Question

Two concentric circular coils, one of small radius $r_1$ and the other of large radius $r_2$, such that $r_1 \ll r_2$, are placed co-axially with centres coinciding. The mutual inductance $M$ of the arrangement is proportional to ______.

• A. $\frac{r_1}{r_2}$
• B. $\frac{r_2}{r_1}$
• C. $\frac{r_1^2}{r_2}$
• D. $\frac{r_2^2}{r_1}$

Hint:

In mutual inductance problems, always calculate the \textbf{Field} of the larger/longer coil first and find the \textbf{Flux} through the smaller one.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Mutual inductance ($M$) relates the flux linked with one coil due to the current in another: $\Phi_1 = MI_2$.
Step 2: Detailed Explanation:
Let current $I_2$ flow through the large outer coil. The magnetic field at the center is $B_2 = \frac{\mu_0 I_2}{2r_2}$. Since $r_1 \ll r_2$, this field is approximately uniform over the area of the small coil. The flux $\Phi_1$ linked with the small coil is $B_2 \times A_1$: $$\Phi_1 = \left( \frac{\mu_0 I_2}{2r_2} \right) \times (\pi r_1^2)$$ Comparing this to $\Phi_1 = MI_2$, we find $M = \frac{\mu_0 \pi r_1^2}{2r_2}$. Therefore, $M \propto \frac{r_1^2}{r_2}$.
Step 3: Final Answer:
The correct option is (c).