Question

A total charge \(q\) is uniformly distributed over a circular ring of radius \(R\). The ring is rotated about its axis with angular velocity \(\omega\). Find the magnetic dipole moment of the ring.

Answer 1

Step 1: The rotating charged ring constitutes a current loop. The current \(I\) is given by the charge passing a point per unit time: \[ I = \frac{q}{T}, \] where \(T\) is the time period of one rotation. 
Step 2: The time period \(T\) is related to angular velocity \(\omega\) as: \[ T = \frac{2\pi}{\omega}. \] Step 3: Substitute \(T\) into current expression: \[ I = \frac{q}{2\pi/\omega} = \frac{q \omega}{2\pi}. \] Step 4: Magnetic dipole moment \(\mu\) for a current loop is: \[ \mu = I \times \text{Area} = I \times \pi R^2. \] Step 5: Substitute \(I\) into \(\mu\): \[ \mu = \frac{q \omega}{2\pi} \times \pi R^2 = \frac{q \omega R^2}{2}. \] Answer: \[ \boxed{\mu = \frac{q \omega R^2}{2}}. \]