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(A) Let a current $i$ flow through the outer loop $B$ of radius $b$.The magnetic field at the center $O$ due to the current $i$ in loop $B$ is given b

Vedclass · Accepted Answer

$\frac{\mu_0 \pi a^2}{2 b}$

Vedclass · Answer

(A) Let a current $i$ flow through the outer loop $B$ of radius $b$.The magnetic field at the center $O$ due to the current $i$ in loop $B$ is given by $B_{center} = \frac{\mu_0 i}{2b}$.Since $b >> a$,we can assume this magnetic field is uniform over the area of the inner loop $A$ of radius $a$.The magnetic flux $\phi$ linked with the inner loop $A$ is $\phi = B_{center} \cdot A_{area} = \left( \frac{\mu_0 i}{2b} \right) (\pi a^2)$.By definition of mutual inductance,$\phi = Mi$.Equating the two expressions for flux: $Mi = \frac{\mu_0 i \pi a^2}{2b}$.Therefore,the mutual inductance $M$ is $M = \frac{\mu_0 \pi a^2}{2b}$.

Two conducting circular loops $A$ and $B$ are placed in the same plane with their centres coinciding as shown in the figure. If $b >> a$,the mutual inductance between them is:

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