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(D) The magnetic field produced by the larger loop of radius $R_1$ at its center is $B = \frac{\mu_0 I}{2 R_1}$.Assuming the smaller loop of radius $R

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$\frac{R_2^2}{R_1}$

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(D) The magnetic field produced by the larger loop of radius $R_1$ at its center is $B = \frac{\mu_0 I}{2 R_1}$.Assuming the smaller loop of radius $R_2$ is placed in this uniform magnetic field,the magnetic flux linked with the smaller loop is $\phi = B \cdot A_2$,where $A_2 = \pi R_2^2$ is the area of the smaller loop.$\phi = \left( \frac{\mu_0 I}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I$.Since $\phi = M I$,comparing the two expressions,we get $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.Therefore,the mutual inductance $M$ is directly proportional to $\frac{R_2^2}{R_1}$.

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 > R_2$,the mutual inductance $M$ between them will be directly proportional to

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