Home exercise 4.1.6: Coaxial cable

$J=\frac{I}{\pi a^2}{a}_z$ for $\rho <a$ and $J=-\frac{I}{\pi (c^2-b^2)}{a}_z$ for $b<\rho <c$, zero elsewhere. The geometry is cylindrically symmetric, so $H=H_{\varphi }(\rho )a_{\varphi }$ and $2\pi \rho H_{\varphi }=I_{\mathrm{enc}}$ on every coaxial circle.

$I_{\mathrm{enc}}=I{\rho }^2/a^2$ (Problem 4.1.5): $H_{\varphi }=\frac{I\rho }{2\pi a^2}$.

$I_{\mathrm{enc}}=I$: $H_{\varphi }=\frac{I}{2\pi \rho }$.

The circle encloses $I$ and the part of the return current between $b$ and $\rho$:

$I_{\mathrm{enc}}=I-I=0$: $H_{\varphi }=0$.

Continuity at $a$ ($I/2\pi a$ from both sides), at $b$ ($I/2\pi b$ from both sides) and at $c$ ($0$ from both sides): no surface currents (Figure 4.1.30). Numbers for $a=1$ mm, $b=3.5$ mm, $c=4$ mm, $I=10$ A: $H=796$ A/m at $\rho =0.5$ mm and again at $\rho =2$ mm, $219$ A/m at $\rho =3.75$ mm, zero outside. Beyond the shell $I_{\mathrm{enc}}=0$: the whole field is trapped in the dielectric between $a$ and $b$ (plus the thin shell). This is why a coaxial cable neither radiates its field nor picks up external ones, the magnetic counterpart of the electrostatic shielding of Application 3.2.2.