Home exercise 4.2.1: Coaxial cable

By symmetry $B=B_{\varphi }(\rho )a_{\varphi }$. Ampère's law on a circle of radius $\rho$, $a<\rho <b$, encloses only the inner current:

Outside the cable ($\rho >b$) the enclosed current is $I-I=0$ and $B=0$.

Take the strip of unit length (along $z$) stretching radially from $\rho =a$ to $\rho =b$ (Figure 4.2.21). $B$ is normal to it, so

This flux links the circuit formed by the inner and the outer conductor once ($N=1$).

Units: ${\mu }_0$ is in H/m and the logarithm is dimensionless. From the energy:

the same. For $b/a=3.6$ (a common RG-type cable) $L/l=2\times 10^{-7}\ln 3.6=256nH/m$. With $C/l=2\pi \epsilon /\ln (b/a)$, Eq. (3.4.29), the product $LC=\mu \epsilon$ does not depend on the radii at all: this is the signal speed $1/\sqrt{\mu \epsilon }$ of Chapter 5. The current inside the inner conductor adds the small internal inductance ${\mu }_0/(8\pi )=50nH/m$ (Exercise 4.2.6).