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, \( I_B - I_A \) for the given Amperian loops is:
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When calculating currents in rotating charge configurations, remember that the velocity of each charge is \( v = R\omega \), and the total current enclosed by an Amperian loop depends on the number of charges and the fraction of the total circle enclosed.
The current enclosed by an Amperian loop due to moving charges is given by:
\[
I = nqv,
\]
where:
- \( n \) is the number of charges,
- \( q \) is the charge,
- \( v \) is the velocity of the charge.
Since the charges are rotating, the velocity of each charge is \( v = R\omega \), where \( \omega \) is the angular velocity and \( R \) is the radius of the circle.
For the larger Amperian loop \( B \), the total current enclosed is the sum of the contributions from all charges. This is:
\[
I_B = N \times q \times R\omega.
\]
For the smaller loop \( A \), the current enclosed is due to only a fraction of the charges. If the loop encloses a small segment, the fraction of the total charge enclosed is proportional to the fraction of the circumference of the circle, so the current enclosed by loop \( A \) is:
\[
I_A = \frac{1}{2\pi} \times N \times q \times R\omega.
\]
Thus, the difference between the currents is:
\[
I_B - I_A = \frac{N}{2\pi} q\omega.
\]
Final Answer: \( \frac{N}{2\pi} q\omega \).
