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A coil of radius ' $r$ ' is placed on another coil (whose radius is $\mathrm{R}$ and current flowing through it is changing) so that their centres coincide $(\mathrm{R} \gg \mathrm{r})$. If both the coils are coplanar then the mutual inductance between them is ( $\mu_0=$ permeability of free space)
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Verified Answer
The correct answer is:
$\frac{\mu_0 \pi \mathrm{r}^2}{2 \mathrm{R}}$
Magnetic field, $B=\frac{\mu_0 I}{2 R}$
Flux passing through the coil,
$$
\begin{aligned}
\phi & =\mathrm{B} \times \pi \mathrm{r}^2 \\
\therefore \quad \phi & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \times \pi \mathrm{r}^2
\end{aligned}
$$
Mutual Inductance, $/ \mathrm{M}=\frac{\phi}{\mathrm{I}}$
$$
\therefore \quad M=\frac{\frac{\mu_0 I}{2 R} \times \pi r^2}{I}=\frac{\mu_0 \pi r^2}{2 R}
$$
Flux passing through the coil,
$$
\begin{aligned}
\phi & =\mathrm{B} \times \pi \mathrm{r}^2 \\
\therefore \quad \phi & =\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}} \times \pi \mathrm{r}^2
\end{aligned}
$$
Mutual Inductance, $/ \mathrm{M}=\frac{\phi}{\mathrm{I}}$
$$
\therefore \quad M=\frac{\frac{\mu_0 I}{2 R} \times \pi r^2}{I}=\frac{\mu_0 \pi r^2}{2 R}
$$
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