A 100 millihenry coil carries a current of 1 ampere. Energy stored in its magnetic field is

0.5 J

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1 J

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0.05 J

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0.1 J

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Solution

The correct option is C 0.05 J
Energy stored U is given by

$U = \frac{1}{2} L i^{2}$
$= \frac{1}{2} \times (100 \times 10^{- 3}) (1)^{2}$
$= 0.05 J$ .

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A 100 millihenry coil carries a current of 1 ampere. Energy stored in its magnetic field is

The correct option is C 0.05 JEnergy stored U is given byU=12Li2=12×(100×10−3)(1)2=0.05J.

The correct option is C 0.05 J
Energy stored U is given by

$U = \frac{1}{2} L i^{2}$
$= \frac{1}{2} \times (100 \times 10^{- 3}) (1)^{2}$
$= 0.05 J$ .