Problem 1:

During some time interval Δt, a fat wire of radius a, carries a constant current I, uniformly distributed over its cross-sectional area.  A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor.
(a)  Find the electric field E in the gap at a distance s < a from the axis and the time t.  (Assume the charge is zero at t = 0). 
(b)  Find the magnetic field B in the gap at a distance s < a from the axis.

Solution:

• Concepts:
  Maxwell's equations
• Reasoning:
  When the charge on the plates of the"capacitor" changes, the changing electric field between the plates induces a magnetic field between the plates.
• Details of the calculation:
  (a)  Let the current flow in the z-direction.
  
  (b)  ∇×B = μ₀j + (1/c²)∂E/∂t.
  

Problem 2:

An infinite straight wire carries the current I(t) = 0 for t < , I(t) = I₀ for t > 0, that is, a constant current I₀ is turned on abruptly at t = 0.  Find the scalar and vector potentials in the Lorentz gauge.

Solution:

• Concepts:
  Potentials in the Lorentz gauge
• Reasoning:
  In the Lorentz gauge the expressions for the potentials are
  Φ(r,t) = [1/(4πε₀)]∫_v' dV' ρ(r', t_r)/|r - r'|,
  A(r,t) = [μ₀/(4π)]∫_v' dV' j(r', t_r)/|r - r'|.
  t_r = t - |r - r'|/c.
• Details of the calculation:
  Φ(r, t) = 0.
  A(r,t) = [μ₀/(4π)]∫_v' dV' j(r', t_r)/|r - r'| = [μ₀/(4π)]∫ dz' k I(z', t_r)/|r - z'k|.
  A(r,t) = A(r,t) k, symmetry dictates that A(r,t) = A(ρ,t).
  
  Define R as shown in the figure.
  A(ρ,t) = [μ₀/(4π)]∫_-∞^∞ dz I(t - R/c)/R.
  For points such that t < R/c, or (ct)² < z² + ρ², or z² > c²t² - ρ², we have I(t - R/c) = 0.
  For points such that t > R/c, or (ct)² > z² + ρ², or z² < c²t² - ρ², we have I(t - R/c) = I₀.
  Therefore A(ρ,t) = [μ₀/(4π)]∫_-a^a dz I₀/(z² + ρ²)^½ = [μ₀/(2π)]∫₀^a dz I₀/(z² + ρ²)^½,
  where a = (c²t² - ρ²)^½.
  A(ρ,t) = [μ₀I₀/(2π)] ln[((a² + ρ²)^½ + a)/ρ]
  = [μ₀I₀/(2π)] ln[((c²t² - ρ²)^½ + ct)/ρ].

Problem 3:

The vector potential A(r,t) of an oscillating dipole p at the origin is A(r,t) = -ik p exp(i(kr - ωt))/(4πε₀rc).
Note:  complex notation, the real part matters.
(a)  Let p = p k.  Use this vector potential to calculate the magnetic field.
(b)  Use Maxwell's equations to calculate the accompanying electric field.
(c)  Find the fields in the radiation zone.

Solution:

• Concepts:
  Maxwell's equations, radiation fields
• Reasoning:
  The radiation field decreases as 1/r.
• Details of the calculation:
  (a)  B = ∇×A.  A = A k = A[(r/r) cosθ  - (θ/θ) sinθ].
  B = (1/r)[∂(rA_θ)/∂r - ∂A_r/∂θ] (φ/φ) = B (φ/φ).
  B = -[ikp/(4πε₀rc)] [-iksinθ + sinθ/r] exp(i(kr - ωt)).
  Taking the real part we have
  B = -[k²p sinθ/(4πε₀rc)] [cos(kr - ωt) - sin(kr - ωt)/(kr)].
  
  (b)  ∇×B = (1/c²)∂E/∂t = -(iω/c²)E,  since E is proportional to exp(-iωt).
  ∇×B = (r/r) (1/(rsinθ))∂(Bsinθ)/∂θ - (θ/θ) (1/r))∂(rB)/∂r
  = -(r/r) [2k⁴p cosθ/(4πε₀c)] [1/(kr)² + i/(kr)³]exp(i(kr - ωt))
  - (θ/θ) [k⁴p sinθ/(4πε₀c)] [[1/(kr)² + i/(kr)³ - i/(kr)]exp(i(kr - ωt))].
  Taking the real part we have
  E_r = [2ck⁴p cosθ/(4πε₀ω)][sin(kr - ωt)/(kr)² + cos(kr - ωt)/(kr)³],
  E_θ = [ck⁴p sinθ/(4πε₀ω)][sin(kr - ωt)/(kr)² + cos(kr - ωt)(1/(kr)³ - 1/(kr)].
  
  (c)  radiation zone:  E and B are proportional to 1/r.
  E_R = E_R (θ/θ),
  E_R = -[ck⁴p sinθ/(4πε₀ω)] cos(kr - ωt)/(kr) = -[ω²p sinθ/(4πε₀c²)] cos(kr - ωt)/r,
  B_R = B_R (φ/φ),
  B_R = -[k²p sinθ/(4πε₀rc)] cos(kr - ωt) =  -[ω²p sinθ/(4πε₀c³)] cos(kr - ωt)/r.
  B_R(r,t) = r×E_R(r,t)/(rc).

Problem 4

(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)(B²/2μ₀ + ε₀E²/2).
  (∂/∂t)(B²/2μ₀ + ε₀E²/2) + (1/μ₀)∇∙(E×B) = -E∙j.
  Interpretation:
  Let u = B²/2μ₀ + ε₀E²/2 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 = ∮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 = 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 = I²R_{unit length}.
  R_{unit length} = 1/(σπa²).
  Therefore thermal energy dissipated per unit length = field energy flowing into the wire per unit length.]