Problem Set 10 Solutions

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1 Massachusetts Institute of Technology Department of Physics Physics 87 Fall 25 Problem Set 1 Solutions Problem 1: EM Waves in a Plasma a Transverse electromagnetic waves have, by definition, E = Taking the curl of Faraday s Law and using Ampère s Law plus Ohm s law, 2 E = E = µ σe t + ɛ E t For a monochromatic plane wave and the given expression for σ this gives dispersion relation k 2 = µ iωσ ɛ ω 2 = 1 c 2ω2 ω 2 p, ω 2 p = iωσ ɛ = n ee 2 mɛ The high-frequency limit of Griffiths Eq 9161 is ɛ ɛ = 1 Ne2 mɛ ω 2 j f j = 1 ω2 p ω 2 where we have used the sum rule j f j = Z where Z is the number of electrons per molecule Since N is the number density of molecules, the total number density of electrons is n e = N j f j Now, the dispersion relation for a plane wave of permittivity ɛ which includes the effects of all charges is k 2 = µ ɛω 2 which reproduces k 2 = ω 2 ω 2 p/c 2 or ω 2 = ω 2 p + k 2 c 2 b Writing E a = Re{Ẽae i±kaz ωt } e x where a {i,r,t} for the incident, reflected, and transmitted wave, respectively, and applying the boundary conditions as in Griffiths Section 932, we find Ẽ t = 2 Ẽ i 1 + β, Ẽ r = 1 β Ẽ i 1 + β where β = kc/ω and k satisfies the dispersion relation k 2 c 2 = ω 2 ω 2 p < Thus, β = iα where α ω 2 p/ω 2 1 is real and positive 1

2 c The reflection and transmission coefficients are ratios of the time-average Poynting fluxes relative to that of the incident flux For the incident and reflected waves, because they are in vacuum, c B a = ±Re{Ẽae i±kaz ωt } e y where the + sign applies for the incident wave and the sign for the reflected wave Writing the complex amplitude as Ẽi = Ẽi e iδ, we find the incident wave has fields E i = Ẽi coskz ωt + δ e x, c B i = Ẽi coskz ωt + δ e y The reflected wave is similar except now Ẽr = Ẽr e iδr with Ẽr = Ẽi and δ r = δ + arctan[2α/α 2 1] which follows from writing 1 iα/1 + iα as e iφ and showing cosφ = 1 α 2 /1 + α 2 and sinφ = 2α/1 + α 2 Thus, E r = Ẽi coskz ωt + δ r e x, c B r = Ẽi coskz ωt + δ r e y Because the electric and magnetic fields have the same phase up to a minus sign for B r, the time averages for the incident and reflected waves give intensity I a = 1 µ E a B a time avg = 1 2µ c E i 2 for a {i,r} Thus the reflection coefficient is R = I r /I i = 1 The transmitted wave is more complicated: E t = Re{Ẽte i kz ωt } e x, Bt = Re{ k/ωẽte i kz ωt } e y, where k = ω/c β = iωα/c = iκ These imply E t = Ẽt e κz cos ωt + δ t e x, c B t = α Ẽt e κz cos ωt + δ t + π/2 e y where Ẽt = Ẽt e iδt with Ẽt = 2 Ẽi / 1 + α 2 and δ t = δ arctanα Now the π/2 phase shift of the magnetic field which arises because k is imaginary means that the time average of E t B t involves the time average of cos ωt+ δ t cos ωt + δ t + π/2 = sin2ωt 2δ t, which vanishes Thus I t = 1 µ E t B t time avg = T = As expected, R +T = 1 The transmitted plasma wave, instead of propagating, is exponentially damped by the factors e κz If ω > ω p, k would be real and the wave would propagate in the plasma This behavior is very similar to that of EM waves in a waveguide Reflection of low-frequency waves by the earth s ionosphere the upper layers of the atmosphere, which are ionized by cosmic rays makes long-distance radio transmission possible ham radio 2

3 Problem 2: Split-Ring Resonators and Negative Index Materials a For wavelengths much larger than the ring, the magnetic field must be spatially uniform over the ring, so that the magnetic flux through the ring is Φt = Re{ Φe iωt } where Φ = πr 2 µ H The current is constant around the ring but varies with time, It = Re{Ĩe iωt } The Kirchoff loop rule is dφ dt IR = or iω Φ = ĨR Finally, the magnetic moment of the ring is mt = πr 2 It e y = Re{πr 2 Ĩe iωt } e y = Re{ me iωt } e y Combining these equations yields m = iωπr2 2 µ H R b For a thin circular ring, the charges move only in the φ-direction and their velocity is v φ = r φ = I/λq The equation of motion then gives d 2 x dt 2 + γd x dt = I + γi λq e φ = qe φ m The azimuthal electric field follows from the integral form of Faraday s law: 2πrE φ = dφ dt Writing the time dependence as in part a, from the equation of motion and Faraday s law we get which yields iω + γĩ = iωλq2 πr 2 µ H 2πrm m = iωλq2 πr 2 2 µ H 2πrm iω + γ The inertia gives the iω term in the denominator If it is neglected, then this result agrees with part a provided we identify R = 2πrmγ/λq 2 c The electric field in the ring is E φ = E φ + E φ where E φ is transverse ie, E = and E φ is longitudinal ie, E = The first part is simply E φ = 2πr 1 dφ/dt as before With the gap in the ring, nonzero charge density accumulates on either side of the gap, producing a source for 3

4 the longitudinal electric field The current It flows into one side of the gap depositing a charge Qt where It = dq/dt On the other side of the gap It flows out leaving a charge Qt If the gap has thickness d, the electric field in the gap due to the charges is E φ gap = Q/Cd where C is the capacitance of the gap But the charges also produce an electric field in the ring, which has opposite sign so that 2π E φr dφ = The result assuming homogeneity so that E φ is the same everywhere around the ring is E φ = Q/[C2πr d] or, with d r, E φ = Q/C2πr Combining the transverse and longitudinal parts we get where Φ = πr 2 B y 2πrE φ = dφ dt Q C d Repeating the calculation of part b including the Q/C contribution gives current iω + γĩ = λq2 Q iωπr 2 µ H 2πrm C where I = dq/dt yields Ĩ = iω Q This gives iω + γ + iλq2 m = iωλq2 πr 2 2 µ H 2πrmωC 2πrm or Aω 2 H m = ω 2 ω 2 + iγω, A = λq2 πr 2 2 µ 2πrm, ω 2 = λq2 2πrmC = γ RC e The permeability follows from B = µ H + M = µ H where M = N m is the magnetization magnetic dipole moment per unit volume All quantities have time dependence e iωt so the equations hold also between complex amplitudes Taking the y-component and using the complex amplitudes, µ = µ 1 + N m H = µ f The dispersion relation is k = ñω/c where 1 ñ 2 = ɛ µ ɛ µ NAω 2 ω 2 ω 2 + iγω In general, ñ is complex Using the modulus and phase representation for negative permittivity and permeability, ɛ = ɛ e iπ, µ = µ e iπ, and ñ = c ɛ µ 1/2 e iπ = c ɛµ < 4

5 This has a great effect on Snell s law, sin θ t = sin θ i n For sin θ i = 1/ 2 and the three cases considered, the transmitted ray has direction θ t = arcsin 2/3 = 281 for n = 15, θ t = θ i = 45 for n = 1 For n = 5, Snell s law gives sinθ t = 2 which has no real solution In this case the incident wave is completely reflected and there is no transmitted wave Rays crossing into a negative index material cross to the other side of the normal Focusing is then achieved with a concave lens instead of a convex one Problem 3: Griffiths Problem 931 p 412 The electric and magnetic fields for the TEM mode of a coaxial cable are E = A coskz ωt s in terms of cylindrical coordinates s,φ,z e s, B A coskz ωt = e φ, cs a Using the formulae for the divergence in cylindrical coordinates, we get E = 1 s s se s =, B 1 = s Faraday s Law is easily checked, B φ φ = E = E s z e φ = ka s sinkz ωt e φ = B φ t e φ = B t, as is Ampere s law, B = B φ z e s + 1 s s sb φ = ka cs sinkz ωt e s = 1 E s c 2 t e s = 1 E c 2 t In the last line we used ω = kc, the dispersion relation for the TEM mode The boundary conditions 9175 are trivially satisfied because E e s is perpendicular to the waveguide implying E = while B e φ is parallel to the waveguide implying B = 5

6 b The linear charge density follows by integrating Gauss s law over a cylinder of radius s > and length dz centered along the z-axis, Q enc ɛ = λdz 2π = E d a = E s sdzdφ = 2πA coskz ωtdz ɛ yielding λz,t = 2πɛ A coskz ωt Similarly, the linear current density follows by integrating Ampere s law around a circle of radius s > encircling s = : 2π µ I enc = B d l = B φ sdφ = 2π A coskz ωt c yielding Iz,t = 2πA/µ c coskz ωt c From part b, I/λ = 1/ɛ µ c = c If there were only one sign of charge carrier, the current would equal the charge density times the velocity of the charge carriers, Griffiths Eq 513, implying that the charge carriers move at c, which violates relativity However, there are two signs of charge carriers, and if the ions are stationary but the electrons have speed v < c, we have I = en e v and λ = n e n i, where n e and n i are the linear charge densities of electrons and ions, divided by the electron charge Thus I/λ = v/ 1 n e /n i which can easily equal c even if v < c Problem 4: Griffiths Problems 13 and 15 p 42 a With V = and A = qt/4πɛ r 2 e r, the fields are E = V A t = q 4πɛ r 2 e r, B =, from which it follows that J = and ρ = qδ 3 x In other words, these are the potentials of a static point charge, in an unfamiliar gauge b The gauge transformation is A = A+ qt = qt 4πɛ r 4πɛ r e r+ qt 2 4πɛ r e 2 r =, V = V qt t 4πɛ r We now recognize the usual Coulomb gauge potentials of a point charge Problem 5: Griffiths Problem 19 p 426 a The vector potential is given by the same integral as appears in Example 12, As,t = µ 4π e It r dz z s2 + z, 2 6 = q 4πɛ r

7 where t r = t c 1 s 2 + z 2, but now the expression for It r is different: It r = kt r for t r > Thus, A z s,t = µ k/2π ct 2 s 2 [ t c 1 ] s 2 +z 2 s 2 +z dz, s < ct,, 2 s > ct, which for s < ct evaluates to A z s,t = µ k 2π [ t ln u + u 2 1 s ] u2 1 c For s > ct the fields vanish, while for s < ct,, u ct s E = µ k 2π ln u + u 2 1 e z, B µ k = u2 1 e φ for u > 1 2πc A useful check is the case s ct: B φ = µ kt/2πs then agrees with the magnetostatic limit b For It r = q δt r, we have As,t = µ q 4π e δt c 1 s 2 + z 2 z dz s2 + z 2 We must rewrite the delta function using δfz = i δz z i f z i, where z i are the roots of fz In the present case there are two roots, z = ± ct 2 s 2 Evaluating the integral gives A z s,t = µ q c 2π ct 2 s 2 Note the 2π rather than 4π, because the past light cone intersects the string at two points, not one ie there are two roots z i The resulting fields are E = µ q c 3 t 2π [ct 2 s 2 ] 3/2 e z, B µ q cs = 2π [ct 2 s 2 ] 3/2 e φ for s < ct In this case, there is no magnetostatic limit However, we can check that the direction of the Poynting flux makes sense: S = µ 1 E B e s is radially outward, as it must be In part a, the Poynting flux is inward because the fields close to the wire increase steadily with time, and the Poynting flux must be inwards just as it is for a charging capacitor 7

Worked Examples Set 2

Worked Examples Set 2 Q.1. Application of Maxwell s eqns. [Griffiths Problem 7.42] In a perfect conductor the conductivity σ is infinite, so from Ohm s law J = σe, E = 0. Any net charge must be on the