Question

A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.

Answer 1

Answer:

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$

Explanation:

As we know that magnetic field due to torroid is given as

$B = \frac{\mu_0 N i}{2\pi b}$

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

$\phi = B.A$

$\phi = \frac{\mu_0 N i}{2\pi b}(a^2)$

now total flux of N number of coils is given as

$\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}$

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

$L = \frac{\phi}{i}$

$L = \frac{\mu_0 N^2 (a^2)}{2\pi b}$