Calculate the self-inductance for very long solenoid.

Explanation

Let us calculate the self-inductance of a long solenoid of cross-sectional area A and length l, having n turns per unit length.
The magnetic field due to a current I flowing in the solenoid is `B=mu_0nI` (neglecting edge effects, as before)
The total flux linked with the solenoid is `Nphi_B=NBA`
`Nphi_B=(nl)(mu_0nI)(A) " "(because n=N/l)`
`=mu_0n^2 AlI`
where nl is the total number of turns. Thus, the self - inductance is,
`L=(Nphi_B)/I`
`L=mu_0n^2Al`...(1)
If we fill the inside of the solenoid with a material of relative permeability u, (for example soft iron, which has a high value of relative permeability), then,
`L=mu_r mu_0 n^2 Al`...(2) `[because mu_1=mu/mu_0]`
The self-inductance of the coil depends on its geometry and on the permeability of the medium.