Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coincid

Question

Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 \gg R_2$, the mutual inductance $M$ between them will be directly proportional to:

Accepted Answer

This problem corresponds directly to **Example 6.8** in the NCERT textbook .

1.  **Setup:** A current $I_1$ flows through the larger outer loop of radius $R_1$. 
2.  **Magnetic Field:** The magnetic field $B_1$ produced at the center of the large loop is given by $B_1 = \frac{\mu_0 I_1}{2R_1}$. Since $R_1 \gg R_2$, this field is assumed to be uniform over the area of the smaller inner loop.
3.  **Magnetic Flux:** The magnetic flux $\Phi_2$ linked with the smaller loop (radius $R_2$) is the product of the field and the area of the small loop: $\Phi_2 = B_1 A_2 = \left(\frac{\mu_0 I_1}{2R_1}\right) (\pi R_2^2)$.
4.  **Mutual Inductance:** By definition, $\Phi_2 = M I_1$. Comparing the expressions, we get $M = \frac{\mu_0 \pi R_2^2}{2R_1}$.
5.  **Proportionality:** Therefore, $M \propto \frac{R_2^2}{R_1}$.

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