Rankers Physics
Topic: Electromagnetic Induction

40. A 100 millihenry coil carries a current of $1 A$. Energy stored in its magnetic field is (1991)

$0.5 J$
$1 A$
$0.05 J$
$0.1 J$

Solution:

Energy stored in the magnetic field of an inductor is $U = \frac{1}{2} L I^2$.
Here, $L = 100 mH = 0.1 H$ and $I = 1 A$.
$U = \frac{1}{2} \times 0.1 \times 1^2 = 0.05 J$.

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