Obtain an expression for the self-inductance of a long solenoid.

Hint: A solenoid is a long coil of iron on which there are a large number of turns of insulated wire and which results in the formation of magnetic field when current is passed through it.Self-inductance is the property of the current-carrying coil that resists or opposes the change of current flowing through it. This occurs mainly due to the self-induced emf produced in the coil itself.

Complete step by step answer:
In expression for the self-inductance of a long solenoid.
We Consider a solenoid of N turns with length l and area of cross section A. It carries a current I. If B is the magnetic field at any point inside the solenoid, then
Magnetic flux produced can be given as ${{\phi }_{B}}=\overset{\to }{\mathop{B}}\,.\overset{\to }{\mathop{A}}\,=BA\cos \theta $
Here the angle between the magnetic field vector and area vector is ${{0}^{\circ }}$.
So ${{\phi }_{B}}=BA$
Also we have calculated the magnetic field due to solenoid and it is given by,
$B={{\mu }_{0}}nI$.
Here n is the number of turns per unit length.
Magnetic flux per turn $B={{\mu }_{0}}nIA$
Hence, the total magnetic flux $\phi $ linked with the solenoid is given by the product of flux through each turn and the total number of turns.
$\phi ={{\mu }_{0}}nIA\times N$
But $N=nl$
Therefore, $\phi ={{\mu }_{0}}{{n}^{2}}IAl......(1)$
If L is the coefficient of self-induction of the solenoid, then
$\phi =LI......(2)$
From equation (1) and (2),
$L={{\mu }_{0}}{{n}^{2}}Al$

Note: While coming at the expression we have used the assumption that the
solenoid is a long solenoid and the magnetic field lines are continuous straight within the
solenoid. A long solenoid is one whose length is very large as compared to its radius of cross section. The magnetic field inside it is constant at any point.