Question

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

Answer 1

In expression for the self-inductance of a long solenoid.
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 per turn $=B$ $\times$ area of each turn
But, $B=\frac{{\mu }_{0}NI}{l}$
Magnetic flux per turn $=\frac{{\mu }_{0}NIA}{l}$
Hence, the total magnetic flux $\left(\varphi \right)$ linked with the solenoid is given by the product of flux through each turn and the total number of turns.
$\varphi =\frac{{\mu }_{0}NIA}{l}\times N$
i.e. $\varphi =\frac{{\mu }_0{N}^{2}IA}{l}....\left(1\right)$
If $L$ is the coefficient of self induction of the solenoid, then
$\varphi =LI....\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$LI=\frac{{\mu }_0{N}^{2}IA}{l}$
$L=\frac{{\mu }_0{N}^{2}A}{l}$
If the core is filled with a magnetic material of permeability $\mu$,
Then, $L=\frac{\mu N^{2}A}{l}$