Two concentric circular coils having radii ' $r_1$ ' and ' $r_2$ ' $\left(r_2 \ll r_1\right)$ are placed…
Two concentric circular coils having radii ' $r_1$ ' and ' $r_2$ ' $\left(r_2 \ll r_1\right)$ are placed co-axially with centres coinciding. The mutual inductance of the arrangement is ( $\mu_0=$ permeability of free space) (Both coils have single turn)
$\frac{\mu_0 \pi r_2}{2 r_1}$
$\frac{\mu_0 \pi}{2 r_1 r_2}$
$\frac{\mu_0 \pi r_1}{2 r_2}$
$\frac{\mu_0 \pi r_2^2}{2 r_1}$
Solution
The magnetic field at the centre of a loop is given by
$B=\frac{\mu_0 N I}{2 R}$
$\therefore \quad$ Magnetic field produced by ring $A, B_A=\frac{\mu_0 I}{2 r_1}$
$\therefore \quad$ Magnetic flux produced in ring B due to $\mathrm{B}_{\mathrm{A}}$, $\phi_B=B_A A_B$
$\begin{aligned}
& A_B=\pi r_2^2 \\
\therefore \quad & \phi_B \\
= & \frac{\mu_0 I}{2 r_1} \times \pi r_2^2=\frac{\mu_0 \pi r_2^2}{2 r_1} I
\end{aligned}$ Mutual Inductance $\mathrm{M}=\frac{\phi}{\mathrm{I}}$ (from MI $\phi=\mathrm{M}$ )
$\therefore \quad$ We can write,
$M=\frac{\phi_B}{I}=\frac{\mu_0 \pi r_2^2 \cdot I}{2 r_1 \cdot I}=\frac{\mu_0 \pi r_2^2}{2 r_1}$