The magnetic energy stored in an inductor of inductance $4 \mu ext{H}$ carrying a current of $2 ext{A}$ is

Question

The magnetic energy stored in an inductor of inductance $4 \mu	ext{H}$ carrying a current of $2 	ext{A}$ is

Accepted Answer

The magnetic energy $U$ stored in an inductor is given by $U = \frac{1}{2}LI^2$. Given $L = 4 \mu	ext{H} = 4 	imes 10^{-6} 	ext{H}$ and $I = 2 	ext{A}$. Thus, $U = \frac{1}{2} 	imes (4 	imes 10^{-6}) 	imes (2)^2 = 2 	imes 10^{-6} 	imes 4 = 8 	imes 10^{-6} 	ext{J} = 8 \mu	ext{J}$.

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