The ratio of magnetic flux through the solenoid to the current passing through it is called selfinductance of a solenoid. It is given by $L=\; \frac{\phi}{I}$ Energy stored in an inductor: When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy. Let I be the current through the inductor $L$ at any instant $t$. The current rises at the rate $\; \frac{dI}{dt}$. So the induced e.m.f. is $e=\; \frac{-L dI}{dt}$ The work done against induced e.m.f. in $d t$ is $dW\;=P dt\;$ $\;=-e I dt \;\; [P=V I]$ $\;=\; \frac{L dI}{dt} I dt$ $=LIdI$ For total work from 0 to $I_0$ current $=\; \frac{1}{2} L I_0^2$Hence, this work done is stored as the magnetic potential energy $U$ in the inductor$U=\; \frac{1}{2} L I^2$