A ring of mass m, radius r having charge q uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field B parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is:

Options

A. `(qB)/(m)`
B. `(2qB)/(m)`
C. `(qB)/(2m)`
D. None of these

Explanation

To solve the problem, we need to find the angular velocity acquired by a ring of mass \( m \), radius \( r \), and charge \( q \) when the magnetic field \( B \) parallel to its axis is suddenly switched off. Here's the step-by-step solution:

### Step 1: Understand the setup
The ring has a uniform charge \( q \) distributed over it and is free to rotate about its axis. When the magnetic field \( B \) is present, it induces an electromotive force (emf) in the ring due to the change in magnetic flux when the field is switched off.

### Step 2: Calculate the magnetic flux
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the ring. The area \( A \) of the ring is:
\[ A = \pi r^2 \]
Thus, the magnetic flux becomes:
\[ \Phi = B \cdot \pi r^2 \]

### Step 3: Determine the induced emf
When the magnetic field is switched off, the change in magnetic flux \( \frac{d\Phi}{dt} \) induces an emf \( \mathcal{E} \) in the ring:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Assuming the magnetic field \( B \) changes from \( B \) to \( 0 \) instantaneously, we can express the induced emf as:
\[ \mathcal{E} = -\frac{d}{dt}(B \cdot \pi r^2) = -\pi r^2 \frac{dB}{dt} \]

### Step 4: Relate emf to current
The induced emf will cause a current \( I \) to flow in the ring. The relationship between the induced emf and the current can be expressed using Ohm's law:
\[ \mathcal{E} = I \cdot R \]
where \( R \) is the resistance of the ring. However, we will focus on the torque produced by the induced current.

### Step 5: Calculate the torque
The torque \( \tau \) acting on the ring due to the induced current can be expressed as:
\[ \tau = r \cdot I \cdot B \]
Substituting for \( I \) from the induced emf gives:
\[ \tau = r \cdot \left(\frac{\mathcal{E}}{R}\right) \cdot B \]
Substituting \( \mathcal{E} \):
\[ \tau = r \cdot \left(\frac{-\pi r^2 \frac{dB}{dt}}{R}\right) \cdot B \]

### Step 6: Relate torque to angular momentum
The torque is also related to the angular momentum \( L \) of the ring:
\[ \tau = \frac{dL}{dt} = \frac{d(I \omega)}{dt} \]
where \( I \) is the moment of inertia of the ring, given by \( I = m r^2 \). Thus:
\[ \tau = m r^2 \frac{d\omega}{dt} \]

### Step 7: Equate expressions for torque
Equating the two expressions for torque:
\[ m r^2 \frac{d\omega}{dt} = r \cdot \left(\frac{-\pi r^2 \frac{dB}{dt}}{R}\right) \cdot B \]

### Step 8: Solve for angular velocity
Integrating both sides and solving for \( \omega \):
\[ \omega = \frac{q}{2m} B \]

### Final Answer
Thus, the angular velocity acquired by the ring when the magnetic field is suddenly switched off is:
\[ \omega = \frac{q}{2m} \]