42. 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: (2021)
Solution:
Let current $I_1$ pass through the outer loop of radius $R_1$. The magnetic field at its centre is $B_1 = \frac{\mu_0 I_1}{2 R_1}$.
Since $R_1 >> R_2$, this field is nearly uniform over the inner loop of area $A_2 = \pi R_2^2$.
Flux linked is $\Phi_2 = B_1 A_2 = \frac{\mu_0 I_1 \pi R_2^2}{2 R_1} = M I_1$, hence $M \propto \frac{R_2^2}{R_1}$.
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