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JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1
A current of 2A is increasing at a rate of 4A/s through a coil of inductance 2H. The energy stored in the inductor per unit me is
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Correct option: C
16 J/s
Option C (16 J/s) is correct because the energy stored per unit time is equal to the power supplied to the inductor, which is calculated as P=LididtP = Li\frac{di}{dt}.

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