The Field Inside and Outside a Thick Wire

Ampere's law says the loop integral of B around a closed path equals mu0 times whatever current pokes through that loop. Outside a thick wire, any Amperian loop you draw encloses all of the current, so the wire behaves exactly like an infinitely thin one: B = mu0 I/(2 π r), shrinking as you move away. Inside the wire, a loop of radius r only catches the current flowing through the smaller circle of radius r, not the whole wire. Since the current is spread evenly over the cross section, the enclosed fraction is just the ratio of areas, (r/R)², where R is the wire's radius.

That area fraction is the whole story. Enclosed current is I_enc = I (r/R)², and plugging that into Ampere's law gives B = mu0 I r/(2 π R²) for r < R. Notice this is linear in r: double the distance from the axis, double the field, as long as you're still inside the wire. At the surface, r = R, both formulas agree and give B = mu0 I/(2 π R), so the field peaks right at the boundary and falls off on either side.

This is the same Amperian-loop logic used to find E inside a uniformly charged sphere, just swapped from Gauss's law to Ampere's law and from charge density to current density. It also sets up the hollow-shell case: if all the current sits in a thin outer shell, a loop drawn inside that shell encloses zero current, so B = 0 everywhere inside, no matter how close to the shell wall you get.