JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Magnetic Effects of Current

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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) $\frac{N ^{2}}{2 π} q ω$
(B) $\frac{2 π}{N} q ω$
(C) $\frac{N}{2 π} q ω$ ✓ Correct
(D) $\frac{N}{π} q ω$

Answer: (C)

Solution

The N equally spaced point charges, each of value

q

, rotate rigidly on the ring with angular velocity

ω

, completing one revolution in time

T = \frac{2 π}{ω}

. Averaged over this period, this rotating charge distribution behaves as a steady ring current: the total charge

N q

effectively circulates once every period

T

, so the ring current is

I = \frac{N q}{T} = \frac{N q ω}{2 π} .

The small Amperian loop A threads a small segment of this ring current directly, so it links exactly this current:

I_{A} = \frac{N q ω}{2 π} .

The larger Amperian loop B, however, encircles the ring as a whole without being pierced by the current path itself (the charge loop lies entirely within the area it bounds, closing on itself), so it links zero net current:

I_{B} = 0.

Therefore,

I_{A} - I_{B} = \frac{N q ω}{2 π} - 0 = \frac{N}{2 π} q ω .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Magnetic Effects of Current

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