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(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.Since $R_

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

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(D) The magnetic field $B$ at the center of a circular loop of radius $R_1$ carrying current $I_1$ is given by $B = \frac{\mu_0 I_1}{2 R_1}$.Since $R_1$  >  $R_2$, we assume the smaller loop of radius $R_2$ is placed within the magnetic field of the larger loop.The magnetic flux $\phi_2$ linked with the smaller loop is $\phi_2 = B \cdot A_2$, where $A_2 = \pi R_2^2$ is the area of the smaller loop.Substituting the values, we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 R_1} \right) (\pi R_2^2) = \left( \frac{\mu_0 \pi R_2^2}{2 R_1} \right) I_1$.By definition, the mutual inductance $M$ is given by $\phi_2 = M I_1$, so $M = \frac{\mu_0 \pi R_2^2}{2 R_1}$.Thus, $M$ 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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