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

The magnetic energy stored in an inductor of inductance

4 μH

carrying a current of

2 A

is:

$8 μJ$
$4 μJ$
$4 mJ$
$8 mJ$

Solution

Magnetic energy stored in an inductor is given by,

$U = \frac{1}{2} L i^{2} = \frac{1}{2} \times (4 \times 10^{- 6}) \times 2^{2} = 8 \times 10^{- 6} J = 8 μJ$

Asked in: NEET 2023 (All India)

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