The energy stored in a coil of inductance $L$ carrying a steady current $i$ is in the form:

The magnetic energy stored in an inductor of inductance $4\,\mu H$ carrying a current of $2\,A$ is:

Two circuits $A$ and $B$ are connected to identical d.c. sources each of e.m.f. $10 \ V$. Self-inductances of circuits $A$ and $B$ are respectively $L_A = 10 \ H$ and $L_B = 10 \ mH$. The total resistance of each circuit is $40 \ \Omega$. The ratio of energy consumed in circuit $A$ and circuit $B$ to build up the current to steady value is

Define an inductor. Derive the equation for the energy $U = \frac{1}{2}LI^2$ stored in an inductor.

If the current of $I$ gives rise to a magnetic flux $\phi$ through a coil having $N$ turns,then the magnetic energy stored in the medium surrounding the coil is

When a current $I$ flows through a coil,it stores the energy $E_1$ in it. Now the current is reduced to $I/2$,the energy stored in the coil becomes $E_2$. The change in the energy is