(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.Since $r_2 \ll r_

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

(D) The magnetic field at the center of a circular loop of radius $r_1$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2 r_1}$.Since $r_2 \ll r_1$,the magnetic field $B$ is approximately uniform over the area of the smaller loop.The magnetic flux $\phi$ linked with the smaller loop of radius $r_2$ is $\phi = B \times A_2$,where $A_2 = \pi r_2^2$ is the area of the smaller loop.Substituting the values,we get $\phi = \left( \frac{\mu_0 I}{2 r_1} \right) \times (\pi r_2^2) = \frac{\mu_0 \pi r_2^2}{2 r_1} I$.The mutual inductance $M$ is defined by the relation $\phi = M I$.Therefore,$M = \frac{\phi}{I} = \frac{\mu_0 \pi r_2^2}{2 r_1}$.

Class 12 Physics Electromagnetic Induction Mutual Induction

Medium

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 inductance of the arrangement is ($\mu_0 =$ permeability of free space) (Both coils have single turn).

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A
$\frac{\mu_0 \pi r_2}{2 r_1}$
B
$\frac{\mu_0 \pi}{2 r_1 r_2}$
C
$\frac{\mu_0 \pi r_1}{2 r_2}$
D
$\frac{\mu_0 \pi r_2^2}{2 r_1}$

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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 inductance of the arrangement is ($\mu_0 =$ permeability of free space) (Both coils have single turn).

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$A$ small square loop of wire of side $l$ is placed inside a large square loop of side $L$ $(L > l)$. If the loops are coplanar and their centers coincide,the mutual inductance of the system is directly proportional to

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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