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

The magnetic potential energy stored in a certain inductor is $64 \times 10^{-3} \ J$,when the current in the inductor is $80 \ mA$. The inductance of this inductor is: (in $H$)

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

Energy stored in a coil of self-inductance $40 \, mH$ carrying a steady current of $2 \, A$ is.....$J$

Assertion : If the current in a solenoid is reversed in direction while keeping the same magnitude,the magnetic field energy stored in the solenoid remains unchanged.
Reason : Magnetic field energy density is proportional to the magnetic field.

What is the energy stored in the inductor of the given circuit?