## Assignment 2, solutions

#### Problem:

(a)  From Maxwell's equations, derive the conservation of energy equation,
(∂u/∂t) + ∇∙S = -E∙j,
where u is the energy density and S is the Poynting vector.  Rewrite this equation in integral form and explain why it is a statement of energy conservation.
(b)  A long, cylindrical conductor of radius a and conductivity σ carries a constant current I.  Find S at the surface of the cylinder and interpret your result in terms of conservation of energy.

Solution:

• Concepts:
Maxwell's equations, energy density and energy flux in the electromagnetic field
• Reasoning:
We are explicitly instructed to use Maxwell's equations to derive the conservation of energy equation.
• Details of the calculation:
(a)  Maxwell's equations in SI units are
∇∙E = ρ/ε0×E = -∂B/∂t,  ∇∙B = 0,  ×B = μ0j + (1/c2)∂E/∂t.
E∙j
= E∙(1/μ0)(×B) - ε0(∂E/∂t)∙E
= -(1/μ0)∇∙(E×B) + B∙(1/μ0)(×E) - ε0(∂E/∂t)∙E
= -(1/μ0)∇∙(E×B) - (1/μ0)B∙(∂B/∂t) - ε0(∂E/∂t)∙E
= -(1/μ0)∇∙(E×B) - (∂/∂t)((1/2μ0)B2 + (ε0/2)E2).
(∂/∂t)((1/2μ0)B2 + (ε0/2)E2) + (1/μ0)∇∙(E×B) = -E∙j.
Interpretation:
Let u = (1/2μ0)B2 + (ε0/2)E2 be the energy density in the electromagnetic field.
Let S = (1/μ0)(E×B) be the energy flux.
Then (∂u/∂t) + ∇∙S = -E∙j.
V∇∙E  dV = ∮Sn dA,  -∫V ∂u/∂t dV = ∮Sn dA +  ∫V E∙j  dV.
This is a statement of energy conservation.  The rate at which the field energy in a volume decreases equals the rate at which it leaves across the boundaries plus the rate at which it gets converted into other forms.
(b)  In the wire j = σE.  Let j = j k, I = jπa2 = σEπa2.  E = I/(σπa2).
On the surface of the wire E = I/(σπa2) k.
Outside of the wire B = (φ/φ) μ0I/(2πr).  On the surface B = (φ/φ) μ0I/(2πa).
S = (1/μ0)(E×B).
S
=  -(ρ/ρ) I2/(2σπ2a3).
Energy is flowing into the wire and is converted into heat.
[The field energy that is flowing into the wire per unit length is P = S2πa = I2/(σπa2).
The thermal energy dissipated per unit length is P = I2Runit length.
Runit length = 1/(σπa2).
Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]

#### Problem 2:

Consider the circuit shown.

Let V = 50 V, R1 = 10 Ω, R2 = 100 Ω, and L = 50 H.
All circuit elements are ideal and no current flows before the switch is closed.
(a) After the switch is closed at t = 0, find the current I flowing through the switch as a function of time.
(b) After 8 s the switch is opened again.  Right after the switch is opened, what is the voltage across R2 and across the switch?

Solution:

• Concepts:
Transient RL circuits, Kirchhoff's rules
• Reasoning:
We are asked to analyze the transient behavior of an RL circuit.
• Details of the calculation:
(a)  Let I1 flow through R1 and I2 flow through R2.
Assume the directions for I1 and I2 are from top to bottom.
V - R1I1 - LdI1/dt = 0, V - I2R2 = 0, I = I1 + I2.
I2 = V/R2,  dI1/dt = V/L - R1I1. I1(t) = (V/R1)[1 - exp(-R1t/L)].
I = 0.5 A + 5 A[1 - exp(-(0.2/s)t)].
(b)  Right after the switch is opened I = 5 A[1 - exp[(-0.2/s)8 s)] = 4 A of current flow through R2 from the bottom to the top.  The voltage across R2 is 400 V, Vbottom - Vtop = 400 V.  The voltage across the switch is 350 V.

#### Problem 3:

A thin disc with radius R and uniform charge density σ is centered at the origin.  Its normal points along the z-axis.  It is rotating with angular velocity ω k about the z-axis.
(a)  Find the potential and the electric field due to the disc on the z-axis.
(b)  Show that the potential reduces to that of a point charge for large distances from the origin.
(c)  Find the magnetic moment of the disk.
(d)  At large distances from the origin, find the magnitude and direction of the Poynting vector S.
Does the rotating disk produce a radiation field?  Explain.

Solution:

• Concepts:
• Reasoning:
Static charge and current distributions do not produce electromagnetic radiation.
• Details of the calculation:
(a)  The disk lies in the xy-plane with its center at the origin.  The z-axis is the symmetry axis.
dV(z) = k2πrσdr/(r2 + z2)½ is the potential on the axis due to a ring of radius r.  (k = 1/(4πε0))
Integrating from r = 0 to r = R we find V(z) = k2πσ[(R2+z2)½ - z] for a disk of radius R.
Q = σπR2, so V(z) = k(2Q/R2)[(R2+z2)½ - z] = (Q/(2πε0R2))[(R2+z2)½ - z].
Ez = -dV(z)/dz = (Q/(2πε0R2))[1 - z/(R2+z2)½].  Symmetry dictates that E = Ezk.

(b)  (R2 + z2)½ - z = z(1 + R2/z2)½ - z  ≈ z(1 + ½R2/z2} - z = ½R2/z  for z >> R.
V(z) ≈ (Q/(2πε0R2)) ½R2 /z = Q/(4πε0z)  for z >> R.
We have azimuthal symmetry.  The most general solution for r > R is
V(r,θ) = ∑n=0[Bn/rn+1]Pn(cosθ).  We find the Bn my matching at θ = 0, Pn(cosθ) = 1 to V(z) found above.  To first order B1 = kQ and Bn = 0 for n > 1.
Therefore at large distance V(r) = Q/(4πε0r), the potential reduced to that of a point charge, and the corresponding field E(r) = Q/(4πε0r2) (r/r)  also reduces to that of a point charge.

(c)  The magnetic moment of ring with radius r and width dr and surface charge density σ  rotating with angular velocity ω k about the z-axis is dm = dm k,
with dm = IA = σvdr πr2 = σωπr3dr.
The magnetic moment of the disk is m = ∫0Rdm = σωπ∫0Rr3dr = ¼σωπR4 = ¼QωR2.

(d)  At large distances the magnetic field is a dipole field.
B(r) = (μ0/(4π))[3(m∙r)r/r5 - m/r3] = (μ0m/(4πr3))[2 cosθ (r/r) + sinθ (θ/θ)]
= (μ0QωR2/(16πr3))[2 cosθ (r/r) + sinθ (θ/θ)].
S = (1/μ0)E×B =  (1/μ0)  Q/(4πε0r2) (μ0QωR2/(16πr3)) sinθ (φ/φ)
= Q2ωR2sinθ/(64π2ε0r5) (φ/φ).  S encircles the z-axis in the (φ/φ) direction.
There is no radiation field.  E and B do not change with time, we have only static fields.

#### Problem 4:

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:

• Concepts:
The Poynting vector, energy conservation
• Reasoning:
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.
• Details 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> = ½ω2B02π2a4/R is the thermal energy generated per unit time, averaged over one revolution.
(c)  Ewire = (θ/θ)ωB0πa2 sin(ωt)/(2πa) = ½ωB0a sin(ωt).  Bwire = μ0I/(2πb)(φ/φ).
B = B0 + Bwire.
S = (1/μ0)(E×B) = (1/μ0)(Ewire×Bwire) + (1/μ0)(Ewire×B0).
(Ewire×Bwire) 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)|Ewire×Bwire|2πb = ω2B02π2a4 sin2(ωt)/(2πaR).
[(1/μ0)(Ewire×B0) 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)|Ewire×Bwire|2πb = ω2B02π2a4 sin2(ωt)/R.
Averaged over one revolution:  ½ω2B02π2a4/R = <P>.

#### Problem 5:

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:

• Concepts:
Maxwell's equations
• Reasoning:
Place the origin of the coordinate systemt at the center of the charge distribution.  Use Gauss' law and charge conservation to find E(r) and j(r) at a point P.
• Details 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.
(b)  ×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.