A coil carrying a steady current of $4 \mathrm{~A}$ stores $0.2 \mathrm{~J}$ of energy in its magnet

Question

A coil carrying a steady current of $4 \mathrm{~A}$ stores $0.2 \mathrm{~J}$ of energy in its magnetic field. What is the self-inductance of the coil?

Professor PC Thomas Classes · Answer

(b) : The magnetic potential energy stored in an inductor is given by $U = \frac{1}{2} L I^2$. We are given $U = 0.2 \mathrm{~J}$ and $I = 4 \mathrm{~A}$. Rearranging the formula to solve for the self-inductance $L$: $L = \frac{2U}{I^2} = \frac{2 	imes 0.2}{4^2} = \frac{0.4}{16} = 0.025 \mathrm{~H}$. Converting this to millihenries gives $L = 25 \mathrm{mH}$.