Question

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

Answer 1

Total number of turns in solenoid $N=nl$

Magnetic field inside the long solenoid $B={\mu }_{o}ni$

Flux through one turn ${\varphi }_1=BA={\mu }_{o}niA$

Thus total flux through N turns ${\varphi }_t=N{\varphi }_1=nl\times {\mu }_{o}niA={\mu }_o{n}^{2}lAi$

Using ${\varphi }_t=Li$ where $L$ is the self inductance of the coil

$\therefore$ ${\mu }_o{n}^{2}lAi=Li$ $⟹L={\mu }_o{n}^{2}Al$