**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:

Concepts: Static fields and radiation fields | |

Reasoning: Static charge and current distributions do not produce electromagnetic radiation. | |

Details of the calculation: (a) E(r) = Q/(4πε_{0}r^{2}) (r/r),
the field of a point charge Q.(b) B(r) = (μ_{0}/(4π))[3(m∙r)r/r^{5}
– m/r^{3}].Here m = IA k = QωR^{2}/2 k.B(r) = (μ_{0}m/(4πr^{3}))[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/μ_{0})E×B = (1/μ_{0}) Q/(4πε_{0}r^{2})
(μ_{0}QωR^{2}/(8πr^{3})) sinθ (φ/φ)= Q ^{2}ωR^{2}sinθ/(32π^{2}ε_{0}r^{5}) (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:

Concepts: Maxwell's equations | |

Reasoning: 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. | |

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∙dS
= -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.(c) ∇×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) yieldsΦ _{M}(r) = A_{0} = constant, B = 0. |

**Problem 3:**

Assume magnetic charges exist and Maxwell's equations are of the form

**∇**∙**E** = ρ/ε_{0, }**∇**∙**B** = μ_{0}ρ_{m, }-**∇**×**E**
= μ_{0}**j _{m}** + ∂

(a) Write down expressions for the electric field

(b) Write down expressions for the momentum density

(c) Show that there is electromagnetic angular momentum L

Solution:

Concepts: Field energy and field momentum of the electromagnetic field | |

Reasoning: (1/μ _{0})(E×B)
= energy flux, (1/(cμ_{0}))(E×B)
= momentum flux, (1/(c ^{2}μ_{0}))(E×B)
= ε_{0}(E×B) = momentum density.The angular momentum density of the field about the origin then is r ×
ε_{0}(E×B). | |

Details of the calculation: (a) Field of the point charge: E(r) = [1/(4πε_{0})]q_{e}(r
- Rk)/|r - Rk|^{3}.Field of the magnetic monopole: B(r) = [μ_{0}/(4π)]q_{m}r/r^{3}).(b) S(r) = (1/μ_{0})(E(r)×B(r))
= energy flux.S(r)/c = momentum flux.g(r) = S(r)/c^{2} = ε_{0}E(r)×B(r)) = momentum density.g(r) = [μ_{0}ε_{0}q_{m}/(4πr^{3})]
E(r)×r.(g r) = -(φ/φ)[μ_{0}/(4π)^{2}]q_{e}q_{m}Rsinθ/(|r
- Rk|^{3}r^{2}).ℒ(r) = r×g(r) = angular momentum
density of the field about the origin.ℒ(r) = (θ/θ) [μ_{0}/(4π)^{2}]q_{e}q_{m}Rsinθ/(|r - Rk|^{3}r).= ( θ/θ) [μ_{0}/(4π)^{2}]q_{e}q_{m}(R/r)sinθ/(r^{2}
+ R^{2} - 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 = ∫_{V}dV ℒ(r) = -k [μ_{0}/(4π)^{2}]q_{e}q_{m}R2π
∫_{0}^{π}sin^{3}θdθ∫_{0}^{∞}rdr/(r^{2}
+ R^{2} - 2rRcosθ)^{3/2}.Symmetry dictates that the x- and y-components of L integrate to
zero.∫ _{0}^{∞}xdx/(a - bx + cx^{2})^{3/2} = (4a -
2bx)/[(b^{2} – 4ac)(a - bx + cx^{2})^{1/2}] _{0}^{∞}.^{
}As x --> ∞, (4a - 2bx)/[(b^{2} – 4ac)(a - bx + cx^{2})^{1/2}]
--> -2b/((b^{2} – 4ac)√c).∫ _{0}^{∞}xdx/(a + bx + cx^{2})^{3/2} =
(-2b√a – 4a√c )/((b^{2} – 4ac)√(ac)).∫ _{0}^{∞}rdr/(r^{2} + R^{2} - 2rRcosθ)^{3/2}
= (-4R^{2}cosθ – 4R^{2})/(4R^{3}cos^{2}θ –
4R^{3}) = R ^{-1}(1 + cosθ)/(1- cos^{2}θ) = R^{-1}/(1- cosθ).L = -k [μ_{0}/(4π)^{2}]q_{e}q_{m}2π
∫_{0}^{π}sin^{3}θdθ/(1 – cosθ)= -k [μ_{0}/(4π)^{2}]q_{e}q_{m}2π
∫_{-1}^{1}(1 - x^{2})dx/(1 – x)= -k [μ_{0}/(4π)^{2}]q_{e}q_{m}2π
∫_{-1}^{1}(1 + x)dx = -k [μ_{0}/(4π)]q_{e}q_{m},
independent of R.Using vector identities: |

**Problem 4:**

Given are Maxwell's equations in macroscopic
form,

**∇**∙**D** = ρ_{f}, **∇**×**E** = -∂**B**/∂t, **
∇**∙**B** =
0, **∇**×**H** = **j**_{f} + ∂**D**/∂t,

and the constitutive relations for a linear, isotropic medium

**D** = ε**E**, **B** = μ**H**.

Let **j**_{f} = σ_{c}**E** + **j**_{b}, where
σ_{c} is the conductivity of the medium and
**j**_{b}
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 , **j**_{b},
σ_{c} , and
v = (μ_{0}ε_{0})^{-½} in the Lorentz
gauge.

(f) Starting with Maxwell's equations, derive the wave equation for V in the Lorentz
gauge.

Solution:

Concepts: Maxwell's equations, vector calculus | |

Reasoning: Starting from Maxwell's equations, we introduce the electromagnetic potentials and derive the differential equation that they satisfy. | |

Details 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. and V are not unique. A A' and V' are also acceptable
potentials. and E B are invariant under such a transformation.(d) H
=
(1/μ)∇×A, D
= -ε∇V - ε∂A/∂t.(e) ∇×H = j_{f} + ∂D/∂t. ∇×(∇×A)
= μj_{f} + μ∂D/∂t.∇(∇∙A)
- ∇^{2}A = μσ_{c}E + μj_{b}
+ με∂E/∂t= μσ _{c}(-∇V - ∂A/∂t) + με∂(-∇V
- ∂A/∂t)/∂t + μj_{b}.∇ ^{2}A - με∂^{2}A/∂t^{2}
- μσ_{c}∂A/∂t = ∇(∇∙A) + μσ_{c}∇V
+ με∂(∇V)/∂t - μj_{b}.∇^{2}A - (1/v^{2})∂^{2}A/∂t^{2}
- μσ_{c}∂A/∂t = ∇(∇∙A)
+ μσ_{c}∇V + (1/v^{2})∂(∇V)/∂t
- μj_{b}.Lorentz condition:
∇∙A = -(1/v^{2})∂V/∂t .Simplification: ∇ ^{2}A - (1/v^{2})∂/∂t^{2}A^{2}
- μσ_{c}∂A/∂t = μσ_{c}∇V
- μj_{b
}is the wave equation for A in the Lorentz gauge.(f) ∇∙D = ρ_{f}, -ε∇^{2}V - ε∂(∇∙A)/∂t
= -ε∇^{2}V + ε(1/v^{2})∂^{2}V/∂t^{2}
= ρ_{f}.∇ ^{2}V - (1/v^{2})∂^{2}V/∂t^{2}
= -ρ_{f}/εis 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 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) (a^{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>. |