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JEE Main 2025 (24 Jan, Shift 2) · Ampere's Law, Cylinders, Solenoid & Toroid

Ampere's Law, Cylinders, Solenoid & Toroid question from JEE Main 2025 (24 Jan, Shift 2)

Q15JEE Main 2025 · 24 JAN · SHIFT 2Medium

Ampere's Law, Cylinders, Solenoid & Toroid

Figure

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\omega 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, IA−IBI_A - I_B, for the given Amperian loops is

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Answer: (b)

  1. Total current of the rotating ring: I=NqT=Nqω2πI = \dfrac{Nq}{T} = \dfrac{Nq\omega}{2\pi}.
  2. Loop A is threaded once by the ring, so IA=Nqω2πI_A = \dfrac{Nq\omega}{2\pi}.
  3. Loop B is crossed by the ring twice, once in each direction, so the currents cancel: IB=0I_B = 0.
  4. IA−IB=N2πqωI_A - I_B = \dfrac{N}{2\pi}q\omega.
  5. So the answer is (b).
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