Calculate the self-inductance of a very long solenoid.

(N/A) 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 given by:
$B = \mu_{0} n I$ (neglecting edge effects).
The total flux linked with the solenoid is:
$N \phi_{B} = N B A$
Substituting $N = nl$ and $B = \mu_{0} n I$:
$N \phi_{B} = (nl)(\mu_{0} n I)(A) = \mu_{0} n^{2} A l I$
Since the self-inductance $L$ is defined as $L = \frac{N \phi_{B}}{I}$,we get:
$L = \mu_{0} n^{2} A l$ ... $(1)$
If the solenoid is filled with a material of relative permeability $\mu_{r}$,the self-inductance becomes:
$L = \mu_{r} \mu_{0} n^{2} A l$ ... $(2)$
Thus,the self-inductance of the coil depends on its geometry and the permeability of the medium.