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A coil of self-inductance 2 H carries a current of 4 A. The energy stored in its magnetic field is:
A4 J
B16 J
C8 J
D32 J
Answer & Solution
Correct answer: B. 16 J
$U = \tfrac{1}{2} L I^2 = \tfrac{1}{2}(2)(4^2) = 16$ J.
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