Question

Define the term self-inductance of a solenoid. Obtain the expression for the magnetic energy stored in an inductor of self-inductance I, to build up a current I through it.

Answer 1

Self inductance is defined as the induction of a voltage in a current-carrying wire when the current in the wire itself is changing. The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by
$L=\frac{\varphi }{I}$


Energy stored in an inductor : When a current grows through an inductor, a back e.m.f, is set up which opposes source e.m.f. in building up the current. This work done is stored as magnetic potential energy.

Let I be the current through the inductor I. at any instant t.

The current rises at the rate dI/dt.

So the induced e.m.f. is

$e=\frac{-LdI}{dt}$

The work done against induced e.m.f. in time dt is

$dW=Pdt$

= $-eIdt$

=$\frac{LdI}{dt}Idt$

= $LidI$

For total work from $0$ to$I_0$ current

$W=\int dW$

= ${\int }_0^{I_0}LIdI$

= $L{\left[\frac{I_2}{2}\right]}_0^{I_0}$

= $\frac{1}{2}L{I}_0^2$

Hence this work done is stored as the magnetic potential energy U in the inductror

$U=\frac{1}{2}L{I}^2$