Problem 1:
In the derivation of the wave equations for A and f from Maxwell's equations in a vacuum, one gets at one stage
and
.
This is made solvable by use of a gauge choice, using the fact that physics is invariant under a gauge transformation. Write down the Lorentz gauge condition and the resulting PDE's for f and A.
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? We are asked to use the Lorentz condition to simplify the given PDE's. |
| How do they apply? 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. 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 |
| Details of the calculation: None |
Problem 2:
(a) From Maxwell's equations, derive the conservation of energy
equation,
,
where u is the energy density
and S is the Poynting vector.
(b) A long, cylindrical conductor of radius a and conductivity
s 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, 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 derive the conservation of energy equation. |
|
How do they apply? (a) Maxwells equations in SI units are  ,
 .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) = -(1/m0)Ñ×(E´B) - (¶/¶t)((1/2m0)B2 + (e0/2)E2). (¶/¶t)((1/2m0)B2 + (e0/2)E2) + (1/m0)Ñ×(E´B) = -E×j. 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. |
| Details of the calculation: (b) In the wire j = sE. Let j = j k, I = jpa2 = sEpa2. E = I/(spa2). On the surface of the wire E = I/(spa2) k. Outside of the wire B =  m0I/(2pr). On the surface
B =
m0I/(2pa).  .
S = -  I2/(2sp2a3).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 = S2pa = I2/(spa2). The thermal energy dissipated per unit length is P = I2Runit length. Runit length = 1/(spa2), since j = sE, pa2j = pa2sE, I = pa2sV/l = V/Rres, Rres = l/(pa2s). Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.] |