ( '7 l t lA Yw

a

r = 4,2 X l0- ' ' 62

2nd Year Electromagnetism 2012: Homework No. 3.

Lecturer: Steve Cowley

These questions require lectures 5 and 6.

Answers in red

Question 1. Circular Polarisation. (i) Calculate the real part of the complex wave solution toMaxwell’s equations in a vacuum:

E = E0(x+ iy) exp ik(z − ct). (1)

where x and y are unit vectors in the x and y directions respectively.

Real(E) = Real(E0(x+ iy) [cos [k(z − ct)] + i sin [k(z − ct)]]). (2)

Real(E) = E0(cos [k(z − ct)]x− sin [k(z − ct)]y) (3)

(ii) Calculate the magnetic field of the wave.

∂B

∂t= −∇×E = −ikE0[z× (x+ iy)] exp ik(z − ct). (4)

B =E0

c[z× (cos [k(z − ct)]x− sin [k(z − ct)]y)]. (5)

(iii) Describe the field seen by an observer at z = 0. What happens if we replace (x+ iy) with (x− iy).

at z = 0Real(E) = E0(cos [ωt]x+ sin [ωt]y) (6)

with ω = kc, this is a vector rotating in an anticlockwise direction. The magnetic field is perpendicular tothis also rotating antoclockwise. If we replace (x+ iy) with (x− iy) we get a vector rotating in the clockwisedirection. These are the right and left circular polarizations.

(iv) Calculate the motion in the wave of an electron which starts from rest at (0, 0, 0) at time t = 0. Youmay ignore the magnetic force – when is this approximation valid?

since there are no forces in z direction it stays in x− y, z = 0 plane. Integrate

mdv

dt= −eE = −E0(cos [ωt]x+ sin [ωt]y) (7)

after 2 integrations and using initial conditions we get

r =eE0

ω2m((cos [ωt]− 1)x+ (sin [ωt]− ωt)y) (8)

you can easily show that the magnetic force is v/c times smaller so it is ok to ignore it for nonrelativistic

motion – here that is when c≫ eE0

ωm.

Question 2. General 1D Solution of the Wave Equation.

Consider the wave equation for linearly x polarized waves travelling in the ±z directions:

∂2Ex

∂t2= c2

∂2Ex

∂z2(9)

(i) Transform Eq. (2) to the independent variables q = z − ct ands = z + ct and show that:

∂2Ex

∂s∂q= 0 (10)

use (∂

∂z

)t

=

(∂s

∂z

)t

(∂

∂s

)q

+

(∂q

∂z

)t

(∂

∂q

)s

=

(∂

∂s

)q

+

(∂

∂q

)s

and

(∂

∂t

)z

= c

((∂

∂s

)q

−(∂

∂q

)s

)then it is trivial

(ii) Hence show that the general solution of Eq. (2) is:

Ex = E+(q) + E−(s) = E+(z − ct) + E−(z + ct) (11)

where E+ and E− are arbitrary functions.

one integration over s gives

∂Ex

∂q= G(q)

where G(q) is arbitrary then integrating over q gives

Ex = E+(q) + E−(s)

wheredG

dq= E+(q).

(iii) Calculate the general form of the magnetic field in terms of E+ and E−.

∂By

∂t= −∂Ex

∂z

→ By =1

c(E+(q)− E−(s))

(iv) Suppose that at t = 0:

Ex = E0e−αz2

and By = 0 (12)

calculate the fields for t > 0. Describe the solution.

initial bump becomes two travelling pulses going in opposite directions

Ex =E0

2

(e−α(z−ct)2 + e−α(z+ct)2

), By =

E0

2c

(e−α(z−ct)2 − e−α(z+ct)2

)

Question 3. Spectrum.

We define the Fourier transform of Ex by:

Ek(t) =

∫ ∞

−∞Ex(z, t)e

−ikzdz (13)

(i) Calculate Ek(t) for the pulse E+ calculated in Question 2.

Ek(t) =

∫ ∞

−∞

E0

2

(e−α(z−ct)2

)e−ikzdz =

E0√π

2√αe−

k2

4α−ikct

(ii) Calculate the ”spectrum”, i.e. |Ek|2. How does the width of the spectrum (in k) depend on the widthof the pulse in z?

|Ek(t)| =E2

0π

4αe−

k2

2α

the pulse is width ∆z = 1/√α and the spectrum width ∆k =

√2α are reciprocal

Question 4. General Plane Waves.

We may represent a general electromagnetic plane wave by (real part of the complex exponentials:

E = E0e(ik·r−iωt) and B = B0e

(ik·r−iωt) (14)

(i) Show that Faraday’s law: ∂B∂t = −∇×E becomes:

−iωB0 = −ik×E0 (15)

(ii) Hence show that Maxwell’s equations for plane waves in a vacuum become:

ω

kB0 = n×E0 and

ω

kE0 = −c2n×B0 (16)

where n is the unit vector in the direction of k and k = |k|. Hence show that ωk = ±c.

ω2

k2E0 = −c2n× ω

kB0

= −c2n× (n×E0)

= −c2 ((n ·E0)n− (n · n)E0))

= c2E0

where we have used n ·E0 = 0 and n · n = 1.

(iii) Let us define two constant unit vectors e1 and e2 by e2 = n×e1 and e1 = e2×n. Show that n = e1×e2.

e1 × e2 = (e2 × n)× e2 = (e2 · e2)n− (e2 · n)e2 = n

(iv) Show that the real part of E = E0(e1 + ie2)E0e(ik·r−iωt) yields circularly polarized waves.

Question 5. Some Common Waves.

(i) Look up the frequency or wave length of your cell phone carrier signal. How big is the wavelengthcompared to the size of the phone?

3 band 900/1800/1900MHz longest wavelength is 1/3 metre. Larger than phone but not by much.

(ii) What is the wavelength of sodium orange yellow light? How big is a sodium atom?

589nm. atom size is about 0.2nm quite a bit smaller than wavelength.

(iii) What is the typical wavelength of a hospital x-ray?

from 5× 10−11m to 5× 10−13m.

2nd Year Electromagnetism 2012: Homework No. 4.

Lecturer: Steve Cowley Answers in red

Question 1. Energy Conservation in Waves.

This question uses results developed in Lecture 8.

(i) Derive an expression for the Poynting flux for a wave Ex = E+ cos [k(z − ct)]. What is the time averageflux?

since we have E = Exx = E+ cos [k(z − ct)]x and B = Byy = 1cE+ cos [k(z − ct)]y

N = N z =ExBy

µ0z =

E2+

cµ0z cos2 [k(z − ct)]

the time average of the cosine squared is a half so that

N = c[ϵ0E

2+

2] = cW

where W is the time averaged energy density in the waves i.e. the time average of: 12ϵ0E

2x + 1

2µ0B2

y . It isworth noting that the energy density in magnetic and electric field is identical for these EM waves.

(ii) The Rutherford lab has a plan to produce 10 petawatt (1016) laser with a wavelength of approximately910 nanometres. The laser would be pulsed 300J in 3×10−14s. This laser can be focussed onto a small spotto intensities of 1027Wm−2. Calculate/estimate the typical electric field in the focus.

(iii) Compare the electric field in the focus to the the typical electric field seen by an electron in a hydrogenatom. Hint what is the approximate distance of an electron from the nucleus in hydrogen?

from last part we have N = c[ϵ0E

2+

2 ] = 1.3× 10−4E2+ = 1027Wm−2 thus E+ ∼ 2.8× 1015V m−1

(iv) Estimate the typical kinetic energy of an electron in the focus. How does it compare with the rest massenergy of an electron (mec

2)? Can you do this relativistically?

First Newton’s 2nd law with no relativity dropping the magnetic field force gives

medVxdt

= −eEx

for the purpose of estimating we take dVx

dt ∼ ωVx where for this wavelength ω ∼ 2× 1015s−1 then

1

2meV

2x ∼ 1

2me

(eEx

meω

)2

∼ 2.8× 10−8J

the rest mass energy of the electron is mec2 = 8.1 × 10−14J – thus since the kinetic energy is larger than

the rest mass energy the electron is highly relativistic and our estimate is wrong. This is now beyond thesyllabus and your knowledge – but a simple way to estimate is use the fully relativistic equation:

dpxdt

= −eEx

thus px ∼ eE+/ω and that for highly relativistic particles E =√p2c2 +m2

ec4 ∼ pc ∼ eE+c/ω ∼ 6.4×10−11J

still larger than the rest mass energy but this estimate is accurate.

Question 2. Radiation from a Current Sheet.

This question expands on the model developed in class (Lecture 7).

(i) Suppose we take a current sheet source,

Jx(z, t) = I0 cos (ωt)δ(z) (1)

Write down the radiated waves.

from the lecture the electric field in the waves is:

Ex = −1

2

õ0

ϵ0I0 cos k(z − ct) and By =

1

cEx for z > 0

Ex = −1

2

õ0

ϵ0I0 cos k(z + ct) and By = −1

cEx for z < 0

(ii) Write down a formula for the energy flux in the waves. Show that the energy conservation relation(Lecture 8.) is obeyed.

from question 1.

N =E2

+

cµ0z cos2 [k(z − ct)] =

õ0

ϵ0

I204z cos2 [k(z − ct)] for z > 0

N = −√µ0

ϵ0

I204z cos2 [k(z + ct)] for z < 0

the time average loss of energy to radiation per metre squared of sheet current is then the sum of the leftand right fluxes time averaged: √

µ0

ϵ0

I204

the negative of the work done by the electric field on the current per metre squared of current sheet

−∫JxExdz =

∫I0 cos (ωt)δ(z)

1

2

õ0

ϵ0I0 cos k(z − ct) =

õ0

ϵ0

I202

cos2 [ωt]

which when time averaged yields the answer above for the time averaged radiated power.

(iii) In JET, the European fusion experiment in Oxfordshire, we heat the plasma with waves from an antennathat is almost a sheet. The Antenna is about 2 metres square and radiates waves at frequency ∼ 3×108s−1.Calculate the approximate wavelength of the radiated waves.

the wave length is about 1m – there are some corrections for the plasma response that we ognore here

(iv) The antenna on JET puts about 30 megawatts of power into the plasma. Approximating the antennaas a sheet calculate the amplitude of the oscillating current in the antenna.

the power is approximately 7.5MW per metre cubed – using our formula from above we have:

7.5× 106 =

õ0

ϵ0

I204

thus I0 = 283 amps

(v) [HARDER] Suppose we have two current sheets, i.e. :

Jx(z, t) = I0 sin (ωt+ ψ)δ(z − z0) + I0 sin (ωt− ψ)δ(z + z0) (2)

Where the phase, ψ, and z0 are constants. Calculate the radiated fields. What values of the phase andz0 minimize the radiation towards z → −∞? What values simultaneously maximize the radiation towardsz → ∞ and minimize the radiation towards z → −∞?

we can simply superimpose the solutions for each current sheet using the formula derived in class and shiftingthe origin and phase:

for z > z0 Ex =1

2

õ0

ϵ0I0 sin [k(z − z0 − ct)− ψ] +

1

2

õ0

ϵ0I0 sin [k(z + z0 − ct) + ψ]

for z < −z0 Ex = −1

2

õ0

ϵ0I0 sin [k(z − z0 + ct) + ψ]− 1

2

õ0

ϵ0I0 sin [k(z + z0 + ct)− ψ]

to minimize the radiation towards z → −∞ we make the second term in the expression above cancel thefirst i.e.

sin [k(z − z0 + ct) + ψ] = − sin [k(z + z0 + ct)− ψ]

this is true if they have a phase difference of an odd number of π so:

ψ − kz0 =2m+ 1

2π

where m is an integer. For maximum radiation towards z → ∞ we must make the two terms add in phasewhich gives the relation:

ψ + kz0 = nπ

where n is an integer. so for example we could choose m = n = 0 which would yield ψ = −kz0 = π/4. Thisphased antenna would radiate in only one direction.

Question 3. Radiation from an Extended Source – HARD

This question is advanced – but still worth a shot.

(i) We begin by redoing the current sheet of Lecture 7. with complex sources. Let us write a current sheetat z0 as:

Jx(z, t) = I0(z0)e−iωtδ(z − z0) (3)

Where the constant I0(z0) is labelled by the position of the sheet. Show that for z > z0 with this source:

Ex(z, z0, t) = −1

2

õ0

ϵ0I0(z0)e

ik(z−z0−ct) By(z, z0, t) = −1

2µ0I0(z0)e

ik(z−z0−ct) (4)

and that for z < z0 with this source:

Ex(z, z0, t) = −1

2

õ0

ϵ0I0(z0)e

−ik(z−z0+ct) By(z, z0, t) =1

2µ0I0(z0)e

−ik(z−z0+ct) (5)

where we have labelled the fields with an argument z0 to show that they depend on z0.

(ii) We can construct an extended source by adding sheets together – well integrating.

I0(z) =

∫ ∞

−∞I0(z0)δ(z − z0)dz0 (6)

(iii) By integrating the equations for Ex(z, z0, t) and By(z, z0, t) over z0 show that

Ex(z, t) =

∫ ∞

−∞Ex(z, z0, t)dz0 and By(z, t) =

∫ ∞

−∞By(z, z0, t)dz0 (7)

satisfy the 1D equations with an extended current source I0(z)e−iωt i.e.

∂By

∂t+∂Ex

∂z= 0 and µ0ϵ0

∂Ex

∂t+∂By

∂z= −µ0I0(z)e

−iωt (8)

the equations for Ex(z, z0, t) and By(z, z0, t) are

∂By(z, z0, t)

∂t+∂Ex(z, z0, t)

∂z= 0 and µ0ϵ0

∂Ex(z, z0, t)

∂t+∂By(z, z0, t)

∂z= −µ0I0(z)e

−iωtδ(z − z0)

integrating these equations over z0 is easy as we can take the integration under the differentiation withrespect to z and t. Then the above equations fall out.

(iv) Hence show that when z > a and I0(z) = 0 for z > a:

Ex(z, t) =

∫ ∞

−∞−1

2

õ0

ϵ0I0(z0)e

ik(z−z0−ct)dz0 = −1

2

õ0

ϵ0eik(z−ct)

∫ ∞

−∞I0(z0)e

−ikz0dz0 (9)

this z satisfies z > z0 for all z0 where I0 = 0 so we can use Eq. (4) for Ex(z, z0, t) then it falls out. This isjust waves propagating towards infinity from the source, the integral adds all the contributions from bits ofthe source. Describe this solution.

2nd Year Electromagnetism 2012: Homework No. 5.

Lecturer: Steve Cowley

Question 1. Spherical Waves.

Waves from a small source are spherical waves that begin to look like plane waves far from the source. Herewe look at scalar spherical waves – of course EM waves are vector waves.

(i) In three dimensions the wave equation is

∂2ϕ

∂t2= c2∇2ϕ. (1)

We assume spherically symmetric waves thus ϕ = ϕ(r, t). Using the form of ∇2 in spherical polar coordinateswrite down the partial differential equation for ϕ.

Using the spherical form of ∇2 we get∂2ϕ

∂t2= c2

1

r2∂

∂r(r2

∂ϕ

∂r)

(ii) Let ϕ = A(r, t)/r. Show that A obeys the one dimensional wave equation:

∂2A

∂t2= c2

∂2A

∂r2. (2)

differentiating using chain rule we have:

(r2∂ϕ

∂r) = r

∂A

∂r−A

and therefore∂2ϕ

∂t2=

1

r

∂2A

∂t2= c2

1

r2∂

∂r(r2

∂ϕ

∂r) = c2

1

r

∂A

∂r2

hence Eq. (2)

(iii) Write down solutions for A(r, t).

the solutions are the general solutions of the wave equation given in Homework 3 Question 2.

A(r, t) = H(r − ct) +G(r + ct)

where H(....) and G(....) are arbitrary functions.

(iv) Suppose the source of the waves is localised at (or very close to) r = 0. Find the wave solutions thatradiate away from the source.

can only be the outgoing wave solutionA(r, t) = H(r − ct)

where H(....) is an arbitrary function.

(v) Explain physically why the amplitude of the wave decreases as 1/r.

Physically it is because the wave energy is spread over a bigger sphere at larger distances hence the amplitudedrops with distance.

Question 2. Multipole Moments.

For a stationary charge distribution the scalar potential is given by:

V (r) =

∫ρ(r′)

4πϵ0|r− r′|dr′3. (3)

See Lecturer 9.

(i) Show that for r ≫ r′

1

|r− r′|∼ 1

r+

r · r′

r3. . . , (4)

where |r| = r.

same trick used in solution of question 1 problem set 1. use

1

|r− r′|=

1

[(r− r′) · (r− r′)]1/2= [r2 − 2r · r′ + r′2]−1/2 ∼ 1

r+

r · r′

r3. . . ,

(ii) Show that the potential far from the charge distribution (i.e. when r ≫ r′) – has the form:

V (r) =Q

4πϵ0r+

r · p4πϵ0r3

, (5)

and give expressions for the charge Q and the dipole moment p.

substitute Eq. (4) into (3) to get result with

Q =

∫ρ(r′)dr′3

p =

∫r′ρ(r′)dr′3

(iii) Water molecules have a dipole moment – estimate how big it is.

a fraction of an electron charge is moved to the oxygen leaving a positive charge on the hydrogen so veryroughly

|p| ∼ 0.3ed ∼ 0.5× 10−29mC

where d is the size of the molecule.

Question 3. Vector Potential and Waves.

Find the vector and scalar potentials for an Electromagnetic wave traveling in the z direction and polarisedin the x direction.State which gauge you use.

in the Lorentz gauge with V = V (z, t) and A = A(z, t) we have

0 =1

c2∂V

∂t+∂Az

∂z

E = Ex(z − ct)i = −∇V − ∂A

∂t

we can take V = Az = 0 without loss of generality and then

A = Ax(z − ct)i

and

Ex = −∂Ax

∂tBy =

∂Ax

∂z

Question 4. Some Numbers.

(i) Find the approximate wavelength of red light.

(ii) Find the wavelength of the photon emitted in the n = 2 to n = 1 transition in hydrogen. How does itcompare to the size of the hydrogen atom.

(iii) Find the wavelength of BBC radio 4.

(iv) Find the wavelength of the gamma ray photons emitted whenan electron and positron anihilate eachother.

Electricity and Magnetism IIProblem Sheet 6: Vacuum Waves and Radiation

1. Wave-vectors: A sinusoidal wave of wave number k and angular frequency ω may be writtenas Ψ(r, t) = Ψ0 exp i(k.r− ωt) where Ψ0 is a constant vector (e.g. one of the components of theEM field). For such waves, propagating in the directions specified below, write k in terms of itscomponents and hence write down expressions in cartesian coordinates for waves travelling inthe:

k = kxx + kyy + kzz and r = xx + yy + zz

(a) z direction;k = kz → Ψ = Ψ0 exp i(kz − ωt)

(b) −x direction;k = −kx → Ψ = Ψ0 exp i(−kx− ωt)

(c) yz plane at 45 to y and z;k = (k/

√2)y + (k/

√2)z → Ψ = Ψ0 exp i((ky/

√2) + (kz/

√2)− ωt)

(d) xy plane at angle φ from x towards y.k = (k cosφ)x + (k sinφ)y → Ψ = Ψ0 exp i((k cosφ)x+ (k sinφ) y − ωt)

c) d)z

y45˚

k

k/√2

k/√2!˚k!sin!

k!cos!

ky

x

2. Complex Phase: In complex form, an EM wave is described by E(r, t) = E0 exp i(k.r− ωt),where E0 is a constant vector with complex components. If E0 is in the y direction, show thatthe real (physical) wave electric field can be written as E = E0 cos(k.r − ωt + φ)y , wheretanφ = E0i/E0r and E0r and E0i are the real and imaginary parts of the complex field E0.

E = E0eik·r−ωt = E0e

ik·r−ωt y

Easiest is to write E0 = E0r + iE0i = E0(cosφ+ i sinφ) = E0eiφ

where E0r = E0 cosφ and E0i = E0 sinφ

so tanφ = E0i/E0r and E02 = E0r

2 + E0i2

3. Elliptical polarisation:

(a) An elliptically polarised wave propagates in the +x direction in vacuum. The (real physical)electric field of the wave is,

E = E0y cos(kx− ωt) y + E0z sin(kx− ωt) z.

Express this field in the complex form.

E = <[E0yei(kx−ωt) y] + <[−iE0ze

i(kx−ωt) z]

but −i = e−iπ/2 so, E = <[E0yei(kx−ωt) y] + <[E0ze

i(kx−ωt−π/2) z]

(b) Show that the magnetic field of the wave in complex form is

B(x, t) = (1/c)(E0zeiπ/2 y + E0yz)ei(kx−ωt).

1

What is the real physical magnetic field?

Using B = −∇×E, and k = (k, 0, 0), and E(x) only,

B = −

∣∣∣∣∣∣x y z∂∂x 0 00 Ey Ez

∣∣∣∣∣∣ = −<[ikE0ye

i(kx−ωt) z]−<[ikE0zei(kx−ωt−π/2) y]

(note change of sign for y term). Now integrate w.r.t. t,

B = <[(k/ω)E0yei(kx−ωt) z]−<[(k/ω)E0ze

i(kx−ωt−π/2) y]

B = <[(E0y/c)ei(kx−ωt) z]−<[(E0z/c)e

i(kx−ωt−π/2) y]

Real part is,

B = (E0y/c) cos (kx− ωt) z− (E0z/c) cos (kx− ωt− π/2) y

Or,

B = −(E0z/c) sin (kx− ωt) y + (E0y/c) cos (kx− ωt) z

(c) Use these results to show that the time-averaged energy density in the electric field is givenby 〈UE〉 = 1

4ε0(E0y2 + E0z

2) , and that the time-averaged energy density in the magneticfield, 〈UB〉, is equal to 〈UE〉.UE = 1

2ε0E2 = 1

2ε0(E ·E)

= 12ε0(E0y cos(kx− ωt) y + E0z sin(kx− ωt) z) · (E0y cos(kx− ωt) y + E0z sin(kx− ωt) z).

= 12ε0(E

20y cos2(kx− ωt) + E2

0z sin2(kx− ωt)).

So, 〈UE〉 = 12ε0(E

20y〈cos2(kx− ωt)〉+ E2

0z〈sin2(kx− ωt)〉)= 1

2ε0(E20y

12 + E2

0z12) = 1

4ε0(E0y2 + E0z

2)

Similarly, UB = 12(B2/µ0)

So 〈UB〉 = 〈 12µ0

((E0y/c)2 sin2(kx− ωt) + ((E0z/c)

2 cos2(kx− ωt))〉= 1

2µ0c2(E0y

2 12 + E0z/c

2 12) = 1

4ε0(E0y2 + E0z

2) = 〈UE〉where we made use of 1/c2 = ε0µ0

(d) Calculate the time-averaged energy flux and show that 〈N〉 = c〈U〉x , where 〈U〉 is thetotal average energy density. Hence give a physical interpretation to the Poynting flux.

N = 1µ0

(E×B) =

∣∣∣∣∣∣x y z0 E0y cos(kx− ωt) E0z sin(kx− ωt)0 −(E0z/c) sin (kx− ωt) (E0y/c) cos (kx− ωt) z

∣∣∣∣∣∣= x 1

µ0

(E0y

2/c) cos2(kx− ωt) + (E0z2/c) sin2(kx− ωt)

So 〈N〉 = x 1

cµ0

E0y

2 12 + E0z

2 12

= 1

2cε0E0y

2 + E0z2

x = c〈U〉x

Since 〈U〉 = 〈UE〉+ 〈UB〉 = 2× 14ε0(E0y

2 + E0z2)

4. Dipole Antenna: A simple Hertzian antenna, stretching from z = +dl/2 to z = −dl/2, has atime-varying current I flowing in it. In lecture 3, we derived the electric field generated in theradiation field in such an antenna (i.e. assuming that the wavelength of the radiation is muchlarger than the antenna, λ = (2πc/ω) dl);

E =1

4πε0

[I] dl sin θ

rc2θ

2

(a) If the current flowing is given by I = I0 sinωt, show that the radiation field can be expressedas a plane electromagnetic wave of form;

E = E0 cos(k · r− ωt) n

where n is a unit vector normal to the direction of propagation k. Give a value for k andE0.

I = I0 sinωt → I = I0ω cosωt and→ [I] = I0ω cosω(t− r/c)

Put into above expression for E,

E =1

4πε0

I0ω dl sin θ

rc2cosω(t− r/c)θ

Which is a plane wave with n = θ, k = ω/c (note cosx = − cosx), and E0 =1

4πε0

I0ω dl sin θ

rc2

(b) Hence give an expression for the mean energy flux radiated by the antenna, (you may usethe expression derived in question 3 for the Poynting flux).

From above 〈N〉 = c〈U〉k = cε0E02〈cos2 ω(t− r/c)〉r =

1

2cε0

16π2ε20

I02ω2 dl2 sin2 θ

r2c4 3r

(c) By integrating over all angles, show that the total (mean) power radiated can be writtenin the form 〈P 〉 = RradIrms

2, and write Rrad in terms of the wavelength of emission λ. Youwill require the integral

∮sin2 θdΩ = 8π

3 , where the integral is over all solid angle.

〈P 〉 =

∮〈N〉 · dS =

∮1

2

1

16π2ε0

I02ω2 dl2 sin2 θ

r2c3r2dΩ (r · r)

=1

2

1

16 2π2ε0

I02ω2 dl2

r2c3r2 8π

3=I0

2ω2 dl2

12πε0c3

but ω = 2πc/λ, so 〈P 〉 =I0

24π2c2 dl2

12πε0c3λ2=

2π

3ε0c

(dl

λ

)2

Irms2

where we used I0 =√

2Irms. So Rrad =2π

3ε0c

(dl

λ

)2

.

(d) A one metre long antenna is used to send radio waves to taxis at a frequency of 50 MHz.The receivers can detect an electric field of 0.01 Vm−1. Calculate the antenna’s effectiveresistance. Estimate the power needed to cover London with one transmitter, and thecorresponding current required.

ω = 2πf = 3.14× 108 rad s−1 and λ = c/f = 6.0m

Rrad = 2π3ε0c

(1.06.0

)2= 789/36 = 21.9 Ω

Say London has r = 10 km, using E0 =1

4πε0

(√

2Irms)ω dl sin θ

rc2, with θ = π/2 (antenna

most efficient if pointing up).

Gives Irms =4πε0rc

2E0√2ωdl

= 2.25 A.

and 〈P 〉 = 21.9× 35.32 = 111 W

5. Synchrotron Radiation: In Lecture 2 we derived the formula for radiation from an acceleratingcharge. For acceleration a in the z−direction the fields are given by;

E =Q[a] sin θ

4πε0 rc2θ; B =

Q[a] sin θ

4πε0 rc3φ

3

(a) Find the instantaneous energy flux in the waves at a distance r as a function of θ and theretarded acceleration [a].

N =1

µ0(E×B) =

1

µ0

Q[a] sin θ

4πε0 rc2

Q[a] sin θ

4πε0 rc3(θ × φ) =

Q2[a]2 sin2 θ

16π2ε0 r2c3r

(b) Show that the total instantaneous radiated power is:

W =Q2[a]2

6πε0c3(Larmor equation)

Integrating over all angles (as in previous question); W =Q2[a]2

16 2π2ε0 c38π

3=Q2[a]2

6πε0c3

(remember 1/r2 in N cancels with the r2 from the surface element integrated over - powerradiated should be independent of distance).

(c) Use this result to calculate the power of synchrotron radiation radiated by an electron ofspeed v ( c) traveling in a circle in a magnetic field B = Bx.

In a magnetic field F = −e(v ×B) = ma, so a = evB/m which is constant in amplitude(though of course direction of radiation varies).

Hence Wsynch =e4v2B2

6πε0m2c3

(d) Synchrotron radiation cools plasmas and can be a problem for fusion. Estimate the syn-chrotron radiation power from a plasma with 1022 electrons and temperature of 108 K in amagnetic field of 5 T. Approximately how long does it take for the plasma to lose half it’senergy/temperature. Assume that v is sufficiently below c to use the above formula.

Hint. How are the temperature and the mean velocity of the particles related.

1

2m〈v〉2 =

3

2kBT → 〈v〉 =

(3kBT

m

)1/2

= 6.7× 107 ms−1

and Wsynch = 2.0× 10−14 W (= 125 keVs−1) per particle, or Wplasma = 200 GW.

Each particle has =32kBT = 2.0× 10−15, so half its energy would be irradiated in 0.5 · 2×

10−15/2× 10−14 ≈ 0.05 sec

6. Circular Accelerators: When β ⊥ β, the relativistically correct expression for power loss ofcharged particles is given by,

Pobs =q2

4πε0

2

3cγ4β2

(a) The Large Hadron collider accelerates protons up to a final design energy of 7 TeV in a ringof radius r = 4.3 km. Calculate the relativistic factor of the protons γ and hence calculatetheir β = v/c and (1− β).

γ = E/mc2 = 7× 1012/938× 106 = 7460 (units were eV and eV/c2)

γ = (1− β2)−1/2 → β = (1− (1/γ2))1/2 ≈ (1− 12

1γ2

)

Hence β ≈ 1 and 1− β = 9× 10−9

(b) Hence calculate β, and the rate at which a proton irradiates energy. If the accelerator stores3× 1014 protons in the ring, calculate the total energy radiated per second, and compare itto the energy stored by the ring. You might also like to compare the energy in the protonbeam, to the energy of a TGV train (∼400 tons) travelling at 150 km/hour.

∴ β = a/c = (v2/r)/c = β2c/r ' c/r in our case

4

Put in above expression, P =e2

4πε0

2cγ4

3r2= 7.7× 10−12 per electron

or Ptot = 2.3 kJ/s

Energy in beam is 7× 1012 · e · 3× 1014 = 340 MJ

So losses insignificant.

(BTW energy of train at 150km/hour = 42 m/s = 12mv

2 = 350 MJ comparable!)

(c) The LHC used to house an electron-positron collider (LEP). If instead of protons, LHC wereused to accelerate 1014 electrons/positrons to 7 TeV, calculate the total radiated power loss.Can you see why the LHC is a proton collider?

For electrons γ = E/m = 7× 1012/0.511× 106 ≈ 1.4× 107

So rate of loss per electron = 88 W.

Total loss would be 8.8× 1015 W - much more than could ever be supplied!

Ring would need to be bigger (Fermi proposed having a ring around the world), or otherwiseelectrons can only be accelerated to such high energies in a straight line.

5

5th March, 2012

Electricity and Magnetism IIProblem Sheet 7: Dielectrics

1. EM waves in a dielectric: An HIL medium of permittivity ε and permeability µ has zerofree charge density, and zero conductivity. The electric field of a wave propagating in the +xdirection in the medium is given by E(x, t) = E0 cos(kx− ωt) y.

(a) Use Faraday’s law to calculate the magnetic field of the wave. Check that these fields alsosatisfy the Ampere-Maxwell equation.

Faraday’s Law, ∇×E = −B. Note E is only fn of x, so only retain derivatives wrt x.

∇×E =

x y z∂∂x 0 00 Ey 0

= zE0 cos(kx− ωt) z = −kE0 sin(kx− ωt) z = −B

So integrating (and noting we are only looking for oscillating solutions, ∴ no constant ofintegration);

B = (k/ω)E0 cos(kx− ωt) z = (k/ω)E0 cos(kx− ωt) z = (ηE0/c) cos(kx− ωt) z

where we have used c′ = ω/k = c/η.

To check with Ampere-Maxwell Law, ∇×B = µεE

∇×B =

x y z∂∂x 0 00 0 Bz

= (ηE0/c)k sin(kx− ωt) y = (k2E0/ω) sin(kx− ωt) y

(note we cancelled a minus from the curl with one from the differentiation of cos, and weagain used ω/k = c/η).

but µεE = µεωE0 sin(kx− ωt) y = (k2/ω2)ωE0 sin(kx− ωt) y = ∇×B QED

(b) Show that the energy density in the electric and magnetic fields of the wave are equal atevery (x, t). Show that the Poynting vector is given by N = c′U x where U is the totalenergy density.

uE = 12εE

2 = 12εE0

2 cos2(kx− ωt)

uB = 12B

2/µ = 12(1/µ)(ηE0/c)

2 cos2(kx− ωt) = 12(εµ/µ)E0

2 cos2(kx− ωt) = uE

∴ uE + uB = εE2 = εE02 cos2(kx− ωt)

but N = (1/µ)E×B = (1/µ)E0 cos(kx− ωt) y × (ηE0/c) cos(kx− ωt) z

= (µε/µ)(c/η)E20 cos2(kx− ωt) (y × z) = c′U x QED

2. Cerenkov radiation: Energetic particles with velocity with v/c = β are fired into a dielectricof refractive index η. As demonstrated in figure 1, a front of radiation is produced in an opticalshock travelling at an angle θc.Find an expression for θc and hence calculate the minimum energy required for an electron toemit Cerenkov radiation in:

(a) glass (η = 1.5)

(b) a helium cell at atmospheric pressure (η = 1.000036)

1

θcβ

βct

ct/η

Figure 1: Cerenkov radiation

From diagram cos θc = (ct/η)/(βct) = 1/ηβ

Since cos θc < 1, 1/ηβ < 1 → β > 1/η. So for:

(a) η = 1.5, β > 2/3, E = γmc2 > (1− β2)−1/2 · 0.511 MeV = 0.68 MeV

(b) η = 1.000036, β > 0.999964, E = γmc2 > (1− β2)−1/2 · 0.511 MeV = 60.2 MeV

3. Normal incidence on a dielectric: Let the region z < 0 be filled with a dielectric withdielectric constant ε1 (or equivalently refractive index n1) and the region z > 0 be filled with adielectric with dielectric constant ε2 (refractive index n2). The incident wave in the region z < 0has the form:

Ex = Exi cos (ωt− kz)

(a) What are the appropriate boundary conditions at z = 0?

εE⊥, E‖, B⊥, B‖ are all continuous.

(b) Calculate the amplitudes of the transmitted and reflected waves.

From Continuity of E‖, Ei + Er = Et (1)

And Continuity of B‖, Bi −Br = Bt

→ (η1Ei/c)− (η1Er/c) = (η2Et/c) (2)

Adding η1×(1) and (2) gives, t =EtEi

=2η1

η1 + η2Subtracting (2) from η×(1) gives,

r =ErEi

=η1 − η2

2η1

2η1η2 + η1

=η1 − η2η2 + η1

(c) Show that the sum of the (moduli of) energy flux in the reflected and transmitted waves isequal to the energy flux in the incident wave.

Remembering (from e.g. Q1), N = c′U z = (c/η)εE2z = (c/εr1/2)εrε0E

2z = ηcε0E2z.

(NB for dielectric we used η = εr1/2)

So Nr +Nt = η1cε0Er2 + η2cε0Et

2 = η1cε0Ei2(r2 + (η2/η1)t

2),

(note we dropped the directions, since they are all travelling in z, though of course for thereflected wave the flux is in the opposite direction).

= η1cε0Ei2

(∣∣∣∣η1 − η2η2 + η1

∣∣∣∣2 +η2η1

∣∣∣∣ 2η1η1 + η2

∣∣∣∣2)

= η1cε0Ei2

(η1

2 − 2η1η2 + η22

(η1 + η2)2+

4η1η2(η1 + η2)2

),

= η1cε0Ei2 = Ni. QED

2

4. Fresnel coefficients: For the case of a wave incident on a dielectric interface with the electricfield normal to the plane of incidence:

(a) Show that application of the boundary condition on the normal component of the B-field(along with the condition of continuity on the E field) leads to Snell’s law.

From B⊥ continuous, Bi sin θi +Br sin θi = Bt sin θt (see figure in lecture 7 notes).

(where we used θi = θr).

Or η1Ei sin θi + η1Er sin θi = η1 sin θi(Ei + Er) = η2Et sin θt

But from E‖ continuous, Ei + Er = Et (3)

Substituting in the above, η1 sin θi(Et) = η2Et sin θt

η1 sin θi = η2 sin θt

(b) Now using the condition for B‖ at the boundary obtain the Fresnel relations for this case:

t =2η1 cos θi

η1 cos θi + η2 cos θtr =

η1 cos θi − η2 cos θtη1 cos θi + η2 cos θt

Continuity of B‖, Bi cos θi −Bi cos θi = Bt cos θt

→ η1Ei cos θi − η1Ei cos θi = η2Et cos θt (4)

multiplying (3) by η1 cos θi, η1Ei cos θi + η1Ei cos θi = η1Et cos θi (5)

Adding (4) and (5) gives t = Et/Ei =2η1 cos θi

η1 cos θi + η2 cos θt

Subtracting (4) from (5) gives r = Er/Ei =η1 cos θi − η2 cos θt

2η1 cos θit =

η1 cos θi − η2 cos θtη1 cos θi + η2 cos θt

(c) Combine the Fresnel relations with Snell’s law to show that:

r =sin(θt − θi)sin(θt + θi)

and t =2 cos θi sin θtsin(θt + θi)

Snell’s Law, η1 = η2 sin θt/ sin θi, so,

r =η1 cos θi − η2 cos θtη1 cos θi + η2 cos θt

= η2 sin θt cos θi −η2 cos θt sin θi

η2 sin θt cos θi +η2 cos θt sin θi=

sin(θt − θi)sin(θt + θi)

t =2η1 cos θi

η1 cos θi + η2 cos θt=

2η2 sin θt cos θi

η2 sin θt cos θi +η2 cos θt sin θi=

2 cos θi sin θtsin(θt + θi)

QED

(d) Similarly show that for the polarisation of the beams in the plane of incidence, the reflectioncoefficients are given by:

t =2η1 cos θi

η2 cos θi + η1 cos θtr =

η1 cos θt − η2 cos θiη2 cos θi + η1 cos θt

(Again refer to figure in lecture 7 notes).

Take continuity of E⊥ = Ex, Ei cos θi + Er cos θr = Et cos θt (6)

Continuity of B‖ gives, Bi −Br = Bt → η1Ei/c− η1Er/c = η2Et/c (7)

3

multiply (6) by η1 and (7) by c cos θi through-out and using θr = θi

η1Ei cos θi + η1Er cos θi = η1Et cos θtη1Ei cos θi − η1Er cos θi = η2Et cos θi

Adding gives, t = Et/Ei =2η1 cos θi

η2 cos θi + η1 cos θt

Subtracting gives r = Er/Ei =η1 cos θt − η2 cos θi

2η1 cos θit =

η1 cos θt − η2 cos θiη2 cos θi + η1 cos θt

5. Brewster’s Angle and Total internal reflection: State the conditions necessary for both;(i) Brewster’s Angle and (ii) Total internal reflection. Calculate both Brewster’s angle, θB andthe critical angle θc (at which total internal reflection occurs) at the boundary of two dielectricwith refractive index η1 and η2. (You may use the Fresnel coefficients derived above).

(i) At Brewster’s angle rp = 0 (no condition on η):

η1 cos θt − η2 cos θi = 0 → η2 cos θt sin θt −η2 cos θi sin θi = 0 → sin 2θt − sin 2θi = 0

Using a trig formula, this becomes sin(θt − θi) cos(θt + θi) = 0

Since 0 < θi, θt <π2 , (θt + θi) = π/2 or cos θi = sin θt = (η1/η2) sin θi

finally we obtain, tan θB = η2/η1.

(ii) For total internal refection, η2 < η1 and sin θi > η2/η1, so sin θc = η2/η1

Hence calculate:

(a) θB and θc for propagation through both a glass-air interface and an air-glass interface.(η = 1.5 for glass).

For glass to air, θB = tan−1(1/1.5) = 33.7, θc = sin−1(1/1.5) = 41.8,

For air to glass, θB = tan−1(1.5) = 56.3, there is no critical angle as there is no totalinternal reflection.

(b) The maximum permissible entrance angle of a light ray in an optical fibre if the core hasη = 1.6 and the cladding has η = 1.5.

θc = sin−1(1.5/1.6) = 69.6, so any angle up to 90− 69.6 = 20.3 permissible.

So total angular acceptance = ±20.3 = 40.7.

(c) θc for diamond (η = 2.52). Can you see why diamonds are cut with multiple angled faces?

θc = sin−1(1/2.52) = 23.4. Hence it is relatively easy to trap light in a diamond if thefaces are not parallel. This causes it to be reflected a number of times before exiting thediamond, with slightly incident different angles and colours coming out at quite differentplaces, causing it to sparkle!

4

12th March, 2012

Electricity and Magnetism IIProblem Sheet 8: Plasma

1. Evanescence: An EM wave is totally internally reflected at a dielectric interface lying in they-z plane at x = 0. The evanescent E-wave is described by Et = Et e

i(kt·r−ωt) y where kt =iktxx + ktzz. (ktx and ktz are real.)

(a) Use Faraday’s law to show that the evanescent B-field is given by

Bt = (−ktzx + iktxz)(Et/ω)ei(kt·r−ωt)

Starting with B = (1/ω)k×E,

B =1

ω

x y zkx 0 kz0 Ey 0

= (1/ω)x(−kzEy) + z(kxEy) = (−ktzx + iktxz)(Et/ω)ei(kt·r−ωt)

(b) Hence determine the time-averaged Poynting vector in (i) the z-direction, and (ii) thex-direction, to show that in the evanescent wave there is energy flow in the z-direction(parallel to the interface), but not in the x-direction (normal to the interface).

(Hint: remember the Ponyting vector is the product only of the real parts of the waves.)

N =1

µ

x y z0 Ey 0Bx 0 Bz

=1

µx(EyBz)− z(EyBx)

Now, Ey = Et ei(ktzz−ωt)e−ktxx, and

Bx = −ktz(Et/ω)ei(ktzz−ωt)e−ktxx and Bz = iktx(Et/ω)ei(ktzz−ωt)e−ktxx

So Nz = − 1

µ<(Ey)<(Bx) = − 1

µEt cos(ktzz − ωt)e−ktxx · −ktz(Et/ω) cos(ktzz − ωt)e−ktxx

and 〈Nz〉 =ktxµω

Et2 〈cos2(kt · r− ωt)〉e−2ktxx =

ktx2µω

Et2e−2ktxx

And also Nx = 1µ <(Ey)<(Bz) = 1

µ Et cos(ktzz − ωt)e−ktxx ktx(Et/ω) sin(ktzz − ωt)e−ktxx

so that 〈Nx〉 =ktxµω

Et2 〈cos(ktzz − ωt) sin(ktzz − ωt)〉e−2ktxx

= ktx2µω Et

2 〈sin(2(ktzz − ωt))〉e−2ktxx = 0

2. EM waves in plasmas: Consider an em wave propagating in the z direction in vacuum towardsa plasma, which has an electron density n for z > 0 and a boundary at z = 0.

(a) Write out Maxwell’s equations applicable to a plasma in differential form.

(1) ε0∇ ·E = ρ (2) ∇ ·B = 0

(3) ∇×E = −∂B

∂t(4) ∇×B = µ0j + µ0ε0

∂E

∂t

(b) Give an expression for the current density jc in term of the resultant fields for a collisionlessplasma, stating any other appropriate assumptions.

From j = −nev and m∂v

∂t= −eE− mv

τcNote current is assumed to be due to only electrons (due to greater inertia of ions), andeffect of B has been ignored, assuming v c. Also assume collisions can be ignored, then;

1

v = v0e−iωt → m

∂v

∂t= m(−iω)v0e

−iωt = −imωv

Substitute in above, −imωv = −eE → iωmv = eE

So j = −nev =ne2E

−iωm=ine2E

mω

(c) Assuming solutions of the form E = Exei(kz−ωt) x and B = Bye

i(kz−ωt) y, show that forz > 0:

ωBy = kEx; and ω2 = k2c2 + ω2p,

where the plasma frequency is ωp =

(ne2

ε0m

)1/2

.

From ∇×E = −∂B

∂tand ∇×B = µ0j + µ0ε0

∂E

∂t= µ0j + µ0ε0

∂E

∂t

Substituting given values; for Faraday’s Law,

∇×E =

x y z

0 0 ∂∂z

Exei(kz−ωt) 0 0

= ikExei(kz−ωt) ≡ −∂B

∂t= iωBye

i(kz−ωt)

Cancelling iei(kz−ωt) gives, ωBy = kEx.

Now looking at Ampere-Maxwell;

∇×B =

x y z

0 0 ∂∂z

0 Byei(kz−ωt) 0

= −ikByei(kz−ωt)

and µ0j + µ0ε0∂E

∂t= µ0

ine2E

mω+ µ0ε0(−iω)Exe

i(kz−ωt)

equating both sides (and dividing by iei(kz−ωt) again);

−kBy = µ0ne2Exmω

+ µ0ε0(−ω)Ex

but using By = kEx/ω, k.kEx/ω = −µ0ne2Exmω

− µ0ε0(−ω)Ex

multiply through-out by ω and c2 = (µ0ε0)−1,

c2k2 = ω2 − ωp2 with ωp as advertised.

(d) In the region z < 0 the incident wave is Exiei(kz−ωt). There is also a reflected wave with

the same frequency. Thus for z < 0,

Ex = Exiei(kz−ωt) + Exre

−i(kz+ωt).

Derive an expression for By in this region.

From above By = kEx/ω, but note that k changes sign for reflected wave, so;

By = (kExi/ω)ei(kz−ωt) − (kExr/ω)e−i(kz+ωt).

(e) Show that when ω < ωp the electric field in the region z > 0 is given by

Ex = Expe(−kpz−iωt)

and find kp. Find the magnetic field in this region in terms of Exp, ω and kp.

2

For ω < ωp, ck = ±√ω2 − ω2

p = ±iωp√

1− ω2/ωp2,

So, k = i(ωp/c)√

1− ω2/ωp2 = ikp

Hence Ex = Expei(kz−ωt) = Expe

i(ikpz−ωt) = Expe(−kpz−iωt)

Now, By = kEx/ω = i(kpExp/ω)ei(ikpz−ωt),

(f) At the boundary to the plasma, z = 0, the electric field and magnetic field must be contin-uous. Show that this implies:

Exi + Exr = Exp and kExi − kExr = ikpExp.

Hence show that the amplitude of the reflected wave is equal to the amplitude of the incidentwave.

Ignoring exponential terms eiωt cancels, and setting z = 0, we have:

Continuity of E, Exi + Exr = Exp

Continuity of B, kExi − kExr = ikpExp

Eliminate Exp, (k + ikp)Exi − (k − ikp)Exr = 0

→ |Exr| =|k + ikp||k − ikp|

|Exi|

Since both numerator and denominator have amplitude k2 + k2p, |Exr| = |Exi|

3. Power loss in collisionless plasma Write down the expression for the power dissipated by EMfields in a conducting material of volume τ , in terms of the electric field, E and the conductioncurrent density, jc. Hence show that for a sinusoidal EM wave propagating through a collision-free plasma, there is no power dissipation. You may use your expression for jc derived in thelast question. Similarly, using the results of the last question within the plasma, show that thetime average Poynting flux into the plasma is also zero.

P = jc ·E, so 〈P 〉 = 〈<[jc] · <[E]〉 = 〈<[ine2E

mω

]· <[E]〉 =

ne2E20

mω〈sin() cos()〉 = 0.

(Alternately P = 12<jc ·E∗ = 0, since jc is exactly π/2 out of phase of E.)

Similarly E and B are π/2 out of phase, so 〈N〉 = 0

4. Wave velocities and cut-offs: A wave propagating in a collisionless plasma satisfies thedispersion relation ω2 = ω2

p + c2k2, where ωp is the plasma frequency.

(a) Show that the group velocity of the wave is given by vg = c(1 − (ωp/ω)2)1/2 and that thephase (vφ) and group velocities satisfy vgvφ = c2.

c2k2/ω2 = 1− ωp2/ω2. Inverting, ω2/c2k2 = (1− ωp2/ω2)−1

So, vph = ω/k = c (1− ωp2/ω2)−1/2

To find vg, differentiate dispersion relation: 2c2kdk = 2ωdω

vg =∂ω

∂k= c2k/ω = c2/vph = c(1− ωp2/ω2)1/2

So vgvph = c((((((((

((1− ωp2/ω2)−1/2 · c(((((

((((1− ωp2/ω2)1/2 = c2 QED

3

(b) By requiring k to be real for an EM wave to be able to propagate, find an expression for themaximum (critical) density to which a wave of angular frequency w can propagate, (youmay use the expression for the plasma frequency obtained above).

Maximum density reached at normal incidence and when ω = ωp,

→ ω2 =ncre

2

ε0m, so ncr =

ε0mω2

e2.

(c) The density in a discharge plasma is monitored using microwaves. A microwave antennaproducing an em wave of variable frequency, is monitored by a receiver placed on the otherside of the plasma from the antenna, as shown in the diagram. When the discharge isactive, the receiver detects a drop in transmitted power when the frequency falls below 50GHz. Calculate the peak density of the plasma.

ω = 2π50× 109 = 3.14× 1011 s−1, so ncr =ε0mω

2

e2= 3.1× 1019 m−3.

5. Dispersion of Pulsar emissions: Pulsars are magnetised neutron stars that emit pulses ofradio waves of between 1 to 50 milliseconds duration at intervals of a second or less. Therepetition rate of the emission is stable to 1 part in 108. It is believed that the radiationis emitted in a cone that sweeps past the earth as the pulsar rotates (like the light from alighthouse), producing a pulse of radio waves of multiple frequencies. The higher frequenciesarrive first due to dispersion of the em waves in the tenuous inter-stellar plasma. Within monthsof the discovery of pulsars, this information was used to determine their distance.

(a) The density of electrons between the stars is roughly 105 m−3. What is the plasma frequencyin this plasma?

ωp =

(nee

2

ε0m

)1/2

= 1.78× 104 s−1 or fp = 2.8 kHz.

(b) Show that for a pulsar emitting a frequencies f1 and f2, the time delay between arrival ofthe pulses is given by;

∆t ∼=df2p2c

[1

f22 −

1

f12

]where fp = ωp/2π and d is the distance to the pulsar.

Remember that the em energy pulse (or envelope) travels at vwp = vg0 = ηc - the groupvelocity - not vph which of course would be non-sensical since vph = c/η > c in a plasma.

So tarr = d/vg0. Hence ∆t = t2 − t1 =d

vg2− d

vg1=d

c

(1

η2− 1

η1

)∆t =

d

c

((1− fp

2

f22

)−1/2

−(

1− fp2

f12

)−1/2)

but fp f1, f2 so expanding brackets:

∆t ∼=d

c

((1 +

fp2

2f22

)−(1 +

fp2

2f12

))=df2p2c

[1

f22 −

1

f12

]QED

(c) For pulsar CP 0328, three packets of central frequencies 151, 408 and 610 MHz are observed.The 610 MHz pulse arrives first followed by the 408 MHz pulse 0.365 seconds later, withthe 151 MHz a further 4.18 seconds later. Calculate the distance from earth to CP 0328 –check that both time delays give the same distance (almost).

Inverting previous relation, d =2c∆t

fp2

[1

f22 −

1

f12

]−1

From separation between first two pulses, d = 8.2× 1018 m = 864 light years.

From second separation, d = 8.2× 1018 m = 868 light years. (Consistent to 2sf).

6. Grazing-angle x-ray reflections: In the design of an imaging x-ray telescope it is necessaryto reflect 1 keV x-rays off the surfaces of metallic mirrors. Assume that the conductivity of the

4

metal is plasma-like, with a plasma frequency corresponding to a vacuum photon energy of 6eV. What is the maximum allowed angle which the incident x-rays can make with a metallicmirror and still experience total reflection? (You may assume that the x-rays are travelling in avacuum before striking the metal.)

But cos θc = ωp/ω = ~ωp/~ω = 6/1× 103 = 0.006

θc = 89.6 which is the angle of incidence, so the grazing angle = 90− θc = 0.34

5

16th March, 2012

Electricity and Magnetism IIProblem Sheet 9: Conductors and waveguides

1. Reflection from a conductor: An infinite plane electromagnetic wave is incident from vacuumfor z < 0 onto a good conductor for z > 0 where j = σE.

(a) State mathematically the condition for σ so that the displacement current may be ignored.Show that in this limit for the conductor, Faraday’s law and Ampere’s law become:

∇×E = −∂B

∂t; ∇×B = µσE

For |σE| ∣∣∣∣ε∂E∂t

∣∣∣∣ = |εωE| → σ εω

∇×E = −∂B

∂t; ∇×B = µσE +

ε∂E

∂t

(b) Show that; E = Ex0e−(z/δ)+i(z/δ)−iωt x; B = By0e

−(z/δ)+i(z/δ)−iωt y

are solutions to these equations for z > 0, with δ =

(2

µ0σω

)1/2

and |Ex0| =ωδ√

2|By0|.

∇×E =

x y z

0 0∂

∂zEx 0 0

= y∂Ex∂z

= (Ex0/δ)(−1 + i)e−(z/δ)+i(z/δ)−iωt y

and −∂B

∂t= iωBy0e

−(z/δ)+i(z/δ)−iωty

Equating the above two lines, (Ex0/δ)(−1 + i) = iωBy0

→ Ex0√

2ei3π/4 = eiπ/2ωδBy0 → Ex0 = e−iπ/4(ωδ/√

2)By0

or |Ex0| = (ωδ/√

2)|By0| as shown

From Ampere-Maxwell,∇×B =

x y z

0 0∂

∂z0 By 0

= −x∂By∂z

= (By0/δ)(1−i)e−(z/δ)+i(z/δ)−iωt x

and j = σµE = σµEx0e−(z/δ)+i(z/δ)−iωtx

Equating, (By0/δ)(1− i) = (By0/δ)√

2e−iπ/4 = σµEx0 = σµe−iπ/4(ωδ/√

2)By0

(By0/δ)

√2e−iπ/4 = σµ

e−iπ/4(ωδ/√

2)By0 → δ2 = (2/ωσµ) QED

(c) The reflection coefficient for normal incidence onto a surface of refractive index η fromvacuum is given by;

r =1− η1 + η

.

Given an expression for the refractive index of a metal (bearing in mind that it is nowcomplex). Show that the energy reflectivity of the surface is R = |r|2 ≈ 1−

√8ωε0/σ.

(You may assume that µ ' µ0).

1

Inspecting the solution for E, one can see that k = 1+ i/δ, hence η = ck/ω = (c/ωδ)(1+ i).

Writing η = a+ ia, where a = (c/ωδ). Then,

R = |r|2 =

∣∣∣∣1− η1 + η

∣∣∣∣2 =

∣∣∣∣1− a− ia1 + a+ ia

∣∣∣∣2 =

((1− a)2 + a2

(1 + a)2 + a2

)≈(1− 2a+ 2a2

1 + 2a+ 2a2

)where we cancelled the ones above since a 1. Rearranging;

R =

(a2 − aa2 + a

)=

(a− 1

a+ 1

)≈(

(a− 1)2

a2 − 1

)≈(

(a2 − 2a+ 1)

a2 − 1

)= 1− 2

a

where we divided by 2a2 and multiplied top and bottom by (a− 1).

but a =c

ωδ=

1

ε0µ0

1

ω

(σµω

2

)1/2

=

(σ

2ε0ω

)1/2

where we used µ ' µ0.

So finally, R = 1−√

8ε0ω/σ QED

(d) Calculate the reflectivity of a gold mirror, σAu = 4.1 × 107 (Ωm)−1 for red (λ = 630 nm),and blue (λ = 420 nm) light.

For red, ω = 2πc/λ = 3.0× 1015 rad/s. So R = 0.93.

For blue, ω = 2πc/λ = 4.5× 1015 rad/s. So R = 0.80.

(The falling reflectivity at higher frequency contributes partly to the colour of gold).

2. Good conductors: The table below shows the conductivities and relative permitivities εr(where ε = εrε0 , and ε0 = 8.85× 10−12 Fm−1 ) of several substances.

Substance σ (Ωm−1) εrAluminium 4× 107 1Sea Water 3 10

Pure Silicon 0.1 1Distilled Water 10−3 10

Polythene 10−15 2

fc (Hz) σ/εω

7.2× 1017 7.2× 1012 - good5.3× 109 5× 104 - good1.8× 109 1.8× 104 - good1.8× 106 18.0 - poor

9.0× 10−6 9.0× 10−6 - insulator

(a) Calculate the limiting frequency below which each of these becomes a “good conductor”

σ > εω → fc = (ω/2π) =σ

2πεrε0. See table for values of fc.

(b) Which of these substances is a good conductor at 100 kHz, and which a weakly conductingdielectric i.e. a good insulator?

for f = 100× 103, ω = 2π105 = 6.28× 105 rad/s. See table for values of σ/εω

(c) Communications are attempted with a submarine at this frequency (100 kHz). Estimatethe maximum depth to which communication is still possible.

Skin depth δ =

(2

µσω

)1/2

=

(2

4π10−7 · 3 · 6.28× 105

)1/2

= 0.9 ≈ 1m

Power transmitted (∝ E×B) falls as e−2z/δ, so at 10m, power is reduced by ≈ e−20 ' 10−9.Hence submarine has to be quite close to the surface to be able to communicate.

3. Skin depth: The conductivity of copper is σCu = 5.7×107 (Ωm)−1, and µ ' µ0 = 4π10−7 Hm−1.

(a) A screened room is used to prevent electromagnetic interference or “noise” with a frequencyof about 1 MHz from reaching a detector. How thick must the copper wall be to attenuatethe intensity of the noise by a factor 10−3?

Skin depth δ =

(2

µσω

)1/2

=

(2

4π10−7 · 5.7× 107 · 6.28× 106

)1/2

= 66.7µm

2

10−3 = e−6.91 = e−2z/δ → z = 6.91× δ/2 = 0.23 mm

(b) Estimate the frequency at which the resistance of a copper wire of radius 0.1 mm startsto rise above its DC value. What is the corresponding value for a wire of 1 mm radius?

Resistance will increase when δ < r → ω >2

σµr2

For r = 0.1 mm, ω > 2/(4π10−7 · 5.7× 107 · (10−4)2) = 2.8× 106 rad/s = 444 kHz

For r = 1 mm, ω > 2/(4π10−7 · 5.7× 107 · (10−3)2 = 2.8× 104 rad/s = 4.44 kHz

4. Poynting and Resistance:

(a) The electric and magnetic fields of a plane wave propagating in the x direction in a goodconductor of conductivity σ, permittivity ε, and permeability µ are given by:

E(x, t) = E0e−x/δei(x/δ−ωt) y and B(x, t) = (µσ/ω)1/2E0e

−x/δei(x/δ−ωt+π/4) z

where kc =√µσω/2. Show that the time-averaged Poynting vector is

〈N〉 =1

2

√σ/2µω E2

0 e−2x/δ x.

N = 1µE×B (note in material µ0 → µ)

N = (1/µ)(µσ/ω)1/2E02e−2x/δei(x/δ−ωt) ei(x/δ−ωt+π/4) (y × z)

So, 〈N〉 = (σ/µω)1/2E02 e−2x/δ

⟨ei(x/δ−ωt) ei(x/δ−ωt+π/4)

⟩x

〈N〉 = (σ/µω)1/2E02 e−2x/δ 〈cos (x/δ − ωt) cos (x/δ − ωt+ π/4)〉 x

〈N〉 = (σ/µω)1/2E02 e−2x/δ

⟨cos (x/δ − ωt)

(1√2

cos(x/δ − ωt)− 1√2

sin(x/δ − ωt))⟩

x

where we used cos(x+ π/4) = cos(x) cos(π/4)− sin(x) sin(π/4) = 1√2(cos(x)− sin(x))

and using 〈cos2(ωt)〉 = 12 and 〈cos(ωt) sin(ωt)〉 = 0)

〈N〉 = 12√2(σ/µω)1/2E0

2e−2x/δ x QED

(b) Calculate the average power per unit volume, P = 〈jc ·E〉, converted from the EM field ofthe wave into heat. Show that ∂

∂x〈N〉+〈jc ·E〉 = 0. Give an interpretation of this equation.

P = 〈jc ·E〉 = 〈σE ·E〉 = σ〈E0e−x/δei(x/δ−ωt) · E0e

−x/δei(x/δ−ωt)〉

= σE02e−2x/δ〈cos(x/δ − ωt) · cos(x/δ − ωt)〉 = 1

2σE02e−2x/δ

But∂〈N〉∂x

= − 1

2√

2(σ/µω)1/2E0

2(2/δ)e−2x/δ = − 1√

2(µσω/2)1/2(σ/µω)1/2E0

2e−2x/δ =

−1

2σE0

2e−2x/δ = −P

So∂〈N〉∂x

+ 〈jc ·E〉 = 0 QED

Evidently energy is lost from the Poynting flux by Ohmic heating.

5. Waveguides: Consider waves propagating in a vacuum between two (“perfect”) conductingplates at x = 0 and x = a. Let the waves have frequency ω. We consider two dimensional waveswith E and B varying in x and z but not y.

3

(a) Show that Maxwell’s equations in vacuum,

∇ ·E = 0; ∇ ·B = 0;∂B

∂t= −∇×E; µ0ε0

∂E

∂t= ∇×B

yield1

c2∂2E

∂t2= ∇2E.

Take ∇× Faraday, ∇× (∇×E) = −∂(∇×B)

∂t− ∂

∂t

(µ0ε0

∂E

∂t

)Were we used Ampere-Maxwell on the RHS. Using the vector identity ∇ × (∇ × A) =∇(∇ ·A)−∇2A.

∇× (∇×E) = ∇(∇ ·E)−∇2E = − 1

c2∂

∂t

(∂E

∂t

)where c = (ε0µ0)

1/2.

∇2E =1

c2∂2E

∂t2

(b) Show that the one dimensional wave,

Ex = Ex0 cos (kz − ωt)

is a solution if ω = kc and that it satisfies the boundary conditions (for a good conductor)at x = 0 and x = 1.

As usual, E = −ω2Ex0 cos (kz − ωt)x and ∇2E = −k2Ex0 cos (kz − ωt)x

which satisfy the wave equation for ω = ck, and satisfy the boundary conditions since thereare no conditions on E⊥.

(c) Show that the two dimensional wave,

Ey = Ey0 sin(nπx

a

)cos (kz − ωt)

is a solution if ω2 = k2c2 + n2π2c2/a2 and that it satisfies the boundary conditions (for agood conductor) at x = 0 and x = 1 if n is an integer.

E = −ω2Ey0 sin(nπx

a

)cos (kz − ωt)y

∇2E =

(∂2

∂x2+

∂2

∂z2

)E = (−

(nπa

)2− k2)Ey0 sin

(nπxa

)cos (kz − ωt)y

So wave eqn satisfied if ω2 = c2(k2 +

(nπa

)2)as given

(d) Give an expression for the group velocity of the two dimensional waves.

Differentiating dispersion relation; 2ωdω = c22kdk → ∂ω

∂k=

c2

ω/k

But c2k2 = ω2

(1− n2π2c2

a2ω2

)→ ω

k= c

(1− n2π2c2

a2ω2

)−1/2So vg =

∂ω

∂k= c

(1− n2π2c2

a2ω2

)1/2

(e) Show that if n = 1 and ω < πc/a no traveling wave solution of this two dimensional formexists.

Condition for wave to stop travelling is k = 0, and it will decay if k < 0.

4

i.e. if 1− n2π2c2

a2ω2< 0

That is if, ω <nπc

a. For n = 1, this becomes ω <

πc

a.

(f) For ω < πc/a find a solution for Ey that decays exponentially in z.

Since ω < πc/a, then ω < nπc/a for any value of n.

let

(1− n2π2c2

a2ω2

)1/2

= ib. So ck = iωb

and Ey = Ey0 sin(nπx

a

)<[ei(kz−ωt)

]= Ey0 sin

(nπxa

)<[ei(iωbz/c−ωt)

]So Ey = Ey0 sin

(nπxa

)e−ωbz/c cos(ωt)

6. EM radiation propagates down a long, perfectly conducting metallic waveguide with width a,and height b. The EM wave is considered to be of the transverse electric (TE) form, with anE-field only in the y direction, E = E0 sin(mπz/b)ei(kx−ωt) y, where m is an integer.

(a) Use Faraday’s law to show that the corresponding B-field is:

B =

(E0

ω

)[imπ

bcos(mπz

b

)x + k sin

(mπzb

)z

]ei(kx−ωt)

and show that it satisfies the correct boundary conditions at z = 0 and z = b.

E has only y component and is a function of x and z, so

∇×E =

x y z∂

∂x0

∂

∂z0 Ey 0

= −x∂Ey∂z

+ z∂Ey∂x

→ −∂B

∂t= −x (mπ/b)E0 cos(mπz/b)ei(kx−ωt) + z ikE0 sin(mπz/b)ei(kx−ωt)

Integrate, B = x (mπ/b)(E0/−iω) cos(mπz/b)ei(kx−ωt)− z (ik/−iω)E0 sin(mπz/b)ei(kx−ωt)

=

(E0

ω

)[imπ

bcos(mπz

b

)x + k sin

(mπzb

)z

]ei(kx−ωt) QED

The main BC to fulfil in a waveguide is that E should be only perpendicular to anysurface. In this case because there is only a y component, at the boundaries z = 0, b:Ey ∝ cos

(mπzb

)= 0, and at the boundaries y = 0, a, E should only have a y component

(which by definition is already true).

(b) Show that there is a time-averaged Poynting flux in the waveguide only in the positive xdirection, and that for m = 1, it is 〈N〉 = (kE2

0/2ωµ0) sin2(πz/b) x.

N =1

µ0E×B =

1

µ0

x y z0 Ey 0Bx 0 Bz

=1

µ0(x(EyBz)− z(EyBx))

So

〈N〉 =1

µ0x(E2

0/ω)k sin2(mπz

b

)⟨<(ei(kx−ωt))<(ei(kx−ωt))

⟩− 1

µ0z((E2

0/ω)k sin(mπz

b

)cos(mπz

b

)(mπ/b)

⟨<(iei(kx−ωt))<(ei(kx−ωt))

⟩The terms in z are π/2 out of phase and so time-average to zero. In x they are in phase and sothe time average = 1

2 . So,

〈N〉 = (kE20µ0/ω)k sin2

(mπzb

)x as requested.

5