A $$100\,mH$$ coil carries a current of $$1\,A.$$ Energy stored in its magnetic field is
A.
$$0.5\,J$$
B.
$$1\,A$$
C.
$$0.05\,J$$
D.
$$0.1\,J$$
Answer :
$$0.05\,J$$
Solution :
Energy stored in coil is $$E = \frac{1}{2}L{i^2}$$
where, $$L$$ is self-inductance of coil and $$i$$ is current induced. Here, $$L = 100\,mH = 100 \times {10^{ - 3}}H$$ and $$i = 1\,A$$
$$\therefore E = \frac{1}{2} \times \left( {100 \times {{10}^{ - 3}}} \right) \times {\left( 1 \right)^2} = 0.05\,J$$
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