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(A) Let $I_1$ be the current flowing through the coil of radius $r_1$.The magnetic field produced at the center of this coil is given by $B_1 = \frac{

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

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(A) Let $I_1$ be the current flowing through the coil of radius $r_1$.The magnetic field produced at the center of this coil is given by $B_1 = \frac{\mu_0 I_1}{2 r_1}$.Since $r_2 \ll r_1$,the magnetic field $B_1$ is approximately uniform over the area of the smaller coil.The magnetic flux $\phi_2$ passing through the coil of radius $r_2$ is $\phi_2 = B_1 \cdot A_2 = B_1 \cdot \pi r_2^2$.Substituting the value of $B_1$,we get $\phi_2 = \left( \frac{\mu_0 I_1}{2 r_1} \right) \pi r_2^2$.The mutual inductance $M$ is defined as $M = \frac{\phi_2}{I_1}$.Thus,$M = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

Two concentric circular coils having radii $r_1$ and $r_2$ $(r_2 \ll r_1)$ are placed co-axially with centres coinciding. The mutual induction of the arrangement is (Both coils have single turn,$\mu_0 =$ permeability of free space).

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