$N$ 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 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
Answer: (D) $\dfrac{N}{2\pi}q\omega$
The rotating charges form a ring current. In one revolution (time $2\pi/\omega$) all $Nq$ of charge passes any point of the ring:
$$I = \frac{Nq\omega}{2\pi}$$
Loop A wraps around the ring at one place, so the ring's current passes through it once: $I_A = \dfrac{Nq\omega}{2\pi}$.
Loop B is threaded by the ring at two diametrically opposite places, where the charges move in opposite directions. The two contributions cancel: $I_B = 0$.
$$I_A - I_B = \frac{N}{2\pi}q\omega$$
Solution by Sreeraj P, M.Sc Physics