Question

Derive expression for self inductance of a long air-cored solenoid of length L, cross- sectional area A and having number of turns N.

Answer 1

Consider a current(I) is passed through the inductor (L).

A: Cross-sectional area

N : Number of turns

L : Length

Magnetic field generated at inside the solenoid is ,

$B=\frac{{\mu }_{o}NI}{L}$

So, the flux through the coil is a obtained:

${\varphi }_B=N(\overrightarrow{B}.\overrightarrow{A})$ ($\overrightarrow{B}$ and $\overrightarrow{A}$ are along same direction)

${\varphi }_B=\frac{{\mu }_o{N}^{2}IA}{L}$ ...(1)

Now, flux $\left({\varphi }_B\right)$ is related to inductance (L) as

${\varphi }_B=LI$

$\Rightarrow L=\frac{{\varphi }_B}{I}$

From (1) we get,

$\Rightarrow L=\frac{{\mu }_0{N}^{2}A}{L}$