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Ampere's law elegantly handles cylindrical conductor problems. For a solid cylinder of radius $R$ carrying uniform current $I$: inside ($r < R$), $B = \mu_0 Ir/(2\pi R^2)$ — linearly increasing because the enclosed current grows as $r^2$ while the circumference grows as $r$. Outside ($r > R$), $B = \mu_0 I/(2\pi r)$ — identical to a wire. Maximum field occurs at the surface: $B_{\max} = \mu_0 I/(2\pi R)$.

For a hollow cylinder (inner radius $a$, outer radius $b$): in the hollow region ($r < a$), $B = 0$ (no enclosed current). Between walls ($a < r < b$): $B = \frac{\mu_0 I(r^2-a^2)}{2\pi r(b^2-a^2)}$ — the enclosed current fraction is $(r^2-a^2)/(b^2-a^2)$. Outside ($r > b$): same as a thin wire, $B = \mu_0 I/(2\pi r)$.

The $B$ vs. $r$ graph is a signature JEE question: linear rise inside solid conductors, $1/r$decay outside, zero inside hollow regions. For a coaxial cable (inner conductor + outer sheath carrying return current):$B = 0$outside both conductors, non-zero only between them. This principle underlies electromagnetic shielding in coaxial cables.$