Two concentric circular coils,one of small radius $r_1$ and the other of large radius $r_2$,such that $r_1 \ll r_2$,are placed coaxially with centers coinciding. Obtain the mutual inductance of the arrangement.

(N/A) Let a current $I_2$ flow through the outer circular coil. The magnetic field at the center of the coil is $B_2 = \frac{\mu_0 I_2}{2 r_2}$.
Since the inner coil has a very small radius $(r_1 \ll r_2)$,the magnetic field $B_2$ can be considered uniform over the area of the inner coil.
The magnetic flux $\phi_1$ through the inner coil is given by $\phi_1 = B_2 \cdot A_1 = B_2 (\pi r_1^2)$.
Substituting the value of $B_2$,we get $\phi_1 = \left( \frac{\mu_0 I_2}{2 r_2} \right) (\pi r_1^2) = \left( \frac{\mu_0 \pi r_1^2}{2 r_2} \right) I_2$.
By definition,$\phi_1 = M_{12} I_2$,where $M_{12}$ is the mutual inductance.
Therefore,$M_{12} = \frac{\mu_0 \pi r_1^2}{2 r_2}$.
Since the mutual inductance is reciprocal,$M_{12} = M_{21} = M = \frac{\mu_0 \pi r_1^2}{2 r_2}$.