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A coil of self inductance $0.2$ H carries $5$ A. The magnetic energy stored is:
A$2.5$ J, since $U = \tfrac{1}{2}LI^2 = 0.5\cdot 0.2\cdot 25$
B$0.5$ J, ignoring the squaring of the current here
C$25$ J, the simple product of inductance and squared current
D$1$ J, equal to the product of $L$ and $I$ together
Answer & Solution
Correct answer: A. $2.5$ J, since $U = \tfrac{1}{2}LI^2 = 0.5\cdot 0.2\cdot 25$
$U = \tfrac{1}{2}\cdot 0.2\cdot 5^2 = 2.5$ J.
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