Assignment 2, solutions

Problem 1:

A thin ring of radius R made from a dielectric carries a total charge Q uniformly distributed over its circumference.  It lies in the xy-plane centered at the origin and is rotating about the z-axis with angular velocity ω k.
(a)  Write down a lowest-order expression for E(r) for r >> R.
(b)  Write down a lowest-order expression for B(r) for r >> R.
(c)  Does the ring produce electromagnetic radiation?  Why or why not?
(d)  Is the Pointing vector zero or nonzero for r >> R?

Solution:

bulletConcepts:
Static fields and radiation fields
bulletReasoning:
Static charge and current distributions do not produce electromagnetic radiation.
bulletDetails of the calculation:
(a)  E(r) = Q/(4πε0r2) (r/r), the field of a point charge Q.
(b)  B(r) = (μ0/(4π))[3(m∙r)r/r5 m/r3].
Here m = IA k = QωR2/2 k.
B(r) = (μ0m/(4πr3))[2 cosθ (r/r) + sinθ (θ/θ)].
(c)  There is no radiation field since E and B do not depend on time.
(d)  S is not zero.
S = (1/μ0)EB = (1/μ0)  Q/(4πε0r2) (μ0QωR2/(8πr3)) sinθ (φ/φ)
=  Q2ωR2sinθ/(32π2ε0r5) (φ/φ).
S encircles the z-axis.

Problem 2:

A finite spherically symmetrical charge distribution disperses under the influence of mutually repulsive forces. 
Suppose that the charge density ρ(r, t), as a function of the distance r from the center of symmetry and of time, is known. 

(a)  Prove that the curl of the magnetic field is zero at any point.
(b)  Use this to show that B is zero at any point.

Solution:

bulletConcepts:
Maxwell's equations
bulletReasoning:
Place the origin of  the coordinate system he center of the charge distribution. Use Gauss' law and charge conservation to find E(r) and j(r) at a point P.
bulletDetails of the calculation:
(a)  Symmetry and Gauss' law yield E(r) = Qinsider/(4πε0r3).
Symmetry and charge conservation yield
dQinside/dt = -∫Sj∙dS = -j(r) 4πr2j = j(r)(r/r).
B = μ0j + μ0ε0E/∂t = -μ0(dQinside/dt)r/(4πr3) + μ0ε0(dQinside/dt)r/(4πε0r3) = 0.
(c)  B = 0 --> B = -ΦM B = 0 --> ∇2ΦM = 0 everywhere.
We can treat this as a boundary value problem.
Looking for a solution with no angular dependence yields
ΦM(r) = A0 + B0/r.
Requiring ΦM to not blow up at the origin (sinceB = 0) yields
ΦM(r) = A0 = constant,  B = 0.

Problem 3:

Assume magnetic charges exist and Maxwell's equations are of the form
E = ρ/ε0,   B = μ0ρm,   -E = μ0jm + ∂B/∂t,   ∇B = μ0j + μ0ε0E/∂t.
Assume a magnetic monopole of magnetic charge qm is located at the origin, and an electric charge qe is placed on the z-axis at a distance R from it.
(a)  Write down expressions for the electric field E(r) and the magnetic field B(r).  Make a sketch.
(b)  Write down expressions for the momentum density g(r) and angular momentum density (r) of the electromagnetic field.
(c)  Show that there is electromagnetic angular momentum Lz about the z-axis and derive an expression for it.  Show that Lz is independent of R.
Useful vector identity:  (a)n = (1/r)[a - n(an)].   Here n is r/r is the unit radial vector.

Solution:

bulletConcepts:
Field energy and field momentum of the electromagnetic field
bulletReasoning:
(1/μ0)(EB) = energy flux, (1/(cμ0))(EB) = momentum flux,
(1/(c2μ0))(EB) = ε0(EB) = momentum density.
The angular momentum density of the field about the origin then is r ε0(EB).
bulletDetails of the calculation:
(a)  Field of the point charge:  E(r) = [1/(4πε0)]qe(r - Rk)/|r - Rk|3.
Field of the magnetic monopole:  B(r) = [μ0/(4π)]qmr/r3).

(b)  S(r) = (1/μ0)(E(r)B(r)) = energy flux.
S(r)/c = momentum flux.
g(r) = S(r)/c2 = ε0E(r)B(r)) = momentum density.
g(r) = [μ0ε0qm/(4πr3)] E(r)r.
g
(r) = -(φ/φ)[μ0/(4π)2]qeqmRsinθ/(|r - Rk|3r2).
(r) = rg(r) = angular momentum density of the field about the origin.
(r) = (θ/θ) [μ0/(4π)2]qeqmRsinθ/(|r - Rk|3r).
= (θ/θ) [μ0/(4π)2]qeqm(R/r)sinθ/(r2 + R2 - 2rRcosθ)3/2.
(θ/θ) = i cosθcosφ + j cosθsinφ - k sinθ.

|r - Rk|2 = (r sinθ)2 + (r cosθ R)2.
(c)  Find L.
Using an integral table:
L = ∫VdV(r) = -k0/(4π)2]qeqmR2π ∫0πsin3θdθ∫0rdr/(r2 + R2 - 2rRcosθ)3/2.
Symmetry dictates that the x- and y-components of L integrate to zero.
0xdx/(a - bx + cx2)3/2 = (4a - 2bx)/[(b2 4ac)(a - bx + cx2)1/2] 0.
As x --> ∞, (4a - 2bx)/[(b2 4ac)(a - bx + cx2)1/2]  --> -2b/((b2 4ac)√c).
0xdx/(a + bx + cx2)3/2 = (-2b√a 4a√c )/((b2 4ac)√(ac)).
0rdr/(r2 + R2 - 2rRcosθ)3/2 = (-4R2cosθ 4R2)/(4R3cos2θ 4R3)
= R-1(1 + cosθ)/(1- cos2θ) = R-1/(1- cosθ).
L = -k0/(4π)2]qeqm2π ∫0πsin3θdθ/(1 cosθ)
= -k0/(4π)2]qeqm2π ∫-11(1 - x2)dx/(1 x)
= -k0/(4π)2]qeqm2π ∫-11(1 + x)dx = -k0/(4π)]qeqm, independent of R.

Using vector identities:
(r) = [μ0ε0qm/(4πr3)] r E(r) r) = -[μ0ε0qm/(4πr3)] r r E(r)).
r(rE) = r(Er) - Er2  (BAC - CAB rule)
=  -r3(E)r/r,  since (a)n = (1/r)[a - n(a∙n)].
(r) = [μ0ε0qm/(4π)] (E)r/r.
Angular momentum: L = [μ0ε0qm/(4π)] ∫VdV (E)r/r.
-∞Ex ∂/∂x (r/r) dx = Ex (r/r)|-∞ - ∫-∞ (r/r)∂Ex/∂x dx
= - ∫-∞ (r/r)∂Ex/∂x dx  for a finite charge distribution.
Therefore L = -[μ0ε0/(4π)]qmV(r/r)∇∙EdV.
E = (qe0)δ(r - Rk).
L = -[μ0/(4π)]qmqeV(r/r)δ(r-Rk)dV = -[μ0/(4π)]qmqe k.

Problem 4:

Given are Maxwell's equations in macroscopic form,
D = ρfE = -∂B/∂t,  B = 0,  H = jf + ∂D/∂t,
and the constitutive relations for a linear, isotropic medium
D = εEB = μH.

Let jf = σcE + jb, where σc is the conductivity of the medium and jb represents the density of forced currents, such as currents inside a battery.

(a)  What is the relationship between the field quantities E and B and the potentials A and V?
(b)  E and B do not specify A and V uniquely.  Let A = A' - ψ,  V = V' + ∂ψ/∂t.  Show that Maxwell's equations are satisfied by A' and V' if they are satisfied by A and V.
(c)  Briefly discuss the gauge transformation represented in part b and its implications.
(d)  Write the relationship between the fields given in answers to part (a) in terms of H, D, A, and V.
(e)  Find the wave equation for A in terms of V, μ , ε , jb, σc , and  v = (μ0ε0)- in the Lorentz gauge.
(f)  Starting with Maxwell's equations, derive the wave equation for V in the Lorentz gauge.

Solution:

bulletConcepts:
Maxwell's equations, vector calculus
bulletReasoning:
Starting from Maxwell's equations, we introduce the electromagnetic potentials and derive the differential equation that they satisfy.
bulletDetails of the calculation:
(a) B = 0 --> B = A,   (E + ∂A/∂t) = 0 --> E  = -V - ∂A/∂t.

(b)  Let B' = A',   E'  = -V' - ∂A'/∂t.
Then B' = (A + ψ) = A + ψ.
The curl of a gradient is zero.  So B' = A = B.
E'  = -V' - ∂A'/∂t = -(V - ∂ψ/∂t) - ∂(A + ψ)/∂t
= -V + ∂ψ/∂t - ∂A/∂t - ∂ψ/∂t
= -V - ∂A/∂t = E,
since we can reverse the order of the partial derivatives.

(c)  A --> A + ψ,  V --> V - ∂ψ/∂t is called a gauge transformation.
A
and V are not unique.  A' and V' are also acceptable potentials.
E
and B are invariant under such a transformation.

(d)  H = (1/μ)A,  D = -εV - ε∂A/∂t.

(e)   H = jf + ∂D/∂t.  (A) = μjf + μ∂D/∂t.
(∇∙A) - ∇2A = μσcE + μjb + με∂E/∂t
= μσc(-V - ∂A/∂t) + με∂(-V - ∂A/∂t)/∂t + μjb.
2A - με∂2A/∂t2 - μσcA/∂t = (A) + μσcV + με∂(V)/∂t - μjb.
2A - (1/v2)∂2A/∂t2 - μσcA/∂t = (A) + μσcV +  (1/v2)∂(V)/∂t - μjb.
Lorentz condition:  A = -(1/v2)∂V/∂t .
Simplification: 
2A - (1/v2)∂2A/∂t2 - μσcA/∂t = μσcV  - μjb
is the wave equation for A in the  Lorentz gauge.

(f)  D = ρf,  -ε∇2V - ε∂(A)/∂t = -ε∇2V + ε(1/v2)∂2V/∂t2 =  ρf.
2V - (1/v2)∂2V/∂t2 =  -ρf
is the wave equation for V in the  Lorentz gauge.

Problem 5:

A thin wire of radius b is used to form a circular wire loop of radius a (a >> b) and total resistance R.
The loop is rotating about the z-axis with constant angular velocity ωk in a region with constant magnetic field B = B0i
At t = 0 the loop lies in the y-z plane and the point A at the center of the wire crosses the y-axis.

Let the (θ/θ) direction be tangential to the loop and be equal to the positive z direction at point A.
Let the (φ/φ) direction be tangential to the wire and be equal to the direction indicated in the figure.
(a)  Find the current flowing in the loop.  Neglect the self-inductance of the loop.  What is current density J as a function of time?
(b)  Find the thermal energy generated per unit time, averaged over one revolution.
(c)  Write down an expression for the the pointing vector S on the surface of the wire.
(d)  Use S to find the field energy per unit time flowing into the wire, averaged over one revolution.

Solution:

bulletConcepts:
The Poynting vector, energy conservation
bulletReasoning:
The Poynting vector represents the energy flux in the electromagnetic field.  The energy can circulate or flow into an object.  If it flows into an object and is absorbed, energy conservation requires that the field energy is converted into another form of energy.
bulletDetails of the calculation:
(a)  Flux F = BA = B0πa2 cos(ωt).  (Define the normal to the area by the right-hand rule.)
emf = -dF/dt = ωBπa2 sin(ωt),  I = ωBπa2 sin(ωt)/R.  At t = 0 the current at point A flows in the z-direction.
J = (θ/θ) (a2/b2)ωB0 sin(ωt)/R.
(b)  P = I2R = ω2B02π2a4 sin2(ωt)/R.  
<P> = ω2B02π2a4/R is the thermal energy generated per unit time, averaged over one revolution.
(c)  Ewire = (θ/θ)ωB0πa2 sin(ωt)/(2πa) = ωB0a sin(ωt).  Bwire = μ0I/(2πb)(φ/φ).
B = B0 + Bwire.
S = (1/μ0)(EB) = (1/μ0)(EwireBwire) + (1/μ0)(EwireB0).
(EwireBwire) always points towards the center of the wire.  Let us call this direction the -(ρ/ρ) direction.
Field energy flowing into the wire per unit time per unit length:
-∫0S∙(ρ/ρ) bdφ = (1/μ0)|EwireBwire|2πb = ω2B02π2a4 sin2(ωt)/(2πaR).
[(1/μ0)(EwireB0) has the same direction all along a circumference of the wire and therefore there is no net flux into the wire due to this term.]
Total field energy flowing into the wire per unit time
2πa(1/μ0)|EwireBwire|2πb = ω2B02π2a4 sin2(ωt)/R.
Averaged over one revolution:  ω2B02π2a4/R = <P>.