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Question

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

A
0.5 J
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B
0.05 J
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C
1 J
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D
0.1 J
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Solution

The correct option is D 0.05 J
B. 0.05 J

Given,

self-inductance, L=100mH

current, I=1A

The energy stored in magnetic field, K=12LI2

K=12×100×103×12

= 0.05J

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