Practice Exam in Electrodynamics (T3) June 26, 2018

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1 Practice Exam in Electrodynamics T3) June 6, 018 Please fill in: Name and Surname: Matriculation number: Number of additional sheets: Please read carefully: Write your name and matriculation number on every sheet that you hand in. State the exact number of handed in sheets. Only the use of a two-sided handwritten DIN A4 sheet is allowed, whereas books, lecture notes, and calculators are not! You have hours at your disposal. Only readable and comprehensible solutions will be corrected. Exclusively SI units are used. The metric reads η = diag1, 1, 1, 1). Do not write below this line. Comments: Problem 1 Problem Problem 3 / 14 P / 5 P / 5 P Total Grade / 64 P

2 1 Problem: Short questions Total 14 P) i) Is the following quantity, µ A µ, invariant under a gauge transformation of the four vector potential? Explain. 1 P ) Solution: No, because under gauge transformation A µ = A µ + µ λ, we get µ A µ = µ A µ + µ µ λ. 1 P for the explanation) ii) Let X µν = be a tensor defined on Minkowski space, calculate Xµ µ. 3 P ) Solution: X µ ν = and X µ µ = 3. 3 P ) iii) A loop of radius R is placed on the x y plane such that it is concentric to the origin. A magnetic t field B = B 0 T ẑ is turned on inside the ring, where t is the time, B 0 and T are constants. Calculate E ds, where the integration is along one cycle of the ring in the counter clockwise direction. 3 P ) Solution: E ds = E) da 1 P) 1) = t B da 1 P) ) = πr B 0 T 1 P) 3) iv) Calculate ρ F αβ + α F βρ + β F ρα. Simplify as much as possible. P ) Solution: Plug in F µν = µ A ν ν A µ 1 P for writing F µν ) to get 0. 1 P for the final solution) v) Consider a static electromagnetic problem. For a bounded region in space, can V = V 0 e y /a, a xye y /a, z/a) be an electric or a magnetic field? If so, find the corresponding charge density or current density V 0 and a are constants). 5 P ) Solution: V = V 0 a xe y /a + 4a 4 xy e y /a + a 1 ). P) V = 0. P) Can be an electric field with charge density ρ = ɛ 0 V 0 a xe y /a + 4a 4 xy e y /a + a 1 )1 P). Exam in Electrodynamics Page 1 of 6 June 6, 018

3 Problem: Moving charged wire Consider an infinitely long and infinitesimally thin wire with constant linear charge density λ localized on the z-axis in Cartesian coordinates. i) What is the charge density ρ and the current density j in this problem? Show that the continuity equation holds. Solution: ρ = λδx)δy) 1P); j = 0 1P). Since ρ is time independent, we have t ρ + j = 0 1P). ii) Show that the four-potential ) r A 0 r) = C ln x 0 and A i = 0, with x 0 real and r = x + y, solves Maxwell s equations. What is the relation between the constants C and λ? Hint: Use that x + y) lnr/x 0 ) = πδx)δy).) Solution: Since we work with the four potential, we can focus on the inhomogeneous Maxwell s equations. The others are automatically satisfied. Using the hint we find E = c A 0 t A) = πcδx)δy), t E = 0. The equation involving the current is therefore trivially satisfied. From E = ρ ɛ 0 we find C = λ πcɛ 0 P). For the three trivial Maxwell equations: 3P). iii) Write down the corresponding electric and magnetic fields. Solution: B = 0 1P), E = φ = λ πɛ 0r n rp), where n r is the normal vector in the r-direction. iv) Now we want to consider the same charged wire, but moving with constant velocity v in the x-direction. Give the Lorentz transformation relating the old problem to the new problem. Solution: γ γβ 0 0 Λ α γβ γ 0 0 β) =, where β = v c and γ = 1 1 β P). v) Calculate the new four-potential. Solution: A µ r ) = Λ µ νa ν r)1p)= γc ln r x 0 ), βc ln r x 0 )), 0, 0) with r = x vt) + y P). vi) Calculate the electric and magnetic field in this new setup. Solution: E = γe P). B = A = γβcy r n z P). Exam in Electrodynamics Page of 6 June 6, 018

4 vii) Calculate E B in the new reference frame. What can you say about this quantity in general? Solution: E B = 0 in the rest frame, since B = 0. P). In general this quantity is Lorentz invariant and therefore, since it was zero in the rest frame it is zero everywhere. 1P) viii) Is there a reference frame in which E = 0? If yes, find it. If no, explain why. Solution: No, since the invariant quantity E this quantity to be negative P). B is positive in every frame. But E = 0 implies Exam in Electrodynamics Page 3 of 6 June 6, 018

5 3 Problem: Wave propagation between conducting planes Consider two infinte planes, that lie in the vacuuum parallel to the y, z)-plane and are made out of a perfectly conducting material. One plane lies at x = 0 and the other at x = L. i) Which boundary conditions must the E-field and B-field satisfy at x = 0 and x = L? Solution: At the conducting plates we have E = 0 and B = 0. Thus, we deduce E y x=0,x=l = 0 = E z x=0,x=l and B x x=0,x=l = 0 1P) 4) ii) Derive the differential equations x + λ ) E 0) x) = 0 and x + λ ) B 0) x) = 0 for the fields E 0) x) and B 0) x), where λ is an appropriate constant, which depends on ω, k y and k z. Solution: Using the wave equations E = 0 and B = 0 1P) we find 0 = 1c ) ) t E = x + ω k y kz E 0 e ikyy+kzz ωt) 1P) 5) as well as 0 = 1c ) ) t B = x + ω k y kz B 0 e ikyy+kzz ωt) 1P) 6) so that x + λ ) B 0 = 0 and x + λ ) E 0 = 0 7) with λ = ω k y k z 1P). iii) Find an expression for E x and E y as a function of E z and B z by starting with the dynamical Maxwell equations. Solution: Using the dynamical Maxwell equations E = t B and B = 1 te 1P) we compute iω B0 x By 0 = ik ye 0 z ik z E 0 y ik z E 0 x x E 0 z 1P) 8) as well as i ω E0 x Ey 0 = ik yb 0 z ik z B 0 y ik z B 0 x x B 0 z 1P) 9) and therefore ωb 0 x = k y E 0 z k z E 0 y 10) iωb 0 y = ik z E 0 x x E 0 z 11) Exam in Electrodynamics Page 4 of 6 June 6, 018

6 Solving 10) and 11) for B 0 x and B 0 y we get ω E0 x = k y B 0 z k z B 0 y 1) i ω E0 y = ik z B 0 x x B 0 z 13) B 0 x = 1 ω k ye 0 z k y E 0 y) 14) and 1P for both equations) B 0 y = 1 ω k ze 0 x + i x E 0 z) 15) Inserting 15) into 1) we get ω E0 x = k y B 0 z k z ω k ze 0 x + i x E 0 z) = k y Bz 0 k z ω E0 x k z ω xez) 0 1P) 1 ) ω ω k z Ex 0 = k y Bz 0 i k z ω xez 0 16) which yields E 0 x = ω 1 k z ikz x E 0 z k y ωb 0 z) 1P) 17) Furthermore, inserting 14) in 13) we deduce ω E0 y = i k z ω k ye 0 z k z E 0 y) x B 0 z = i k yk z ω E0 z i k z ω E0 y x Bz 0 1P) i 1 ) ω ω k z Ey 0 = i k yk z ω E0 z x Bz 0 18) yielding E 0 y = ω 1 k z iωx B 0 z + k y k z E 0 z) 1P) 19) iv) From now until the end of the exercise consider a E z = 0. Rewrite the equations for E x and E y that you found in iv) for this case. Solution: For TE-waves it holds E z = 0. Therefore, equations 17) and 19) become E 0 x = E 0 y = ω ω 1 k y ωb 0 kz z 0) 1 iω x B 0 kz z 1) 1P for both equations) v) Make the ansatz B 0) z x) = B 0 cos k x x) and use the boundary conditions that you found in i) to determine the allowed values for k x. Exam in Electrodynamics Page 5 of 6 June 6, 018

7 Solution: Using boundary condition 4) in 1) we directly infer x=0,x=l = 0 x Bz 0 x=0,x=l = 0 ) E 0 y This suggests the Ansatz B 0 z = B 0 cosk x x). If we insert this into ) this yields k x B 0 sink x x) x=0,x=l = 0 1P) k x L πz 1P) k x π Z 1P) 3) L vi) What is the minimal frequency ω min, such that there is no wave propagation for values smaller than ω min? Solution: According to part vi) one has k x = nπ L for some n Z. Using this in 7) for the z- component we find ω 0 = x λ)bz 0 = π n ) L ky kz Bz 0 1P) 4) which yields a nontrivial solution iff ω π n L ky kz = 0 ω π n L = ky + kz 0 1P) 5) Thus, since kx + ky 0 we deduce that in order for 5) to be possible the frequency ω of the wave needs to satisfy ω c πn := ωn) min n N 1P) 6) L That is, for a given n Z the minimal possible frequency below which no propagation can take place nay more is given by ω n) min = nπ L. vii) Consider now the same problem as before, but with the additional boundary condition E x, y, z, t) = E x, y + qd, z, t) for all q Z, where d is some fixed length. What is the spectrum of ω, assuming that k z = π L constant)? is fixed i.e. L is a Solution: In part i) we made the Ansatz E expik y y). Now, due to the periodicity assumption in the y-direction this implies that expik y y + qd))! = expik y y) 1P) which yields e ik yd ) q = 1 q Z 1P) ky d πz k y π d Z 1P) 7) Furthermore since k z = π L this then gives ω n k x ky kz = 0 ω = cπ L + 4m d + 1 L for any n, m Z 1P) 8) Exam in Electrodynamics Page 6 of 6 June 6, 018