Question

Two planar concentric rings of metal wire having radii $\text{r}_1$ and $\text{r}_2$ ( $\text{r}_1 > \text{r}_2$ ) are placed in air. The current $\text{I}$ is flowing through the coil of larger radius. The mutual inductance between the coils is given by ( $\mu_0 =$ permeability of free space)}

• A. $\frac{\mu_0 \pi \text{r}_1^2}{2\text{r}_2}$
• B. $\frac{\mu_0 \pi \text{r}_2^2}{2\text{r}_1}$
• C. $\frac{\mu_0 \pi (\text{r}_1+\text{r}_2)^2}{2\text{r}_1}$
• D. $\frac{\mu_0 \pi (\text{r}_1-\text{r}_2)^2}{2\text{r}_2}$

Hint:

For concentric coils, $M \propto \frac{(\text{Small Radius})^2}{\text{Large Radius}}$.

Answer 1

The Correct Option is B

Step 1: Concept
Mutual Inductance $M$ is defined by $\phi_2 = MI_1$, where $\phi_2$ is the flux through the smaller coil due to current in the larger coil.

Step 2: Meaning
Magnetic field at the center of the larger coil is $B_1 = \frac{\mu_0 I}{2r_1}$.

Step 3: Analysis
Flux through the smaller coil is $\phi_2 = B_1 \times A_2 = \frac{\mu_0 I}{2r_1} \times \pi r_2^2$.
Comparing with $\phi_2 = MI$, we get $M = \frac{\mu_0 \pi r_2^2}{2r_1}$.

Step 4: Conclusion
The mutual inductance is $\frac{\mu_0 \pi \text{r}_2^2}{2\text{r}_1}$. Final Answer: (B)