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NEET PHYSICSEasy

The magnetic energy stored in an inductor of inductance 4μH4 \mu\text{H} carrying a current of 2A2 \text{A} is

1

4μJ4 \mu\text{J}

2

4mJ4 \text{mJ}

3

8mJ8 \text{mJ}

4

8μJ8 \mu\text{J}

Step-by-Step Solution

The magnetic energy UU stored in an inductor is given by U=12LI2U = \frac{1}{2}LI^2. Given L=4μH=4×106HL = 4 \mu\text{H} = 4 \times 10^{-6} \text{H} and I=2AI = 2 \text{A}. Thus, U=12×(4×106)×(2)2=2×106×4=8×106J=8μJU = \frac{1}{2} \times (4 \times 10^{-6}) \times (2)^2 = 2 \times 10^{-6} \times 4 = 8 \times 10^{-6} \text{J} = 8 \mu\text{J}.

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