Question

Prove that the inductance of parallel wires of length l in the same circuit is given by \[L=\left(\frac{\mu_{0}l}{\pi}\right)\ln(d/a)\], where [a] is the radius of a wire and [d] is the separation between the wire axes.

Answer 1

Consider two long parallel wires carrying equal currents in opposite directions, forming a circuit. The magnetic fields between the wires reinforce each other. At a distance [r] from either wire, the field due to that wire is \[B(r)=\frac{\mu_{0}I}{2\pi r}\].

Neglecting end effects and taking \[d\gg a\], the flux contribution from one wire across the region between the conductors is approximately \[\Phi_{1}=\int_{a}^{d}B(r),l,dr=\frac{\mu_{0}Il}{2\pi}\ln \left(\frac{d}{a}\right)\]. The other wire contributes an equal amount, so the total flux linked with the circuit is \[\Phi=2\Phi_{1}=\frac{\mu_{0}Il}{\pi}\ln \left(\frac{d}{a}\right)\].

By definition, inductance is flux linkage per current. Therefore, \[L=\frac{\Phi}{I}=\left(\frac{\mu_{0}l}{\pi}\right)\ln \left(\frac{d}{a}\right)\]. Hence proved. This is the usual approximation for long wires with \[d\gg a\].