Question

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:

• A. \( \frac{2\pi}{N} q\omega \)
• B. \( \frac{N^2}{2\pi} q\omega \)
• C. \( \frac{N}{\pi} q\omega \)
• D. \( \frac{N}{2\pi} q\omega \)

Hint:

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.

Answer 1

The Correct Option is D

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: \( \frac{N}{2\pi} q\omega \).