The Magnetic energy stored in an inductor of inductance 4µH carrying a current of 2 A is :

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8 µJ

4 µ J

4 mJ

8 mJ

answer is A.

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Detailed Solution

Magnetic energy,
$\begin{array}{l} E = \frac{1}{2} {Li}^{2} \\ = \frac{1}{2} 4 \times 10^{- 6} \times 2^{2} \\ = 8 \times 10^{- 6} J = 8 μJ \end{array}$

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