JEE Main 2025 Jan 24 Shift 2, Physics Q11: Ampere's Circuital Law

JEE Main2025Jan 24, Shift 2Physics

$N$ 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 whereas 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
$\frac{N}{2 π} q ω$
D
$\frac{N}{π} q ω$

Solution

The rotating ring has period $T = \frac{2 π}{ω}$ . In one period each charge passes any given fixed point in space exactly once, so the total charge per unit time crossing a fixed cross-section of the ring is

$I = \frac{total charge crossing}{T} = \frac{N q}{2 π / ω} = \frac{N q ω}{2 π}$

Loop A encloses one cross-section of the moving charge stream and is pierced by it exactly $N$ times per revolution. Hence $I_{A} = \frac{N q ω}{2 π}$ .

Loop B encloses the entire ring; charges complete closed orbits inside loop B without ever crossing its boundary, so the net enclosed current $I_{B} = 0$ .

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

Correct option: (3)

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JEE Main 2025 Jan 24 Shift 2, Physics Q11: Ampere's Circuital Law

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