Question

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

Answer 1

Self Inductance of a coil is the magnetic flux linked with the coil when the current through coil is $IA$

$\varphi =li$ where $L$ is the constant of proportionality .

Let a source of emf be connected to an inductor $L$

With increase in current ,the opposite emf ,$e=-L\frac{di}{dt}$

Work done $dw=\left|e\right|idt=Li\frac{di}{dt}dt=Lidi$

$W={\int }_0^{i}Lidi=\frac{1}{2}L{i}^2$

Energy stored in magnetic inductor,$U=\frac{1}{2}L{i}^2$