An inductor may store energy in

Solution :
When the magnetic flux linked with a coil changes, an induced emf acts in the coil which is given by $$e = - \frac{{d\phi }}{{dt}}$$
The magnetic flux linked with a coil carrying a current $$i,$$ is proportional to $$i.$$
$$\eqalign{ & {\text{or}}\,\,\phi \propto i\,\,{\text{or}}\,\,\phi = Li \cr & \therefore e = - \frac{{d\phi }}{{dt}} = - L\frac{{di}}{{dt}} \cr} $$
The work done in maintaining the current for time $$dt$$
$$ = - ei\,dt = L\frac{{di}}{{dt}}i\,dt$$
and the total work done while the current $${i_0}$$ is being established
$$\eqalign{ & W = \int_0^t {L\frac{{di}}{{dt}}} i\,dt = \int_0^{{i_0}} {Li} \,dt \cr & = \frac{1}{2}Li_0^2 \cr} $$
Thus, an inductor may store energy in its magnetic field.
NOTE
The expression $$\left( {W = \frac{1}{2}Li_0^2} \right)$$   reminds us of $$\frac{1}{2}m{v^2}$$  for mechanical kinetic energy of a particle of mass $$m,$$ and shows that $$L$$ is analogus to $$m$$ (i.e. $$L$$ is electrical inertia, which opposes the growth and decay of current in circuit).