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(D) The magnetic field $B_{1}$ at the center of a circular loop of radius $R_{1}$ carrying current $I_{1}$ is given by $B_{1} = \frac{\mu_{0} I_{1}}{2

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$\frac{R_{2}^{2}}{R_{1}}$

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(D) The magnetic field $B_{1}$ at the center of a circular loop of radius $R_{1}$ carrying current $I_{1}$ is given by $B_{1} = \frac{\mu_{0} I_{1}}{2 R_{1}}$.The magnetic flux $\phi_{2}$ linked with the smaller loop of radius $R_{2}$ is $\phi_{2} = B_{1} A_{2}$,where $A_{2} = \pi R_{2}^{2}$ is the area of the smaller loop.Since $R_{1} >> R_{2}$,we can assume the magnetic field $B_{1}$ is uniform over the area of the smaller loop.The mutual inductance $M$ is defined as $M = \frac{\phi_{2}}{I_{1}}$.Substituting the expressions,we get $M = \frac{B_{1} A_{2}}{I_{1}} = \frac{(\frac{\mu_{0} I_{1}}{2 R_{1}}) (\pi R_{2}^{2})}{I_{1}} = \frac{\mu_{0} \pi R_{2}^{2}}{2 R_{1}}$.Thus,$M \propto \frac{R_{2}^{2}}{R_{1}}$.

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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