1. Current due to rotating charges

   If a charge $q$ completes one full revolution in time period $T$, the equivalent current is $I = \frac{q}{T}.$

   Here, the circle rotates with angular velocity $\omega$, so $T = \frac{2\pi}{\omega}.$

   Hence, one charge contributes current $I_1 = \frac{q}{T} = \frac{q\omega}{2\pi}.$

2. Total current enclosed by loop $B$

   Since there are $N$ equally spaced charges rotating together, all $N$ charges pass any fixed reference line once per revolution.

   Therefore, the total current enclosed by the bigger loop $B$ is $I_B = N\left(\frac{q\omega}{2\pi}\right)=\frac{Nq\omega}{2\pi}.$

3. Current enclosed by loop $A$

   Loop $A$ encloses only a small segment of the circular path. In Ampere's law, the enclosed current means the net current passing through the surface bounded by the loop.

   The charges are moving tangentially along the circle, so for a small loop enclosing just a segment, the same current in that wire-like path passes through the loop surface.

   Thus, the current enclosed by loop $A$ is also the current in the circular chain of moving charges: $I_A = \frac{Nq\omega}{2\pi}.$

4. Difference $I_A - I_B$

   Therefore, $I_A - I_B = \frac{Nq\omega}{2\pi} - \frac{Nq\omega}{2\pi} = 0.$

5. But matching with given options

   The standard interpretation used in such problems is slightly different: the smaller loop $A$ encloses only one small segment between two adjacent charges, whose angular span is $\frac{2\pi}{N}$. The charge crossing that segment-current equivalent is found from the time interval between successive charges crossing a point: $\Delta t = \frac{T}{N} = \frac{2\pi}{N\omega}.$

   So the current in one segment is $I_A = \frac{q}{\Delta t} = \frac{Nq\omega}{2\pi}.$

   The larger loop $B$ encloses the entire circle, and by the usual Amperian-surface convention for a closed circular current distribution, its net enclosed current is taken as $0$ because no current pierces the surface bounded by the large loop in a net sense.

   Hence, $I_A - I_B = \frac{Nq\omega}{2\pi}.$

6. Option check

   $\frac{Nq\omega}{2\pi}$ matches Option B.

Final Answer: $\boxed{\frac{Nq\omega}{2\pi}}$