Charge `Q` is uniformly distributed on a thin insulating ring of mass `m` which is initially at rest. To what angular velocity will the ring be accelerated when a magnetic field `B`, perpendicular to the plane of the ring, is switched on ?

Options

A. (a) `(QB)/(2m)`
B. (b) `(3QB)/(2m)`
C. ( c) `(QB)/(m)`
D. (d) `(QB)/(4m)`

Explanation

To solve the problem, we need to analyze the situation when a magnetic field \( B \) is applied perpendicular to the plane of a thin insulating ring with charge \( Q \) uniformly distributed on it. The ring has a mass \( m \) and is initially at rest.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a thin insulating ring of radius \( r \) with charge \( Q \) uniformly distributed.
- The ring is initially at rest, and we apply a magnetic field \( B \) perpendicular to its plane.

2. **Induced Electric Field**:
- When the magnetic field is switched on, it changes with time, which induces an electric field \( E \) in the ring due to Faraday's law of electromagnetic induction.
- The induced electric field \( E \) around the ring can be expressed as:
\[ E = -\frac{d\Phi_B}{dt} \]
- The magnetic flux \( \Phi_B \) through the ring is given by:
\[ \Phi_B = B \cdot A = B \cdot \pi r^2 \]
- Therefore, the induced electric field becomes:
\[ E = -\frac{d(B \cdot \pi r^2)}{dt} = -\pi r^2 \frac{dB}{dt} \]

3. **Force on the Charge**:
- The force \( F \) on the charge \( Q \) due to the electric field \( E \) is given by:
\[ F = Q \cdot E = Q \cdot \left(-\pi r^2 \frac{dB}{dt}\right) = -Q \pi r^2 \frac{dB}{dt} \]

4. **Torque on the Ring**:
- The torque \( \tau \) acting on the ring due to this force is given by:
\[ \tau = r \cdot F = r \cdot \left(-Q \pi r^2 \frac{dB}{dt}\right) = -Q \pi r^3 \frac{dB}{dt} \]

5. **Angular Acceleration**:
- The angular acceleration \( \alpha \) of the ring can be related to the torque by using the moment of inertia \( I \) of the ring:
\[ \tau = I \cdot \alpha \]
- For a thin ring, the moment of inertia \( I \) is given by:
\[ I = m r^2 \]
- Thus, we have:
\[ -Q \pi r^3 \frac{dB}{dt} = m r^2 \alpha \]
- Rearranging gives:
\[ \alpha = -\frac{Q \pi r}{m} \frac{dB}{dt} \]

6. **Finding Angular Velocity**:
- The angular velocity \( \omega \) can be found by integrating the angular acceleration over time:
\[ \omega = \alpha \cdot t = -\frac{Q \pi r}{m} \frac{dB}{dt} \cdot t \]
- Therefore, the final expression for the angular velocity when the magnetic field is switched on is:
\[ \omega = -\frac{Q \pi r}{m} \frac{dB}{dt} \cdot t \]

### Final Answer:
The angular velocity \( \omega \) of the ring when the magnetic field \( B \) is switched on is given by:
\[ \omega = -\frac{Q \pi r}{m} \frac{dB}{dt} \cdot t \]