Question

A positive charge q is distributed over a circular ring of radius a. It is placed in a horizontal plane and is rotated about its axis at a uniform angular speed $\omega$. A horizontal magnetic field B exists in the space. The torque acting on the ring due to the magnetic force is ________.

• A. $\frac{1}{2}q\omega a^{2}B$
• B. $\frac{1}{2}q\omega aB$
• C. $\frac{1}{2}q\omega^{2}aB$
• D. $q\omega a^{2}B$

Hint:

The magnetic moment of a rotating ring is always half of the product of charge, angular velocity, and radius squared.

Answer 1

The Correct Option is A

Step 1: Concept
A rotating charge constitutes a current $I = q\omega / 2\pi$. The magnetic moment is $\mu = I \times \text{Area}$.
Step 2: Analysis
Current $I = q / T = q\omega / 2\pi$. Magnetic moment $\mu = (q\omega / 2\pi) \times \pi a^{2} = \frac{1}{2}q\omega a^{2}$.
Step 3: Calculation
Torque $\vec{\tau} = \vec{\mu} \times \vec{B}$. Since $\vec{\mu}$ is vertical (along axis) and $\vec{B}$ is horizontal, the angle is 90$^\circ$. $\tau = \mu B \sin(90^\circ) = \frac{1}{2}q\omega a^{2}B$.
Step 4: Conclusion
Hence, the torque is $\frac{1}{2}q\omega a^{2}B$.
Final Answer:(A)