JEE Main 2025 · 24 January · evening · Single correct · +4 / −1
N equally spaced charges each of value q , are placed on a circle of radius R . The circle rotates about its axis with an angular velocity ω as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, I_A-I_B, for the given Amperian loops is
- (a) N²/(2 π) q ω
- (b) N/(2 π) q ω
- (c) (2 π)/N q ω
- (d) N/π q ω
Answer and solution
Answer (b) N/(2 π) q ω
Solution
Physics sir · 4 steps
Count how many times the rotating ring pierces each loop, and in which direction it goes through.
- Current of the ring. All N charges pass any fixed point once in every revolution. One revolution takes T = 2π/ω, so the ring carries a steady current I = Nq/T.
I = Nq/(2π/ω) = Nqω/2π
- Loop A. The ring threads through loop A only once. So the whole ring current crosses its surface: I_A = Nqω/2π.
- Loop B. Loop B cuts the ring at two places. The current crosses its surface once in one direction and once in the opposite direction, so the two contributions cancel and I_B = 0.
- So. I_A - I_B = Nqω/2π - 0 = N/2πqω, which is option (b).
The idea to keep. Enclosed current is the net current through the loop's surface. If a wire passes through the same loop twice in opposite directions, its contributions cancel to zero.
Answer from the NTA final answer key. Solution by PrepWiser.