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

`L = (\mu_0 N^2 (a^2))/(2\pi b)`

Step-by-step explanation:

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

`B = (\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 = (\mu_0 N i)/(2\pi b)(a^2)`

now total flux of N number of coils is given as

`\phi_(total) = (\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 = (\phi)/(i)`

`L = (\mu_0 N^2 (a^2))/(2\pi b)`