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NEET PHYSICSEasy

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

A

8 μJ8 \text{ }\mu\text{J}

B

4 μJ4 \text{ }\mu\text{J}

C

4 mJ4 \text{ mJ}

D

8 mJ8 \text{ mJ}

Step-by-Step Solution

The magnetic potential energy (UU) stored in an inductor is given by the formula: U=12LI2U = \frac{1}{2} L I^2

Given: Inductance (LL) = 4 μH=4×10−6 H4 \text{ }\mu\text{H} = 4 \times 10^{-6} \text{ H} Current (II) = 2 A2 \text{ A}

Calculation: U=12×(4×10−6)×(2)2U = \frac{1}{2} \times (4 \times 10^{-6}) \times (2)^2 U=12×4×10−6×4U = \frac{1}{2} \times 4 \times 10^{-6} \times 4 U=8×10−6 JU = 8 \times 10^{-6} \text{ J} U=8 μJU = 8 \text{ }\mu\text{J}

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