Two concentric circular coils having radii $r_1$ and $\mathrm{r}_2\left(\mathrm{r}_2 \ll…
Two concentric circular coils having radii $r_1$ and $\mathrm{r}_2\left(\mathrm{r}_2 \ll \mathrm{r}_1\right)$ are placed co-axially with centres coinciding. The mutual induction of the arrangement is (Both coils have single turn, $\mu_0=$ permeability of free space)
$\frac{\mu_0 \pi r_2^2}{2 r_1}$
$\frac{\mu_0 \pi r_2}{2 r_1}$
$\frac{\mu_0 \pi r_2^2}{r_1^2}$
$\frac{\mu_0 \pi r_2}{r_1}$
Solution
Let $I_1$ be the current through the coil whose radius is $r_1$.
$\therefore \quad$ Magnetic field at the centre of the coil,
$B_1=\frac{\mu_0 I_1}{2 r_1}$
Magnetic flux passing through the coil of radius $\mathrm{r}_2$ is
$\begin{aligned}
& \phi_2=\mathrm{B}_1 \cdot \pi \mathrm{r}_2^2 \quad(\because \phi=\mathrm{B} . \mathrm{A}) \\
& =\frac{\mu_0 \mathrm{I}_1}{2 \mathrm{r}_1} \cdot \pi \mathrm{r}_2^2
\end{aligned}$
$\therefore \quad$ Mutual inductance of the arrangement,
$\begin{aligned}
\mathbf{M} & =\frac{\phi_2}{\mathrm{I}_1} \\
& =\frac{\mu_0 \mathrm{I}_1 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1 \mathrm{I}_1} \\
& =\frac{\mu_0 \pi \mathrm{r}_2^2}{2 \mathrm{r}_1}
\end{aligned}$