Classical Electrodynamics - Jackson Problem 5.18.pdf

PHY 5346 Homework Set 9 Solutions – Kimel 2. 5.18 a) From the results of Problem 5.17, we can replace the problem stated by the system where I∗ is equidistant from the interface and is equal to I∗ = μr−1 μr+1 I. The radius of each current loop is a. Now from Eq. (5.7) F⃗on I = I ∫ dl⃗ × Br⃗ dl⃗ × B⃗ = dl⃗ × B⃗r + dl⃗ × B⃗θ = dlBr −θ̂ + dlBθr̂ By symmetry, only the z − component survives, so, from the figure dl⃗ × B⃗ ⋅ ẑ = dlBr a 4d2 + a2 + dlBθ 2d 4d2 + a2 So Fz = 2πaI 4d2 + a2 aBr + 2dBθ with Br and Bθ given by Eqs. (5.48) and (5.49) and cosθ = 2d 4d2+a2 , r = 4d2 + a2 , and I → I∗. c) To determine the limiting term, simply let r → 2d and take the lowest non-vanishing term in the expansion of the magnetic flux density. Fz = πaI d aBr + 2dBθ Fz = πaI d a μ0I∗a 4d a 2d2 + 2d − μ0I ∗a2 4 1 2d3 − a 2d Fz → − 3πμ0 32 a4I × I∗ d4 The minus sign shows the force is attractive if I and I∗ are in the same direction. This same result can be gotten more directly, using Fz = ∇zmBz  with m = πa2I, and (from Eq. (5.64)) Bz = μ0 4π 2m∗ z3 with m∗ = πa2I∗, and z = 2d Fz = μ0 4π 2πa2I∗πa2I − 3 2d4 = − 3πμ0 32 a4I × I∗ d4 with agrees with out previous result.