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A circular coil of radius ' $\mathrm{R}$ ' has ' $\mathrm{N}$ ' turns of a wire. The coefficient of self-induction of the coil will be ( $\mu_0=$ permeability of free space)
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Verified Answer
The correct answer is:
$\frac{\mu_0 N^2 \pi R}{2}$
Magnetic field at the center of the coil is given by
$$
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
$$
Magnetic flux linked to the coil is
$$
\begin{aligned}
& \phi=\mathrm{NBA}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{I}}{2 \mathrm{R}} \cdot \pi \mathrm{R}^2=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{RI}}{2} \\
& \mathrm{~L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{R}}{2}
\end{aligned}
$$
$$
\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{2 \mathrm{R}}
$$
Magnetic flux linked to the coil is
$$
\begin{aligned}
& \phi=\mathrm{NBA}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{I}}{2 \mathrm{R}} \cdot \pi \mathrm{R}^2=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{RI}}{2} \\
& \mathrm{~L}=\frac{\phi}{\mathrm{I}}=\frac{\mu_0 \mathrm{~N}^2 \pi \mathrm{R}}{2}
\end{aligned}
$$
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