Consider a long solenoid of length $l$ and radius $r$. If $n$ is the number of turns per unit length and $\mu_0$ is the permeability of free space, the inductance of the solenoid is :

The magnetic field inside a long solenoid is given by $B = \mu_0 n I$, where $n$ is the number of turns per unit length and $I$ is the current.

The magnetic flux through a single turn of the solenoid is $\phi = B A = (\mu_0 n I)(\pi r^2)$.

The total number of turns in the solenoid of length $l$ is $N = n l$.

The total magnetic flux linkage is $\Phi = N \phi = (n l)(\mu_0 n I \pi r^2) = \mu_0 \pi n^2 r^2 l I$.

The self-inductance $L$ is defined by the relation $\Phi = L I$.

Substituting the value of $\Phi$, we get $L = \mu_0 \pi n^2 r^2 l$.

Answer: $\mu_0\pi n^2r^2l$