In the figure, the magnetic energy stored in the coil is:

If the current in an $LR$ circuit is reduced to half,what will be the energy stored in it?

$A$ coil has an inductance of $2 \, H$ and resistance of $4 \, \Omega$. $A$ $10 \, V$ potential difference is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be $.......... \times 10^{-2} \, J$.

The expression for the magnetic energy stored in a solenoid of length $L$, in terms of magnetic field $B$ and area $A$ 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 $H$)

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.