Question

A charge ‘q’ is spread uniformly over an isolated ring of radius ‘R’. The ring is rotated about its natural axis with angular speed \( \omega \). The magnetic dipole moment of the ring is:

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

Hint:

A rotating charge distribution generates a magnetic moment. The moment depends on the charge, angular velocity, and radius of the ring.

Answer 1

The Correct Option is C

Step 1: Define Magnetic Dipole Moment For a rotating charged ring, the magnetic dipole moment is given by: \[ M = I A \] where \( I \) is the current and \( A \) is the area.
Step 2: Compute Current Total charge on the ring: \[ Q = q \] The time period of rotation: \[ T = \frac{2\pi}{\omega} \] Thus, current: \[ I = \frac{q}{T} = \frac{q \omega}{2\pi} \]
Step 3: Compute Magnetic Dipole Moment The area of the ring is: \[ A = \pi R^2 \] \[ M = I A = \left(\frac{q \omega}{2\pi}\right) \times (\pi R^2) \] \[ M = \frac{q \omega R^2}{2} \]