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 as shown in the figure. A bigger Amperian loop B encloses the whole circle where as a smaller Amperian loop A encloses a small segment. The difference between enclosed currents, , for the given Amperian loops is

• A. Nqw/(2π)
• B. qw/(2π)
• C. 2Nqw/(π)
• D. Nqw/(π)

Answer 1

Answer: (A)

Nqw/(2π)

Answer 2

The problem asks for the difference between the enclosed currents for two Amperian loops, A and B, around a rotating circle of charges.

1. Current due to Rotating Charges: When a charge q rotates with angular velocity ω, it completes f = ω / (2π) revolutions per second. The current generated by a single rotating charge is given by: $I_{single} = qf = \frac{q\omega}{2\pi}$

2. Current Enclosed by Loop B ($I_B$): Loop B encloses the whole circle. This means it encloses all N equally spaced charges. Since all N charges are rotating, the total current generated by them is the sum of the currents due to each charge. $I_B = N \times I_{single} = N \times \frac{q\omega}{2\pi} = \frac{Nq\omega}{2\pi}$

3. Current Enclosed by Loop A ($I_A$): Loop A encloses a "small segment". In the context of discrete charges, "enclosing a small segment" typically implies that the loop is small enough that it does not encompass any of the discrete charges at any given instant, or that the net current passing through its enclosed area is negligible compared to the total. If the loop A does not enclose any of the discrete charge paths, then no current passes through the area bounded by loop A. Therefore, the current enclosed by loop A is: $I_A = 0$

4. Difference between Enclosed Currents: The difference between the enclosed currents is $I_B - I_A$: $I_B - I_A = \frac{Nq\omega}{2\pi} - 0 = \frac{Nq\omega}{2\pi}$