Question

Derive expression for the self inductance of a solenoid. What factors affect it?

Answer 1

Self Inductance of a Current Carrying Solenoid: Consider a solenoid of radius $r$ metre, length $l$ metre, having total number of turns $N$ and carrying a current $I$ ampere. The magnetic field produced on its axis inside the solenoid is

$B=\frac{{\mu }_{0}NI}{l}N{A}^{-1}{m}^{-1}$

If it is assumed that the solenoid is very long, the intensity of magnetic field $B$ can be assumed to be uniform inside it at each point, then the total magnetic flux linked with the solenoid is

$B=\frac{{\mu }_{0}m}{4\pi }\left[\frac{d^2+l^2+2dl-(d^2+l^2-2dl)}{{\left\{(d-l)(d+l)\right\}}^2}\right]$

$\varphi =B\times$ total effective area of solenoid

$=B\times \left(NA\right)$

where $A=\pi r^2$ is the area of cross section of solenoid.

$\therefore \varphi =\frac{{\mu }_{0}NI}{l}\times NA=\frac{{\mu }_0{N}^{2}AI}{l}$ weber

But by definition $\varphi =LI$ where $L$ is the self-inductance of solenoid, therefore

Self inductance $L=\frac{\varphi }{I}=\frac{{\mu }_0{N}^{2}A}{l}$ henry

If inside the solenoid instead of air or vacuum there is a substance of relative permeability

$L=\frac{{\mu }_0{\mu }_r{N}^{2}A}{l}$ henry

Thus, the self inductance of a solenoid depends on the following factors:

(i) On the number of turns in the solenoid: The self inductance increases on increasing the number of turns.

(ii) On the area of cross section of solenoid (i.e., on the radius of solenoid): The self inductance increases on increasing the area of cross section.

(iii) On the length of solenoid: On increasing the length of solenoid, its self inductance decreases.

(iv) On the relative permeability of the substance placed inside the solenoid: If a soft iron rod is placed inside the solenoid, its self inductance increases.

Remember that if the number of turns per unit length in the solenoid is $n$, then $n=N/l$ and the self inductance of solenoid is

$L={\mu }_0{n}^{2}lA$ henry.