Home exercise 4.1.5: Non-uniform current in a wire

With $dS=\rho d\rho d\varphi a_z$,

As in Problems 4.1.4–4.1.6, $H=H_{\varphi }(\rho )a_{\varphi }$ and $2\pi \rho H_{\varphi }=I_{\mathrm{enc}}(\rho )$ on the circle of radius $\rho$.

The enclosed current is the integral up to $\rho$ (primed variable for the integration):

$I_{\mathrm{enc}}=I$, so $H_{\varphi }=I/2\pi \rho =J_0{b}^2/3\rho$.

Both branches give $J_{0}b/3=I/2\pi b$ at $\rho =b$: continuous, no surface current. Inside, $H_{\varphi }$ now grows as ${\rho }^2$ instead of $\rho$ (Problem 4.1.5), because the current is concentrated towards the surface (Figure 4.1.29). Units: $J_0{\rho }^2/b$ is (A/m${}^2$)(m) $=$ A/m. Numbers: $b=5$ mm and $J_0=10^7$ A/m${}^2$ give $I=2\pi \times 10^7\times (0.005{)}^2/3=524$ A and $H_{\max }=J_{0}b/3=16.7$ kA/m ($B=20.9$ mT) at the surface, and at $\rho =b/2$, $H=4.17$ kA/m, a quarter of the maximum. A current density that increases towards the surface is a crude model of the skin effect, and the limiting case of a surface current gives $H=0$ inside and $I/2\pi \rho$ outside (Problem 4.1.6 with $a\to b$).