Question

Define self-inductance of a coil. Show that magnetic energy required to build up the current $I$ in a coil of self inductance $L$ is given by $\frac{1}{2}L{I}^2$.

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 potential difference across an inductor with rate of change of current $\frac{dI}{dt}$ across it is

$V=L\frac{dI}{dt}$

Hence work done to build up current $I=\int Vdq={\int }_0^I\left(L\frac{dI}{dt}\right)\left(Idt\right)$

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