JEE Physics 2023 Question with Solution

Given: A small circular loop of radius $R$ is placed inside a large square loop of side $L$, with $L \gg R$. The loops are coplanar and their centers coincide.

Find: The mutual inductance $M$.

Use the magnetic field produced by the large square loop at its center and calculate the flux through the small circular loop.

From the solution:

$$\phi = Mi$$

Also,

$$\phi = BA$$

For the small circular loop, the area is

$$A = \pi R^2$$

Using the field expression stated in the solution,

$$\phi = \pi R^2 \left( \frac{4\mu_0}{4\pi} \cdot i \cdot \frac{L}{2} \right) \sqrt{2}$$

Hence,

$$M = \frac{\phi}{i} = \frac{2\sqrt{2} \, \mu_0 R^2}{L}$$

Therefore, the correct option is C.

Given: Mutual inductance is to be found using the flux linked with the small loop due to current $i$ in the large loop.

Find: Which option matches $M$.

The hint indicates the required method:

1. Find the magnetic field due to the larger loop.
2. Multiply by the area of the smaller loop to get flux.
3. Divide the flux by current to obtain mutual inductance.

Using

$$M = \frac{\phi}{i}$$

and

$$\phi = BA$$

with

$$A = \pi R^2$$

the solution simplifies directly to

$$M = \frac{2\sqrt{2} \, \mu_0 R^2}{L}$$

So the mutual inductance varies as $R^2$ and inversely as $L$, which matches option C.