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

Answer 2

Step 1: Understand the problem setup.
The problem involves \( N \) equally spaced charges, each of value \( q \), placed on a circle of radius \( R \). The circle rotates about its axis with an angular velocity \( \omega \). There are two Amperian loops: - Loop \( B \) is a larger loop that encloses the whole circle. - Loop \( A \) is a smaller loop that encloses a small segment of the circle.

Step 2: Determine the current due to rotating charges.
Each charge, \( q \), moves in a circular path with velocity \( v = \omega R \) due to the rotation. The current associated with each charge is the charge passing through a point per unit time. The current for each charge is:
\[ I_{\text{charge}} = \frac{q}{T}, \] where \( T \) is the period of revolution, and \( T = \frac{2\pi}{\omega} \). Thus, the current due to each charge is:
\[ I_{\text{charge}} = \frac{q}{\frac{2\pi}{\omega}} = \frac{q\omega}{2\pi}. \]

Step 3: Total current enclosed by loop \( B \).
Loop \( B \) encloses all \( N \) charges, so the total current enclosed by loop \( B \) is:
\[ I_B = N \times I_{\text{charge}} = N \times \frac{q\omega}{2\pi} = \frac{Nq\omega}{2\pi}. \]

Step 4: Total current enclosed by loop \( A \).
Loop \( A \) encloses a small segment of the circle. If the segment corresponds to a fraction \( f \) of the total circumference, then the number of charges enclosed by loop \( A \) is \( fN \). Therefore, the current enclosed by loop \( A \) is:
\[ I_A = fN \times \frac{q\omega}{2\pi} = f \times \frac{Nq\omega}{2\pi}. \]

Step 5: Calculate the difference between enclosed currents.
The difference in the enclosed currents is:
\[ I_B - I_A = \frac{Nq\omega}{2\pi} - \frac{fNq\omega}{2\pi} = \frac{Nq\omega}{2\pi}(1 - f). \] For a small segment, if the fraction \( f \) is small, the difference between the currents simplifies to:
\[ I_B - I_A = \frac{Nq\omega}{2\pi}. \]

Final Answer:
The difference between the enclosed currents for the two Amperian loops is:
\[ \boxed{\frac{Nq\omega}{2\pi}}. \]