Question

Define self-inductance of a coil. Derive the expression for magnetic energy stored in the inductor of inductance $L$, carrying current $l$.

Answer 1

Self-inductance of a coil is defined as the ratio of the total flux linked with the coil to the current flowing through it.

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

When the current is varied, the flux linked with the coil changes and an $e.m.f.$ is induced in the coil. It is given as

$\epsilon =-\frac{d\left(N{\varphi }_B\right)}{dt}=-L\frac{dI}{dt}$

The self-induced $e.m.f.$ is also called back $e.m.f.$ as it opposes any change in current in the circuit.

So, work needs to be done against back $e.m.f.$ in establishing current. This work done is stored as magnetic potential energy. The rate of doing work is given as

$\frac{dW}{dt}=\left|\epsilon \right|I=LI\frac{dI}{dt}$(neglecting negative sign)

Thus, the total work done in establishing current from $0$ to $I$ is

$w=\int dW=\underset{0}{\overset{I}{\int }}LIdI=\frac{1}{2}L{I}^2$