Problems

Problem 1:

(a)  Write down Maxwell's equations in free space and in the presence of the current density j(r,t) and charge density r(r,t).  Introduce the electromagnetic potentials and derive the differential equation that they satisfy.
(b)  State the Lorentz condition and show the simplification found thereby.
(c)  Define a gauge transformation and describe its effect on the electromagnetic fields.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations, scalar and vector potential, the Lorentz condition, gauge transformations
Why do they apply?
Starting from Maxwell's equations, we are asked to introduce the electromagnetic potentials and derive the differential equation that they satisfy.
How do they apply?
(a)  Maxwell's equations in SI units are
,  .
We find
,  .
Therefore
,  and   
are the differential equations satisfied by the potentials.
Using

we write
.
Details of the calculation:
(b)  Lorentz condition:  (we pick an expression for Ñ×A.)
Simplification: 
, 
are now the differential equations satisfied by the potentials.
f, Ax, Ay, Az now satisfy the inhomogeneous wave equation.
(c)  A and f are not unique.
A --> A + Ñy,  f --> f - ¶y/t is called a gauge transformation.
E and B are invariant under such a transformation.

Problem 2:

Consider Maxwell's differential equations in an ideal linear, isotropic, homogeneous dielectric medium with frequency-independent dielectric constant e and magnetic susceptibility m.
(a)  Derive the conservation theorem for electric charge.
(b)  Uncouple the equations for the Cartesian components of E and B to find 6 uncoupled second-order partial differential equations of the form Ñ2f(x,y,z,t) = a2 2f(x,y,z,t)/t2.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations in macroscopic media
Why do they apply?
We are supposed to manipulate Maxwell's equations to derive the charge conservation theorem and the wave equation for E and B.
How do they apply?
Maxwell's equations in an ideal linear, isotropic, homogeneous dielectric medium are
,.
You may replace E by D/e and B by mH.
Ñ×(Ñ´B) = 0 --> Ñ×jf = -e(/t)Ñ×E = -(/t)rf.
Ñ×jf = -(/t)rf   -->   conservation of charge.
Details of the calculation:
(b)  Assume rf and jf are zero in the medium.
Ñ´(Ñ´E) = Ñ(Ñ×E) - Ñ2E = -(/t)(Ñ´B) = -me¶2E/t2.
Ñ2E = -me¶2E/t2.
Ñ´(Ñ´B) = Ñ(Ñ×B) - Ñ2B = me(/t)(Ñ´E) = -me¶2B/t2.
Ñ2B = -me¶2B/t2.

Problem 3:

(a)  By applying charge conservation to a volume V, derive the integral form of the equation of charge conservation.  Deduce the equation of continuity, Ñ×j + (/t)r = 0.
(b)  Show that the equation of continuity is a consequence of Maxwell's equations.
(c)  A region containing a charge Q0 is produced at time t = 0 inside a conductor of conductivity s .  By considering the flow of current across the surface of the region, show that at subsequent times the charge inside the region is Q(t) = Q0exp(-st/e0).
HINT:  Use Gauss's Law and Ohm's Law in the form j = sE.
(d)  Explain why you expect Q ® 0 as t ® ¥ .

Solution:

Concepts, principles, relations that apply to the problem:
Gauss' theorem, Maxwell's equations, properties of a conductor
Why do they apply?
(a)  If we have a conserved quantity in a volume V, then the rate at which it flows out of the volume must equal the rate it decreases inside the volume. 
For charge this is expressed as òclosed_Aj×ndA = -(/t)òVrdV for any volume V.
Gauss' theorem then yields òVÑ×jdV = -(/t)òVrdV for any volume V.
Therefore Ñ×j = -¶r/t.
(b)  Maxwell's equations relate the sources and the fields.
How do they apply?
Ñ´B = m0j + (1/c2)E/t (SI units), Ñ×(Ñ´B) = 0 --> m0Ñ×j + (1/c2)(/t)Ñ×E = 0.
Ñ×E = r/e0, m0Ñ×j + (1/(e0c2))(¶r/t) = 0.
m0e0 = 1/c2, Ñ×j + (¶r/t) = 0.
Details of the calculation:
(c)  òVÑ×jdV = -(/t)òVrdV.
For a conductor we have j = sE.
òVsÑ×EdV = òV(sr/e0)dV = (s/e0)Q = -(Q/t).
Q = Q0exp(-st/e0).
(d)  In a conductor charges are free to move.  They will move and distribute themselves over the surface until the electric field inside the conductor is zero.
Qinside ¹ 0 implies Einside ¹  0, therefore we expect Q(t) = 0 as as t ® ¥ .

Problem 4:

(a)  Find the magnetic field associated with the potential A(r,t) = (b/2)n´r.
(b)  Find the charge and current distributions that would lead to
A(r,t) = (bt/r3)r,  f(r,t) = 0.
(c)  Determine the charge distribution that will give rise to the potential
f(r,t) = W0exp(-ar)/r.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations, scalar and vector potential, Gauss' law
Why do they apply?
E = -A/t - Ñf, B = Ñ´A, (SI units).
Given the potentials we are asked to find the fields.
How do they apply?
(a)  Ax = (b/2)(nyz - nzy), Ay = (b/2)(nzx - nxz), Az = (b/2)(nxy - nyx).
Bx = Az/y - Ay/z = (b/2)(nx + nx) = bnx, By = bny,  Bz = bnz.
B = bn, a uniform field in the n direction of strength b.
Details of the calculation:
(b)  E = -A/t - Ñf = (b/r3)r.
This is the field of a point charge at the origin with q = 4pe0b.
B = Ñ´A = 0, Ñ´B = m0j + (1/c2)E/t = 0, this implies that j = 0.
(c)  E(r) = -Ñf(r) = W0(a exp(-ar)/r + exp(-ar)/r2)(r/r).
E(r) is radial, we have a spherically symmetric charge distribution.
òsurface of sphere of radius RE×ndA = E(R)4pR2 = 4pW0(1 + aR)exp(-aR) = Qinside/e0.
As r --> 0, Qinside --> 4pe0W0, there is a point charge at the origin.
For r ¹ 0, r(r) = -e0Ñ2f(r) = -e0Ñ2(W0exp(-ar)/r)
= -e0W0(1/r2)(/r)[r2(/r)(exp(-ar)/r)].
r(r) = -e0W0a2exp(-ar)/r.
Therefore r(r) = e0W0(4pd(r) - a2exp(-ar)/r).

Problem 5:

For a single charge, the rate of doing work by external fields B and E is qv×E, in which v is the velocity of the charge. 
(a)  Find the corresponding expression for a continuous distribution of charge and current and interpret it physically.
(b)  Use Maxwell's equations to express the result from part a in terms of the fields alone.
(c)  From your result in part b, verify Poynting's theorem .  
Find expressions for the terms u and S on the left-hand side, interpret those terms and ultimately the physical significance of the theorem that    represents.

Solution:

Concepts, principles, relations that apply to the problem:
Maxwell's equations, energy density and energy flux in the electromagnetic field
Why do they apply?
We are explicitly instructed to use Maxwell's equations to convert the expression obtained in part (a) into one that contains the fields alone.  When work is done on a charge by the field, field energy is converted into other forms of energy.  Energy conservation applies.
How do they apply?
(a)  Let rdV = dq be the amount of charge in a volume element dV.
dW/dt = dqv×E = rv×EdV = j×EdV = rate at which work is done by the field on the charges in dV.
(b)  E×j = E×(1/m0)(Ñ´B) - e0(E/t)×E = -(1/m0)Ñ×(E´B) - B×(1/m0)(Ñ´E) - e0(E/t)×E
= -(1/m0)Ñ×(E´B) - (1/m0)B×(1/m0)(B/t) - e0(E/t)×E
= -(1/m0)Ñ×(E´B) - (/t)((1/2m0)B2 + (e0/2)E2) 
Therefore
(/t)((1/2m0)B2 + (e0/2)E2) + (1/m0)Ñ×(E´B) = -E×j.
Details of the calculation:
Interpretation:
Let u = (1/2m0)B2 + (e0/2)E2 be the energy density in the electromagnetic field.
Let S = (1/m0)(E´B) be the energy flux.
Then (u/t) + Ñ×S = -E×j.
.
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.

Problem 6:

By considering the Poynting vector comment as quantitatively as possible on the energy flow in the following situations:
(a)  a long cylindrical wire of conductivity s carrying a current I.
(b)  a stationary electric charge q sitting on top of a magnetic dipole with magnetic moment m.
(c)  an electromagnetic wave incident normally on the flat surface of a dielectric of permittivity e.

Solution:

Concepts, principles, relations that apply to the problem:
The Poynting vector, energy conservation
Why do they apply?
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.
How do they apply?
(a)  In the wire j = sE.  Let j = j k, I = jpR2 = sEpR2.  E = I/(spR2).
On the surface of the wire E = I/(spR2) k.
Outside of the wire B = m0I/(2pr).  On the surface B = m0I/(2pR).
.  S =  - I2/(2sp2R3).
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 = S2pR = I2/(spR2).
The thermal energy dissipated per unit length is P = I2Rres per unit length.
Rres per unit length = 1/(spR2), since j = sE,  pR2j = pR2sE,
I = pR2sV/l = V/Rres,  Rres = l/(pR2s).
Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]
Details of the calculation:
(b)  Assume the point charge and the magnetic dipole are located at the origin with m = m k.
E = [1/(4pe0)] (q/r2) ,  B = [m0/(4p)] (m/r3)(2cosq + sinq ).
, S =  1/(16p2e0) (mq/r5) sinq.
Field energy circulates about the z-axis in the direction.
(c)  For a plane wave traveling into the k direction E and B are perpendicular to each other and to the k direction, B = E/v = E(me)1/2 (SI units), and <E´B> = k (E2(me)1/2)/2.  Here e and m are characteristics of the material the wave is traveling in.
At the interface part of the field energy is reflected and part of the field energy is transmitted.
For normal incidence we have t12 = 2n1/(n2 + n1) = transmission coefficient.
The transmittance is given by T = (n2/n1)|t12|2 = (<S>t×n)/(<S>i×n).
Here n is a unit vector normal to the interface.  In this problem n = k.
<S>i×n = (E2(m0e0)1/2)/(2m0) = (E2(e0/m0)1/2)/2.
<S>t×n = 4n1n2/(n2 + n1)2(<S>i×n) = (2n1n2/(n2 + n1)2)(E2(e0/m0)1/2)
Here n1 = c(m0e0)1/2 = 1 and  n2 = c(me)1/2.
<S>t = (2n1n2/(n2 + n1)2)(E2(e0/m0)1/2) k = (2n2/(n2 + 1)2)(E2(e0/m0)1/2) k. 
Energy conservation requires that for the reflected part we have
<S>r = <S>i - <S>t.

Problem 7:

A parallel plate capacitor consists of two circular metal plates of radius a, separated by a distance d.  The insulating material between the plates has a dielectric permittivity e.  The capacitor is part of the circuit shown below.

At time t = 0 we close the switch.  As a function of time, and ignoring edge effects, compute the following quantities for all spatial points within the dielectric material:

(i) the electric field,
(ii) the magnetic field,
(iii) the Poynting vector,
(iv) the energy density.

Finally, (v) give the total energy stored in the capacitor as t ® ¥.  Express your answers to (i) - (v) in terms of the circuit elements C, R, and V.

Solution:

Concepts, principles, relations that apply to the problem:
Transient fields, the Poynting vector
Why do they apply?
We are asked to solve for the transients in a circuit and comment on the energy density and the energy flux.
How do they apply?
For a parallel plate capacitor we have  C = eA/d.  Here A = pa2,  C = epa2/d.
When the switch is closed we have from Kirchhoff's loop rule V - IR - Q/C = 0.
Dividing by R we have
dQ/dt + Q/(RC) - V/R = 0.  Try a solution of the form Q = A(1-e-lt).
Ale-lt + A/(RC) - Ae-lt/(RC) - V/R = 0 for all t.
This implies A/(RC) = V/R,  A = VC and Al = A/(RC), l = 1/(RC).
Q = VC(1-e-t/(RC)).
The voltage across the capacitor as a function of time is Q/C, and in the dielectric we have
E = k Q/(dC) as a function of time.

(i)  E = (V/d)(1-e-t/(RC)) k.
(ii)  Ñ´B = m0e¶E/t,  B = B,  2prB = pr2m0e¶E/t,  B = [m0erV/(2dRC)]e-t/(RC).
(iii)  ,
S = (V/d)(1-e-t/(RC))[erV/(2dRC)]e-t/(RC)(-) = -[erV2/(2d2RC)](1-e-t/(RC))e-t/(RC).
Field energy is flowing into the capacitor from the sides.
(iv)  u = (1/2m0)B2 + (e/2)E2 = m0[erV/(dRC)]2e-2t/(RC)/8 + e(V/d)2(1-e-t/(RC))2/2.
(v)  As t ® ¥  u = e(V/d)2/2,  U = pa2eV2/(2d) = CV2/2 = total energy stored in the capacitor.
Details of the calculation:
Check:  .

Problem 8:

(a)  Assume Find the electric field everywhere.  Is E conservative?
(b)  Now assume a uniform, circular conducting ring, of radius R1  > R, is placed in the field you found in (a), concentric with the origin.  The conductance of the ring is s.  Place voltmeter leads on the ring (see figure below), separated by an angle q.   Plot the reading of the voltmeter as a function of q.  Assume an ideal voltmeter, with leads placed so that no flux cuts the circuit defined by (leads + ring).

Solution:

Concepts, principles, relations that apply to the problem:
Faraday's law, conservative and non-conservative electric fields.
Why do they apply?
We are supposed to find the induced electric field.  An induced field is non-conservative.
How do they apply?
(a)  , , .
A perpendicular distance r from the z-axis we have 
F/t = 2prE = -pr2B/t for r < R, F/t = 2prE = -pR2B/t for r > R.
(Note: r is a cylindrical coordinate here.)
E = -(r/2)B/t  for r < R,  E = -(R2/(2r))B/t  for r > R.
E is an induced field and not conservative.
Details of the calculation:
(b)  In the ring we have E = -(R2/(2R1))B/t.
E + A/t = -Ñf.
B = Ñ´A, B = B k for r < R,  B = 0 for r >R.
Let A = Af(r), then B = (1/r)(/r)(rAf(r)) for r < R,  Af(r) = Br/2.
For r > R, 0 = (1/r)(/r)(rAf(r)),  Af(r) = const/r.
Since Af(r) is continuous, const = BR2/2.
At r = R1 we have Af(R1) = BR2/(2R1),  Af(r)/t|R1 = (B/t)R2/(2R1).
E + A/t = 0,  Ñf = 0.  V = constant, DV = 0.
The ideal voltmeter reads zero, independent of the angle q.

Problem 9:

A long coaxial cable consists of two concentric cylindrical conducting sheets of radii R1 and R2 respectively (R2 > R1).  The two conductors are connected to a battery which maintains a voltage V0 between them.  They carry equal and opposite currents I.
(a)  Use Ampere's law to calculate the magnetic field everywhere.
Plot the magnetic field as a function of r (use cylindrical coordinates).
(b)  Use Gauss' law to show that the electric field between the two conductors is
E(r) = Cr/r2.  Express C in terms of R1, R2, and V0.  (Here r is a cylindrical coordinate.)
(c)  Calculate the Poynting vector.  Show that the power carried by the cable is P = V0I.

Solution:

Concepts, principles, relations that apply to the problem:
Gauss' law, Ampere's law, the Poynting vector
Why do they apply?
We are instructed to use Gauss' law and Ampere's law and to calculate the Poynting vector.
How do they apply?

(a)  For a circular loop G of radius r, concentric with the z-axis and lying in the x-y plane  we have
2prB = m0Ithrough G.  B = B.
r > R2:  B = 0.
r < R1:  B = 0.
R1 < r < R2:  B = m0I/(2pr).
(b)  For a Gaussian surface in the form of a cylinder with radius r and length L and its axis on the z-axis we have
2prErL = linsideL/e0.  Er = linside/(2pe0r)(r/r)
For a current to flow in the conductor in a static situation we need a voltage.
.
Er = V0/(ln(R2/R1))(r/r2).
If the conductors have no resistance, the electric field only has a radial component.
Details of the calculation:
(c)  Assume E = V0/(ln(R2/R1))(r/r2), i.e. the conductors have no resistance.
Then , S = k IV0/[2pr2 ln(R2/R1)].
.
P is the power flowing between the conductors into the k direction.

Problem 10:

(a)  Consider two concentric spherical metal shells of radius a and b, a < b.  There is a charge +q on the inner and a charge -q on the outer sphere.  A magnetic dipole with dipole moment m is in the center of the inner sphere.  Find the angular momentum associated with the electromagnetic field of the system.

Solution:

Concepts, principles, relations that apply to the problem:
The Poynting vector, energy momentum and angular momentum density in the electromagnetic field.
Why do they apply?
is the energy flux.  The momentum flux is S/c.  (Photon energy: hn, photon momentum: hn/c.)
The momentum flux equals the moment density times c.
When S is circulating the angular momentum density will not be zero.
How do they apply?
Assume magnetic dipole are located at the origin with m = m k.
Then B = [m0/(4p)](m/r3)(2cosq + sinq ).
The electric field is nonzero only between the shells.
E = [1/(4pe0)](q/r2) between the shells.
, S = [1/(16p2e0)](mq/r5) sinq.
Field energy circulates about the z-axis in the direction.
The  momentum density in the field is g = S/c2,
and the angular momentum density is r´g.
r´(r´z) = (r×z)r - (r×r)z = rz cosq r - r2 z.
Details of the calculation:
.
The x and y components of L vanish from symmetry.
.