Problem 1:

Consider the magnetic vector potentials A₁ = (-By, 0, 0) and A₂ = -½(By, -Bx, 0).
(a)  Show that both vector potentials are associated with the same magnetic field B.
(b)  Construct a gauge transformation ψ(r) which connects the two representations of the vector potential.

Solution:

• Concepts:
  ∇∙B = 0 --> B = ∇×A.
• Reasoning:
  A is not unique.  A' = A + ∇ψ + C, with ψ an arbitrary scalar field and C an arbitrary constant vector is also a vector potential for the same field.  In magnetostatics we choose ∇∙A = 0.
• Details of the calculation:
  (a)  ∇×A = (∂A_z/∂y - ∂A_y/∂z,  ∂A_x/∂z  -  ∂A_z/∂x,  ∂A_y/∂x -  ∂A_x/∂y).
  ∇×A₁ = -(∂A_x/∂y) k = B k.
  ∇×A₂ = (∂A_y/∂x -  ∂A_x/∂y) k = B k.
  (b)  A₂ - A₁ = ∇ψ = ½(By, Bx, 0),  ψ = ½Byx.

Problem 2:

The Aharanov-Bohm experiment is illustrated in the figure below.

It is a two-slit electron scattering experiment where a solenoid is placed in the region behind the screen and between the two classical paths that electrons passing through the slits would follow to reach a point on the screen.  The long and thin solenoid confines the magnetic field to regions that the electrons should not pass through.  In terms of the cylindrical coordinates in Fig (b), the magnetic field may be assumed given by

inside solenoid:              B_r = 0,  B_φ = 0,  B_z = B.
outside solenoid:            B_r = 0,  B_φ = 0,  B_z = 0.

(a)  Show that a vector potential given in the cylindrical coordinates by
inside the solenoid:        A_r = A_z = 0,   A_φ = Br/2.
outside the solenoid:     A_r = A_z = 0,  A = BR²/(2r),
leads to the magnetic field components inside and outside the solenoid given above.

Thus, in the Aharanov-Bohm experiment the electrons never experience a finite magnetic field but they may encounter a non-zero vector potential outside the solenoid.

(b)  What does this result, and that in the Aharanov-Bohm experiment the interference pattern is observed to be shifted when current is flowing in the solenoid, say about the relative importance of the magnetic field and the vector potential in classical and quantum mechanics?

Solution:

• Concepts:
  ∇∙B = 0 --> B = ∇×A.
• Reasoning:
  We can evaluate ∇×A in Cartesian or cylindrical coordinates.
• Details of the calculation:
  (a)  In Cartesian coordinates:
  A_x = -A_φsinφ,  A_y = A_φcosφ,  A_z = 0.
  B = ∇×A = (∂A_z/∂y - ∂A_y/∂z,  ∂A_x/∂z  -  ∂A_z/∂x,  ∂A_y/∂x -  ∂A_x/∂y).
  
  Inside:  A_x = -½Br(y/r) = -½By,  A_y = ½Bx,  A_z = 0.
  B = (0, 0, B_z),  B_z = ∂A_y/∂x - ∂A_x/∂y = ½B + ½B = B.
  Outside: A_x = -½BR²y/(x² + y²),  A_y = ½BR²x/(x² + y²),    A_z = 0.
  B = (0, 0, B_z),  B_z = ∂A_y/∂x -  ∂A_x/∂y = ½BR²(1/r² - 2x²/r⁴ + 1/r² - 2y²/r⁴)
  = ½BR²(2/r² - 2/r²) = 0.
  [In cylindrical coordinates we use
  B_z = (∂(rA_φ)/∂r - ∂A_r/∂φ)/r,  B_r = (∂A_z/∂φ - r∂A_φ/∂z)/r,  B_φ = ∂A_r/∂z - ∂A_z/∂r,
  to obtain the same result.]
  
  (b) In classical mechanics electrons would be acted on by the Lorentz force, which requires a finite magnetic field, and the vector potential has no effect except through the magnetic field.  But in quantum mechanics the Aharonov-Bohm experiment shows that the vector potential can cause physical effects even if the corresponding magnetic field vanishes.

Problem 3:

The concentric cylindrical shells of a cylindrical capacitor have radii a and b > a,  respectively, and height h >> b.  The capacitor charge is Q, with +Q on the inner shell of radius a, and -Q on the outer shell of radius b (see figure).  The whole capacitor rotates about its axis with angular velocity ω = 2π/T.  Neglect edge effects.

(a)  Find the capacitance of the capacitor. 
(b)  Evaluate the magnetic field B generated by the rotating capacitor over all space.
(c)  Find the direction and magnitude of the Poynting vector.

Solution:

• Concepts:
  Electrostatics, magnetostatics, and the Poynting vector
• Reasoning:
  The Poynting vector represents the energy flux in the electromagnetic field.  The energy can circulate and not flow into or out of an object.
• Details of the calculation:
  Use cylindrical coordinates (r, φ, z), with the z-axis coinciding with capacitor axis.
  (a)  From Gauss' law: E(r) = e_r Q/(2πε₀hr) between the cylinders.
  V_ab = ΔV = -∫_b^aE(r)dr = [Q/(2πε₀h)]∫_a^b(1/r)dr = [Q/(2πε₀h)](lnb - lna) = [Q/(2πε₀h)]ln(b/a).
  C = Q/ΔV = 2πε₀h/ln(b/a).
  
  (b)  Assume ω = ωe_z, with ω = 2π/T  >  0.  The surface current density of the inner cylinder due to the capacitor rotation is K = σv e_φ with K = [Q/(2πah)]ωa = Q/(hT).
  The surface current density of the inner cylinder due to the capacitor rotation is -K with
  K = [Q/(2πbh)]ωb = Q/(hT).
  The two cylindrical shells are equivalent to two solenoids with current densities K = nI, where n is the number of turns per unit length. 
  The magnetic field produced by a long solenoid has magnitude B = μ₀nI, with the direction given by the right-hand rule, inside, B = 0 outside.
  Using the principle of superposition we find
  B = 0, r < a and r > b,
  B = -μ₀K e_z = -μ₀ Q/(hT) e_z =   a < r < b.
  
  (c)  The Poynting vector S = (1/μ₀)(E×B) is the energy flux in the electromagnetic field.
  Here S = -Q²/(2πε₀h²rT)(e_r × e_z) = Q²/(2πε₀h²rT) e_φ.
  Field energy circulates about the z-axis in the e_φ direction between the cylindrical shells.

Problem 4:

Assume magnetic charges exist and Maxwell's equations are of the form
∇∙E = ρ/ε_0,   ∇∙B = μ₀ρ_m,   -∇×E = μ₀j_m + ∂B/∂t,   ∇×B = μ₀j + μ₀ε₀∂E/∂t.
Assume a magnetic monopole of magnetic charge q_m is located at the origin, and an electric charge q_e is placed on the z-axis at a distance R from it.
(a)  Write down expressions for the electric field E(r) and the magnetic field B(r).  Make a sketch.
(b)  Write down expressions for the momentum density g(r) and angular momentum density ℒ(r) of the electromagnetic field.
(c)  Show that there is electromagnetic angular momentum L_z about the z-axis and derive an expression for it.  Show that L_z is independent of R.
Useful vector identity:  (a∙∇)n = (1/r)[a - n(a∙n)].   Here n is r/r is the unit radial vector.

Solution:

• Concepts:
  Field energy and field momentum of the electromagnetic field
• Reasoning:
  (1/μ₀)(E×B) = energy flux, (1/(cμ₀))(E×B) = momentum flux,
  (1/(c²μ₀))(E×B) = ε₀(E×B) = momentum density.
  The angular momentum density of the field about the origin then is r × ε₀(E×B).
• Details of the calculation:
  (a)  Field of the point charge:  E(r) = [1/(4πε₀)]q_e(r - Rk)/|r - Rk|³.
  Field of the magnetic monopole:  B(r) = [μ₀/(4π)]q_mr/r³).
  
  (b)  S(r) = (1/μ₀)(E(r)×B(r)) = energy flux.
  S(r)/c = momentum flux.
  g(r) = S(r)/c² = ε₀E(r)×B(r)) = momentum density.
  g(r) = [μ₀ε₀q_m/(4πr³)] E(r)×r.
  g(r) = -(φ/φ)[μ₀/(4π)²]q_eq_mRsinθ/(|r - Rk|³r²).
  ℒ(r) = r×g(r) = angular momentum density of the field about the origin.
  ℒ(r) = (θ/θ) [μ₀/(4π)²]q_eq_mRsinθ/(|r - Rk|³r).
  = (θ/θ) [μ₀/(4π)²]q_eq_m(R/r)sinθ/(r² + R² - 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 = ∫_VdV ℒ(r) = -k [μ₀/(4π)²]q_eq_mR2π ∫₀^πsin³θdθ∫₀^∞rdr/(r² + R² - 2rRcosθ)^3/2.
  Symmetry dictates that the x- and y-components of L integrate to zero.
  ∫₀^∞xdx/(a - bx + cx²)^3/2 = (4a - 2bx)/[(b² - 4ac)(a - bx + cx²)^½] ₀^∞.
  As x --> ∞, (4a - 2bx)/[(b² - 4ac)(a - bx + cx²)^½]  --> -2b/((b² - 4ac)√c).
  ∫₀^∞xdx/(a + bx + cx²)^3/2 = (-2b√a - 4a√c )/((b² - 4ac)√(ac)).
  ∫₀^∞rdr/(r² + R² - 2rRcosθ)^3/2 = (-4R²cosθ - 4R²)/(4R³cos²θ - 4R³)
  = R^-1(1 + cosθ)/(1- cos²θ) = R^-1/(1- cosθ).
  L = -k [μ₀/(4π)²]q_eq_m2π ∫₀^πsin³θdθ/(1 - cosθ)
  = -k [μ₀/(4π)²]q_eq_m2π ∫_-1¹(1 - x²)dx/(1 - x)
  = -k [μ₀/(4π)²]q_eq_m2π ∫_-1¹(1 + x)dx = -k [μ₀/(4π)]q_eq_m, independent of R.

  Using vector identities:
  ℒ(r) = [μ₀ε₀q_m/(4πr³)] r × E(r) × r) = -[μ₀ε₀q_m/(4πr³)] r × r × E(r)).
  r×(r×E) = r(E∙r) - Er²  (BAC - CAB rule)
  =  -r³(E∙∇)r/r,  since (a∙∇)n = (1/r)[a - n(a∙n)].
  ℒ(r) = [μ₀ε₀q_m/(4π)] (E∙∇)r/r.
  Angular momentum: L = [μ₀ε₀q_m/(4π)] ∫_VdV (E∙∇)r/r.
  ∫_-∞^∞E_x ∂/∂x (r/r) dx = E_x (r/r)|_-∞^∞ - ∫_-∞^∞ (r/r)∂E_x/∂x dx
  = -∫_-∞^∞ (r/r)∂E_x/∂x dx  for a finite charge distribution.
  Therefore L = -[μ₀ε₀/(4π)]q_m∫_V(r/r)∇∙EdV.
  ∇∙E = (q_e/ε₀)δ(r - Rk).
  L = -[μ₀/(4π)]q_mq_e∫_V(r/r)δ(r-Rk)dV = -[μ₀/(4π)]q_mq_e k.