Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 24 January, Evening Shift — Question 56

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 $\omega$ 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, $I_{A}-I_{B}$ , for the given Amperian loops is images

Question figure

Option A:
$\frac{\mathrm{N}^{2}}{2 \pi} \mathrm{q} \omega$
Option B:
$\frac{2 \pi}{N} q \omega$
Option C:
$\frac{N}{2 \pi} q \omega$
Correct
Option D:
$\frac{N}{\pi} q \omega$

Answer: C

Step-by-step solution

$\mathrm{I}_{\mathrm{A}}=\frac{\mathrm{Nq}}{\frac{2 \pi}{\omega}}$

$I_{A}=\frac{N q \omega}{2 \pi}, I_{B}=0$

$I_{A}-I_{B}=\frac{N q \omega}{2 \pi}$ images

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 24 January, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Ampère's Law and its Applications

N equally spaced charges each of value q , are placed on a circle of… | JEE Main 2025 PYQ with Solution · DhiX AI