A) 0.1 J
B) 0.05 J
C) 0.05 J
D) 1 J
Correct Answer: B
Solution :
Energy stored\[E=\frac{1}{2}L{{I}^{2}}\] \[=\frac{1}{2}\times 100\times {{10}^{-3}}\times {{(1)}^{2}}=0.05J\]
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