Q 12-06-080JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
A coil has an inductance of $2\ \text{H}$ and resistance of $4\ \Omega$. A $10\ \text{V}$ is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be ______ $\times10^{-2}$ J.
Numerical value type. Enter your answer.
Answer: 625
$I=\dfrac{10}{4}=2.5\ \text{A}$; $U=\tfrac12LI^2=\tfrac12\times2\times6.25=6.25\ \text{J}=625\times10^{-2}\ \text{J}$.
Solution by Sreeraj P, M.Sc Physics