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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

A

1.389 H

B

138.88 H

C

0.138 H

D

13.89 H

Step-by-Step Solution

Energy stored in inductor U=12LI2U = \frac{1}{2}LI^2. Given U=25×10−3 JU = 25 \times 10^{-3} \text{ J} and I=60×10−3 AI = 60 \times 10^{-3} \text{ A}. Substituting values: 25×10−3=12×L×(60×10−3)225 \times 10^{-3} = \frac{1}{2} \times L \times (60 \times 10^{-3})^2. Solving for LL: L=25×2×106×10−33600=50036=13.89 HL = \frac{25 \times 2 \times 10^6 \times 10^{-3}}{3600} = \frac{500}{36} = 13.89 \text{ H}.

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