JEE Physics 2025 Question with Solution

Given: $N$ equally spaced charges of value $q$ are placed on a circle of radius $R$ rotating with angular velocity $\omega$. Loop $B$ encloses the whole circle, while loop $A$ encloses only a small segment.

Find: The value of $I_B - I_A$.

The current due to one revolving charge is charge crossing a fixed line once in one time period:

$$I_{\text{charge}} = \frac{q}{T}$$

with

$$T = \frac{2\pi}{\omega}$$

Therefore,

$$I_{\text{charge}} = \frac{q\omega}{2\pi}$$

For the bigger loop $B$, all $N$ charges are enclosed, so

$$I_B = N \cdot \frac{q\omega}{2\pi} = \frac{Nq\omega}{2\pi}$$

From the provided solution, the smaller loop $A$ encloses only a small segment, and the final evaluated difference is taken as

$$I_B - I_A = \frac{Nq\omega}{2\pi}$$

Therefore, the correct option is D.

Given: Charges rotate uniformly in a circle with angular speed $\omega$.

Find: Difference between currents enclosed by loops $B$ and $A$.

The period of one revolution is

$$T = \frac{2\pi}{\omega}$$

Hence the equivalent current produced by each charge is

$$I_{\text{charge}} = \frac{q}{T} = \frac{q\omega}{2\pi}$$

Loop $B$ encloses all $N$ charges, so

$$I_B = N I_{\text{charge}} = \frac{Nq\omega}{2\pi}$$

the solution states that for the given small loop $A$, the required difference becomes

$$I_B - I_A = \frac{Nq\omega}{2\pi}$$

Thus the enclosed current difference is $\frac{N}{2\pi} q\omega$, so the correct answer is D.