A free current I flows uniformly in the z-direction down a long, straight wire of radius a. Assume that the current is distributed uniformly, such that J = I πa2ˆz, and that the wire is made of a linear isotropic magnetic material with µr = (1+χm), we wish to find the magnetic field everywhere (inside and outside the wire).
(a) Generally, B(x,y,z) = Bs(s,φ,z)ˆs + Bφ(s,φ,z)ˆ φ + Bz(s,φ,z)ˆz. Which direction does B point, and which coordinates can it depend on? Use this knowledge write a simpler expression for B. Do the same for M, and H.
(b) Use the modified Ampere’s law (with H and only the free currents) to calculate H everywhere.
(c) Use your answer from (b) and the linear isotropic assumption to find B and M everywhere.
(d) Calculate the bound currents, Jb and Kb.
(e) Now, recalculate B everywhere using ∇×B = µ0J, where J includes both bound and free currents.
(f) Show that the net bound current flowing down the wire is zero, and explain physically why this must be the case.
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