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**Hint:**Biot- Savart law helps to find the magnetic field inside the solenoid. Each turn $n$ is the solenoid has the length$l$. Then applying the Biot- Savart law we get the magnetic field inside the solenoid as

$B = \dfrac{{{\mu _0}nI}}{l}$

Where, ${\mu _0}$ is the permeability of free space, $n$is the number of turns and $I$is the current in the solenoid and $l$ is the length.

And the magnetic flux is proportional to the current through the solenoid. Thus the proportionality constant is the coefficient of self- inductance of the solenoid.

**Complete Step by step solution**

We are considering a solenoid with $n$ turns with length $l$ . The area of cross section is $A$. The solenoid carriers current $I$ and $B$ is the magnetic field inside the solenoid.

The magnetic field $B$ is given as,

$B = \dfrac{{{\mu _0}nI}}{l}$

Where, ${\mu _0}$ is the permeability of free space, $n$ is the number of turns and $I$ is the current in the solenoid and $l$ is the length.

The magnetic flux is the product of the magnetic field and area of the cross section.

Here the magnetic flux per turn is given as,

$\phi = B \times A$

Substituting the values in the above expression,

$\phi = \dfrac{{{\mu _0}nI}}{l} \times A$

Hence there is $n$ number of turns, the total magnetic flux is given as,

\[

\phi = \dfrac{{{\mu _0}nI}}{l} \times A \times n \\

\phi = \dfrac{{{\mu _0}{n^2}IA}}{l}..............\left( 1 \right) \\

\]

If $L$ is the coefficient of self-inductance of the solenoid, then

$\phi = LI...........\left( 2 \right)$

Comparing the two equations we get,

\[

LI = \dfrac{{{\mu _0}{n^2}IA}}{l} \\

L = \dfrac{{{\mu _0}{n^2}A}}{l} \\

\]

**So the expression for the coefficient of self-inductance is \[\dfrac{{{\mu _0}{n^2}A}}{l}\].**

**Note**The flux through one solenoid coil is $\phi = B \times A$ where the area of the cross section of each turn is $A$.

For the $n$ number of turns then the total magnetic flux is $\phi = n \times B \times A$. The magnetic field is uniform inside the solution.

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