If the current flowing through a coil is reduced by $50 \%$,then the energy stored in the coil will

Energy required to establish a current of $4 \ A$ in a coil of self-inductance $L = 200 \ mH$ is . . . . . . . (in $J$)

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

$(a)$ Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field $B$,area $A$,and length $l$ of the solenoid.
$(b)$ How does this magnetic energy compare with the electrostatic energy stored in a capacitor?

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

$A$ coil of inductance $2\, H$ having negligible resistance is connected to a source of supply whose voltage is given by $V = 3t$ volt (where $t$ is in seconds). If the voltage is applied when $t = 0$,then the energy stored in the coil after $4\, s$ is ............ $J$.