(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/ε_{0},**∇**×**E**= -∂**B**/∂t**, ∇∙B**= 0,**∇**×**B**= μ_{0}**j**+ (1/c^{2})∂**E**/∂t.=

E∙j**E∙**(1/μ_{0})(**∇**×**B**) - ε_{0}(∂**E**/∂t)**∙E**= -(1/μ

_{0})**∇∙**(**E**×**B**) +**B∙**(1/μ_{0})(**∇**×**E**) - ε_{0}(∂**E**/∂t)**∙E**= -(1/μ

_{0})**∇∙**(**E**×**B**) - (1/μ_{0})**B∙**(∂**B**/∂t) - ε_{0}(∂**E**/∂t)**∙E**

= -(1/μ_{0})**∇∙**(**E**×**B**) - (∂/∂t)((1/2μ_{0})B^{2}+ (ε_{0}/2)E^{2}).

(∂/∂t)((1/2μ_{0})B^{2}+ (ε_{0}/2)E^{2}) + (1/μ_{0})**∇∙**(**E**×**B**) = -**E∙j.**Interpretation:

Let u = (1/2μ_{0})B^{2}+ (ε_{0}/2)E^{2}be the energy density in the electromagnetic field.

Let**S**= (1/μ_{0})(**E**×**B**) be the energy flux.

Then (∂u/∂t) +**∇∙S**= -**E∙j**.

∫_{V}**∇∙E**dV = ∮**S**∙**n**dA, -∫_{V}∂u/∂t dV = ∮**S**∙**n**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**= σ**E**. Let**j**= j**k**, I = jπa^{2}= σEπa^{2}. E = I/(σπa^{2}).

On the surface of the wire**E**= I/(σπa^{2})**k**.

Outside of the wire**B**= (**φ**/φ) μ_{0}I/(2πr). On the surface**B**= (**φ**/φ) μ_{0}I/(2πa).

**S**= (1/μ_{0})(**E**×**B**).= -(

S**ρ**/ρ) I^{2}/(2σπ^{2}a^{3}).

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 = I^{2}/(σπa^{2}).

The thermal energy dissipated per unit length is P = I^{2}R_{unit length}.

R_{unit length}= 1/(σπa^{2}).

Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]

Consider the circuit shown.

Let V = 50 V, R_{1} = 10 Ω, R_{2} = 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 R_{2} 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 I_{1}flow through R_{1}and I_{2}flow through R_{2}.

Assume the directions for I_{1}and I_{2}are from top to bottom.

V - R_{1}I_{1}- LdI_{1}/dt = 0, V - I_{2}R_{2}= 0, I = I_{1}+ I_{2}.

I_{2}= V/R_{2}, dI_{1}/dt = V/L - R_{1}I_{1}. I_{1}(t) = (V/R_{1})[1 - exp(-R_{1}t/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 R_{2}from the bottom to the top. The voltage across R_{2}is 400 V, V_{bottom}- V_{top}= 400 V. The voltage across the switch is 350 V.

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:

Static fields and radiation fields - 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/(r^{2 }+ z^{2})^{½}is the potential on the axis due to a ring of radius r. (k = 1/(4πε_{0}))

Integrating from r = 0 to r = R we find V(z) = k2πσ[(R^{2}+z^{2})^{½ }- z] for a disk of radius R.

Q = σπR^{2}, so V(z) = k(2Q/R^{2})[(R^{2}+z^{2})^{½ }- z] = (Q/(2πε_{0}R^{2}))[(R^{2}+z^{2})^{½ }- z].

E_{z}= -dV(z)/dz = (Q/(2πε_{0}R^{2}))[1 - z/(R^{2}+z^{2})^{½}]. Symmetry dictates that**E**= E_{z}**k**.

(b) (R^{2 }+ z^{2})^{½ }- z = z(1 + R^{2}/z^{2})^{½}- z ≈ z(1 + ½R^{2}/z^{2}} - z = ½R^{2}/z^{ }for z >> R.

V(z) ≈ (Q/(2πε_{0}R^{2})) ½R^{2}/z = Q/(4πε_{0}z) for z >> R.

We have azimuthal symmetry. The most general solution for r > R is

V(r,θ) = ∑_{n=0}^{∞}[B_{n}/r^{n+1}]P_{n}(cosθ). We find the B_{n}my matching at θ = 0, P_{n}(cosθ) = 1 to V(z) found above. To first order B_{1}= kQ and B_{n}= 0 for n > 1.

Therefore at large distance V(**r**) = Q/(4πε_{0}r), the potential reduced to that of a point charge, and the corresponding field**E**(**r**) = Q/(4πε_{0}r^{2}) (**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 d**m**= dm**k**,

with dm = IA = σvdr πr^{2}= σωπr^{3}dr.

The magnetic moment of the disk is m = ∫_{0}^{R}dm = σωπ∫_{0}^{R}r^{3}dr = ¼σωπR^{4}= ¼QωR^{2}.

(d) At large distances the magnetic field is a dipole field.**B**(**r**) = (μ_{0}/(4π))[3(**m∙r**)**r**/r^{5}-**m**/r^{3}] = (μ_{0}m/(4πr^{3}))[2 cosθ (**r**/r) + sinθ (**θ**/θ)]

= (μ_{0}QωR^{2}/(16πr^{3}))[2 cosθ (**r**/r) + sinθ (**θ**/θ)].

**S**= (1/μ_{0})**E**×**B**= (1/μ_{0}) Q/(4πε_{0}r^{2}) (μ_{0}QωR^{2}/(16πr^{3})) sinθ (**φ**/φ)

= Q^{2}ωR^{2}sinθ/(64π^{2}ε_{0}r^{5}) (**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.

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 ω**k**
in a region with constant magnetic field **B** = B_{0}**i**.

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 =**B**∙**A**= B_{0}πa^{2}cos(ωt). (Define the normal to the area by the right-hand rule.)

emf = -dF/dt = ωBπa^{2}sin(ωt), I = ωBπa^{2}sin(ωt)/R. At t = 0 the current at point A flows in the z-direction.

**J**= (**θ**/θ)^{2}/b^{2})ωB_{0}sin(ωt)/R.

(b) P = I^{2}R = ω^{2}B_{0}^{2}π^{2}a^{4}sin^{2}(ωt)/R.

<P> = ½ω^{2}B_{0}^{2}π^{2}a^{4}/R is the thermal energy generated per unit time, averaged over one revolution.

(c)**E**_{wire}= (**θ**/θ)ωB_{0}πa^{2}sin(ωt)/(2πa) = ½ωB_{0}a sin(ωt).**B**_{wire}= μ_{0}I/(2πb)(**φ**/φ)**.**

**B**= B_{0}+ B_{wire}.

**S**= (1/μ_{0})(**E**×**B**) = (1/μ_{0})(**E**_{wire}×**B**_{wire}) + (1/μ_{0})(**E**_{wire}×**B**_{0}).

(**E**_{wire}×**B**_{wire}) 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:

-∫_{0}^{2π}**S**∙(**ρ**/ρ) bdφ = (1/μ_{0})|**E**_{wire}×**B**_{wire}|2πb = ω^{2}B_{0}^{2}π^{2}a^{4}sin^{2}(ωt)/(2πaR).

[(1/μ_{0})(**E**_{wire}×**B**_{0}) 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/μ_{0})|**E**_{wire}×**B**_{wire}|2πb = ω^{2}B_{0}^{2}π^{2}a^{4}sin^{2}(ωt)/R.

Averaged over one revolution: ½ω^{2}B_{0}^{2}π^{2}a^{4}/R = <P>.

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) = Q_{inside}**r**/(4πε_{0}r^{3}).

Symmetry and charge conservation yield

dQ_{inside}/dt = -∫_{S}**j**∙d**S**= -j(r) 4πr^{2},**j**= j(r)(**r**/r).

**∇**×**B**= μ_{0}**j**+ μ_{0}ε_{0}∂**E**/∂t = -μ_{0}(dQ_{inside}/dt)**r**/(4πr^{3}) + μ_{0}ε_{0}(dQ_{inside}/dt)**r**/(4πε_{0}r^{3}) = 0.

(b)**∇**×**B**= 0 -->**B**= -**∇**Φ_{M}._{ }**∇**∙**B**= 0 --> ∇^{2}Φ_{M}= 0 everywhere.

We can treat this as a boundary value problem.

Looking for a solution with no angular dependence yields

Φ_{M}(r) = A_{0}+ B_{0}/r.

Requiring Φ_{M}to not blow up at the origin (since**∇**∙**B**= 0) yields

Φ_{M}(r) = A_{0}= constant, B = 0.