JEE MAIN - Physics (2025 - 24th January Evening Shift - No. 1)
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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 $\omega$ 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
$\frac{\mathrm{N}^2}{2 \pi} \mathrm{q} \omega$
$\frac{\mathrm{N}}{2 \pi} \mathrm{q} \omega$
$\mathrm{\frac{2 \pi}{N} q \omega}$
$\frac{\mathrm{N}}{\pi} \mathrm{q} \omega$
Explanation
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$$\begin{aligned} & I_A=\frac{N q}{\frac{2 \pi}{\omega}} \\ & I_A=\frac{N q \omega}{2 \pi}, I_B=0 \\ & I_A-I_B=\frac{N q \omega}{2 \pi} \end{aligned}$$
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