Übungen zur Elektrodynamik (T3)

Transcription

1 Arnold Sommerfeld Center Ludwig Maximilians Universität München Prof. Dr. vo Sachs SoSe 8 Übungen zur Elektrodynamik T3 Übungsblatt Bearbeitung: Juni - Juli 3, 8 Conservation of Angular Momentum Consider the following arrangement of a cylindrical capacitor of length l and an infinitely long cylindrical coil with winding density n placed coaxially between the capacitor plates. The inner capacitor plate of radius a has the charge +Q, the outer radius b charge Q, each distributed equally across the corresponding plate. A variable current flows through the coil. Neglect edge effects.. The current flowing through the coil is slowly switched off. Calculate the angular momentum of the two capacitor cylinders after the current has been switched off. Hint: The magnetic field in the interior of the infinitely long solenoid of winding density n is given by B = µ nẑ. n the exterior it is B =. Note that a changing magnetic field induces an electric field! Solution: The momentum density is p = ɛ µ S = ɛ E B and the torque τ = r F. Furthermore, we shall use cylindrical coordinates and the normalised basis ê r, ê φ, ê z. For the moment we shall neglect the electric field due to the capacitor as that one is radial and thus does not impart any rotational force, i.e. torque, onto the cylinders Outside the coil there is no magnetic field and therefore nothing changes as the current is turned off. nside the coil we have E = t B By of the cylindrical symmetry we may assume that E does not depend φ. Furthermore, as we disregard boundary effects and the magnetic field is homogeneous inside the capacitor we assume that E does not depend on z. Since B points along the z-axis we have for r R i.e. E φ r = r r r rre φ = Ḃz sḃz ds = Ḃz r as B z is constant inside the cylinder and for r R E φ r = µ n R r r = Ḃz r = µ n r by the interface condition of the tangential component of E, as this solves E =. Therefore the torque on the inner cylinder along the z-axis is τ a = aqe φ a = µ nq a

2 note that this is indeed the total torque and not just a density as the Q is the charge of the whole cylinderx and its angular momentum L a = τdt = µ nq a and the torque on the outer cylinder is dt = µ nq a = µ nq a and its angular momentum τ b = bqe φ b = µ nq R L b = τdt = µ nq R and thus the total angular momentum is dt = µ nq R = µ nq R L = µ nq a R. Compare the total angular momentum of the two capacitor plates with the electromagnetic angular momentum before switching off the current. Hint: The momentum density of the electromagnetic field is given by c S. Solution: The electric field due to the central cylinder is E ρ = πɛ Therefore in between the innermost and middle coil where it does not vanish the Poynting vector is given by S = E B = Q µ µ πɛ lρ µ nê ρ ê z = Q πɛ lρ nê φ and thus the total angular momentum is l L = π R a Q l ρ ρɛ µ Sρdρdφdz = µ Qn a R Wave Guide A coaxial cable is a type of cable that has an inner conductor, surrounded by a tubular insulating layer which is again surrounded by a tubular conducting shield. i Show that there are no possible TEM modes in a hollow tube. Solution: We have E z = and B z = and thus E = E xx, y E y x, y e ikz wt and B = B xx, y B y x, y e ikz wt from E = c tb it follows that x E y = y E x and thus, for simply connected spaces in dimensions, there is a potential Φx, y with Φx, y = E x, y everywhere inside the tube. From

3 E = it follows further that Φx, y =. Using the boundary condition, i.e. that Φ = const on the conductor, it follows that the only solution can be Φ = const everywhere and thus E x, y =. Alternatively you can argue that from Maxwell s equations one can derive a two dimensional wave equation which is only satisfied for vanishing fields since Er = = because of symmetry. ii Show that in a coaxial cable there can appear TEM-modes. Solution: n the coaxial cable we have an inner conductor of radius R and an outer tubular conductor of radius R. We use the potential Φx, y from part one, but this time Φx, y = is only valid between the two conducting surfaces, i.e. for R < r < R. t is Ex, y = x Φx, y y Φx, y with constant potential on the surface of the conductors n polar coordinates the Laplace equation is due to symmetry it reduces to with the solution where and thus A = e ikz wt Φr = R Φ and Φr = R Φ r + r r r + r φ r Φ Φ lnr lnr r Φr = r Φr = A lnr + B Φr, φ =, B = Φ lnr Φ lnr lnr lnr Ex, t = A r e re ikz wt Bx, t = A r e φe ikz wt Consistency Check: From ΦR = A lnr + B it follows B = Φ A lnr and thus Φr = A ln r R + Φ. f we let R we see that ΦR for A and thus A =. Then Φ must be constant and again we end up with the result from part i. iii What is the corresponding dispersion relation ωk for such a cable? Solution: From the lecture we know that E x x, y = i ω ω c k c yb z x, y + k x E z x, y it follows for TEM that either E x = or ω c k =. Since E x it is ωk = ck. As Tomas pointed out, this is a potential in a non-simply connected region. t does not follow directly from E = that such a potential exists. What does exist is a potential that changes by a constant when going around the inner disk. But since the potential must be constant on the inner disk, it does not change in this case.

4 3 Green s Function of the D Alembertian The Green s function Gx of a differential operator D x is defined via D x Gx = δx. Calculate the Green s function of the of the d Alembert operator : [ ] Gt, x = c t Gt, x = δtδx Proceed as follows:. Perform a Fourier transformation and solve the resulting equation for Gω, k.. For the inverse transformation carry out the ω integral first. You should consider the cases t > and t < separately. Hint: You have to do a complex contour integral. Use the residue theorem and think about which contours you have to integrate along. 3. Calculate the k integral and bring the result into a simple form. Hint: Use spherical coordinates! Solution: As we expect the Greens function to dependent on the distance x µ x µ only, we redefine the variables to x µ x µ x µ. The equation for the greens function then is: The Fourier transformation of this equation is: k ω Gω, k = c Gt, x = δtδx 3 Gω, k = k ω c = Gt, x = e ikx ωt π 4 d 3 kdω k ω c e ik x e iωt = π 4 k + ω c k ω c The function in the ω integral has singularities at ω = ±ck with residues res ±ck dωd 3 k e ik x ωt = c e ickt k ω c k. There are two cases that must be considered when performing the contour integral. n the case when t > the exponential diverges in the upper half of the complex ω plane and the contour has to be closed in the lower half. The opposite is true for t <. The first case gives the retarded Green s function.

5 This is calculated using the upper results: G r t, x = θt π 4 = cθt π 3 = cθt π x = cθt 8π x e ik x πi c e ickt e ickt d 3 k k ik x sinckt e d 3 k k + = cθt δct x 4πx sinkx sincktdk e ikct x e ikx+ct dk n the last step the fact was used that x > and t >. Therefore the second term in the integral vanishes. The advanced Greens function is then: G a t, x = cθ t δct + x 4πx