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Question

Energy stored in a coil of self inductance $$40\,mH$$ carrying a steady current of $$2\, A$$ is


A
0.8J
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B
8J
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C
0.08J
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D
80J
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Solution

The correct option is B $$0.08 \,J$$
Given:
$$L=40mH\\i=2A$$

$$U = \dfrac{1}{2} Li^2 $$

$$\Rightarrow U = \dfrac{1}{2} \times 40 \times 10^{-3} \times (2)^2 = 0.08 \,J$$

Physics
NCERT
Standard XII

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