Exam # 1 Review Session PHYS 436 - Fall 2010 Electromagnetic Fields II Instructor: Prof. Nadya Mason Date: Sept. 20th, 2010

Problem 1

An infinitely long wire with linear charge density −λ lies along the z axis. An insulating cylindrical shell of radius R and moment of inertia I per unit length is concentric with the wire, and can rotate freely about the z axis. The surface charge density on the shell is σ = λ/(2πR) and is uniformly distributed. The cylinder is immersed in an external magnetic field Bext = Bzˆ, and is initially at rest. Starting at t = 0, the external magnetic field is gradually switched off.

a) Using conservation of angular momentum, determine the final angular velocity ω of the shell.

b) Check your answer by using Faraday’s law.

Solution:

a) Initially the cylindrical shell is at rest, so all the angular momentum is stored in the electromagnetic field, and the corresponding density can be calculated with the formula

` = µ00(r × S) , (1)

where S = 1 E × B is the Poynting vector. Using Gauss’s law, we easily obtain µ0  λ sˆ  − , for s < R E = 2π0 s (2)  0 , for s > R

Notice that the surface charge density on the shell has been chosen to cancel the electric field produced by the wire. Hence, initially the Poynting vector is λB Si = φˆ , for s < R (3) 2π0µ0s

and the total angular momentum (per unit length) stored initially in the field is

Z R Z R φˆ Z R R2λB Li = 2π `i sds = λB ssˆ × sds = λBzˆ sds = zˆ . (4) 0 0 s 0 2

When the magnetic field is turned off, the total angular momentum is the sum of the mechanical momentum of the rotating shell plus the angular momentum stored in the solenoidal magnetic field due to the current on the shell: Lf = Lf,em + Lf,mech (5) Exam # 1 Review Session (PHYS 436 - Electromagnetic Fields II) 2 H From Amp`ere’slaw B · dl = µ0Ienc and the right hand rule, we know that the azimuthal current λω density jφ = Q/T = (2πRσ)/(2π/ω) = 2π produces a magnetic field µ λω B = 0 zˆ , (s < R) (6) f 2π

inside the cylinder. Taking advantage of the result (4), replacing B → Bf we learn that the angular momentum per unit length stored in the fields is now

µ R2λ2ω L = 0 zˆ . (7) f,em 4π

Similarly, the mechanical angular momentum (per unit length) of the rotating shell is

Lf,mech = Iωzˆ , (8)

so that  µ R2λ2  L = I + 0 ωzˆ . (9) f 4π

Ignoring friction and radiation effects, the total angular momentum is conserved and we find

R2λB L = L ⇒ ω = (10) i f 2 2  µ0R λ  2I 1 + 4πI

b) From Faraday’s law, H E·dl = −dΦ/dt, we know that the change of magnetic flux Φ in thez ˆ direction will induced an azimuthal electric field given by s dB E = − z φˆ . (11) ind 2 dt

Hence, the torque (per unit length) exerted by the electric field on the cylindrical shell is

R2λ dB τ = Rsˆ × [E (R)2πRσ] = RλE (R)ˆz = − z zˆ . (12) ind ind 2 dt

Therefore, dL dω R2λ τ = mech = I zˆ ⇒ Iω = − (B − B) (13) dt dt 2 f

where we have integrated in the time coordinate. Using the value of Bf calculated in (6), we conclude

R2λ µ λω  R2λB Iω = − 0 − B ⇒ ω = (14)  2 2  2 2π µ0R λ 2I 1 + 4πI

as before. Exam # 1 Review Session (PHYS 436 - Electromagnetic Fields II) 3 Problem 2 (Griffiths 9.12)

Find all elements of the Maxwell stress tensor for a monochromatic plane wave traveling in the z direction and linearly polarized in the x direction, i.e. E E(z, t) = E cos(kz − ωt + δ)ˆx , B(z, t) = 0 cos(kz − ωt + δ)ˆy . (15) 0 c

←→ Comment on the form of your answer (remember that T represents the momentum flux density).

Solution:

First, we recall that the components of Maxwell’s stress tensor are given by     1 2 1 1 2 Tij = 0 EiEj − δijE + BiBj − δijB . (16) 2 µ0 2

Since only Ex and By are nonzero, it is clear that all the off-diagonal components vanish. A quick calculation then shows     1 2 1 1 2 Txx = 0 ExEx − E + − B = 0 (17) 2 µ0 2     1 2 1 1 2 Tyy = 0 − E + ByBy − B = 0 (18) 2 µ0 2     1 2 1 1 2 2 2 Tzz = 0 − E + − B = −0E0 cos (kz − ωt + δ) = −u , (19) 2 µ0 2

where u is the energy density. That only the Tzz component is non-vanishing is consistent with the fact that the momentum of the fields points in thez ˆ direction, and it is being transported in thez ˆ direction as well.

Problem 3

The electric and magnetic fields can be written in terms of a scalar potential V and a vector potential A as ∂A B = ∇ × A , E = −∇V − . (20) ∂t

Maxwell’s equations in vacuum then imply ∂ 0 = ∇2V + (∇ · A) (21) ∂t  ∂2A  ∂V  0 = ∇2A − µ  − ∇ ∇ · A + µ  . (22) 0 0 ∂t2 0 0 ∂t Exam # 1 Review Session (PHYS 436 - Electromagnetic Fields II) 4 Find a condition that would make both V and the components of A satisfy decoupled wave equations. Assuming a monochromatic wave solution, write down this condition in terms of the amplitudes of the corresponding waves.

Solution:

Inspecting the equations, it is clear that the desired condition is ∂V ∇ · A + µ  = 0 . (23) 0 0 ∂t

The corresponding equations for A and V are now

∂2V 0 = ∇2V − µ  (24) 0 0 ∂t2 ∂2A 0 = ∇2A − µ  (25) 0 0 ∂t2

and a monochromatic wave solution is then of the form

i(k·x−ωt) V0 i(k·x−ωt) A = A0e ,V = √ e , (26) µ00

√ where V0 is a constant, A0 is a constant vector, and we have pulled out a coefficient 1/ µ00 = c in the definition of V for later convenience. Replacing this solution into (23) we find

√ ω 0 = k · A − µ  ωV ⇒ 0 = k · A − V (27) 0 0 0 0 0 c 0

The constraint (23) is known as the “Lorentz gauge” condition. Notice that the physical fields E and B ∂Λ are invariant under the simultaneous transformation A → A + ∇Λ and V → V − ∂t , where Λ = Λ(x, t) is a function of space and time. This freedom is known as “gauge invariance” and makes it possible to reach the Lorentz gauge.