A long straight solid wire of radius 'a' carries a steady current I uniformly distributed over its c

Question

A long straight solid wire of radius 'a' carries a steady current I uniformly distributed over its cross-section. What is the ratio of the magnetic field at $r = a/2$ to the magnetic field at $r = 2a$?

Professor PC Thomas Classes · Answer

Inside the wire ($r < a$), $B_{in} = \frac{\mu_0 I r}{2\pi a^2} \implies B(a/2) = \frac{\mu_0 I}{4\pi a}$. Outside the wire ($r > a$), $B_{out} = \frac{\mu_0 I}{2\pi r} \implies B(2a) = \frac{\mu_0 I}{4\pi a}$. The ratio is 1:1.