Question

A thin long cylinder of radius $R$ is surrounded by a co-axial hollow cylinder of radius $4R$. Both cylinders form a complete circuit for current $I$. Find self inductance per unit length for the arrangement.

Answer 1

The self inductance per unit length for the arrangement is $\frac{\mu_0}{\pi} \ln(2)$.

Answer 2

The magnetic field in the region $R < r < 4R$ is given by Ampere's Law: $B = \frac{\mu_0 I}{2\pi r}$ The magnetic flux per unit length through this region is: $\frac{\Phi}{l} = \int_R^{4R} B dr = \int_R^{4R} \frac{\mu_0 I}{2\pi r} dr = \frac{\mu_0 I}{2\pi} [\ln r]_R^{4R} = \frac{\mu_0 I}{2\pi} (\ln(4R) - \ln(R)) = \frac{\mu_0 I}{2\pi} \ln\left(\frac{4R}{R}\right) = \frac{\mu_0 I}{2\pi} \ln(4)$ The self-inductance per unit length is $L/l = \frac{\Phi/l}{I}$: $\frac{L}{l} = \frac{\frac{\mu_0 I}{2\pi} \ln(4)}{I} = \frac{\mu_0}{2\pi} \ln(4) = \frac{\mu_0}{2\pi} \ln(2^2) = \frac{\mu_0}{2\pi} (2 \ln 2) = \frac{\mu_0}{\pi} \ln(2)$