Two conducting circular loops of radii R_1 and R_2 are placed in...

Magnetic field at the center of the larger loop of radius $R_1$ carrying current $I$ is given by

$B = \dfrac{\mu_0 I}{2 R_1}$

Since $R_1 > R_2$, the magnetic field can be considered approximately uniform over the entire area of the smaller loop.

Magnetic flux linked with the smaller loop of radius $R_2$ is

$\Phi = B \times A_2 = \left(\dfrac{\mu_0 I}{2 R_1}\right) (\pi R_2^2)$

The mutual inductance $M$ is given by

$M = \dfrac{\Phi}{I} = \dfrac{\mu_0 \pi R_2^2}{2 R_1}$

Therefore, $M \propto \dfrac{R_2^2}{R_1}$

Answer: $\dfrac{R_2^2}{R_1}$