Question

Derive the expression for the self-inductance of a long solenoid of cross sectional area A and length l, having n turns per length.

Answer 1

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.