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(B) The magnetic field $B$ at the center of a circular coil of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.Since $r_1 >

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$\frac{\mu_0 \pi r_2^2}{2 r_1}$

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(B) The magnetic field $B$ at the center of a circular coil of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.Since $r_1 > r_2$, we assume the smaller coil (radius $r_2$) is placed in the magnetic field produced by the larger coil (radius $r_1$).The magnetic flux $\phi$ linked with the smaller coil is $\phi = B \cdot A_2$, where $A_2 = \pi r_2^2$ is the area of the smaller coil.$\phi = \left( \frac{\mu_0 I}{2 r_1} \right) (\pi r_2^2) = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.The mutual inductance $M$ is defined by the relation $\phi = M I$.Comparing the two equations, we get $M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

Two planar concentric rings of metal wire having radii $r_1$ and $r_2$ (with $r_1 > r_2$) are placed in air. The current $I$ is flowing through the coil of larger radius. The mutual inductance between the coils is given by ($\mu_0$ = permeability of free space)

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