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If the current of ' $I$ ' gives rise to a magnetic flux ' $\phi$ ' through a coil having ' $N$ ' turns, then magnetic energy stored in the medium surrounding the coil is
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Verified Answer
The correct answer is:
$\frac{\mathrm{N} \phi \mathrm{I}}{2}$
Inductance of the coil, $\mathrm{L}=\frac{\phi}{\mathrm{I}}$
Energy stored due to single turn,
$$
\mathrm{U}=\frac{1}{2} \mathrm{LI}^2=\frac{1}{2}\left(\frac{\phi}{\mathrm{I}}\right) \mathrm{I}^2=\frac{\phi \mathrm{I}}{2}
$$
Energy stored due to $\mathrm{N}$ turns,
$$
\mathrm{U}^{\prime}=\frac{\mathrm{N} \phi \mathrm{I}}{2}
$$
Energy stored due to single turn,
$$
\mathrm{U}=\frac{1}{2} \mathrm{LI}^2=\frac{1}{2}\left(\frac{\phi}{\mathrm{I}}\right) \mathrm{I}^2=\frac{\phi \mathrm{I}}{2}
$$
Energy stored due to $\mathrm{N}$ turns,
$$
\mathrm{U}^{\prime}=\frac{\mathrm{N} \phi \mathrm{I}}{2}
$$
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