38. The magnetic potential energy stored in a certain inductor is $25 mJ$, when the current in the inductor is $60 mA$. This inductor is of inductance: (2018)
Solution:
The magnetic energy stored is $U = \frac{1}{2} L I^2$.
Given $U = 25 mJ = 25 \times 10^{-3} J$ and $I = 60 mA = 60 \times 10^{-3} A$.
$L = \frac{2U}{I^2} = \frac{2 \times 25 \times 10^{-3}}{(60 \times 10^{-3})^2} = \frac{50 \times 10^{-3}}{3600 \times 10^{-6}} = \frac{50000}{3600} \approx 13.89 H$.
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