The magnetic potential energy ($U$) stored in an inductor is given by the formula: $U = \frac{1}{2} L I^2$ Where: $U$ is the energy stored ($25 \text{ mJ} = 25 \times 10^{-3} \text{ J}$) $I$ is the current flowing ($60 \text{ mA} = 60 \times 10^{-3} \text{ A}$)

• $L$ is the self-inductance of the inductor.

Calculation: Rearranging the formula to solve for $L$: $L = \frac{2U}{I^2}$ $L = \frac{2 \times (25 \times 10^{-3} \text{ J})}{(60 \times 10^{-3} \text{ A})^2}$ $L = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}}$ $L = \frac{50 \times 10^{-3}}{3.6 \times 10^{-3}}$ $L = \frac{50}{3.6} \approx 13.888... \text{ H}$

Rounding to two decimal places, the inductance is $13.89 \text{ H}$.