A uniform circular ring of radius R, mass m has uniformly distributed charge q. The ring is force to rotate about its own axis (which is vertical) without friction. In the space a uniform magnetic field B, directed vertically down ward exists in a cylindrical region. The cylindrical region is co-axial with B and has a radius greater than R. If B increasesa at a constant rate `(dB)/(dt)=alpha`. angular acceleration of the ring will.

Options

A. Directly proportional to R
B. Directly proportional to q
C. Directly proportional to m
D. Independent of R and m

Answer

To solve the problem step by step, we will analyze the situation involving a uniform circular ring with charge in a changing magnetic field.

### Step 1: Understand the Setup
We have a uniform circular ring of radius \( R \), mass \( m \), and charge \( q \). The ring is rotating about its vertical axis in a uniform magnetic field \( B \), which is directed vertically downward. The magnetic field is increasing at a constant rate given by \( \frac{dB}{dt} = \alpha \).

**Hint:** Visualize the setup and identify the forces acting on the ring due to the changing magnetic field.

### Step 2: Induced Electric Field
According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electric field. The induced electric field \( E \) can be calculated using the formula:
\[ E = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux through the area of the ring.

The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A = B \cdot \pi R^2 \]
Since \( B \) is changing with time, we have:
\[ \frac{d\Phi}{dt} = \pi R^2 \frac{dB}{dt} = \pi R^2 \alpha \]
Thus, the induced electric field \( E \) is:
\[ E = -\frac{d\Phi}{dt} = -\pi R^2 \alpha \]

**Hint:** Remember that the induced electric field is related to the rate of change of magnetic flux.

### Step 3: Calculate the Force on the Ring
The electric field induces a force on the charged ring. The linear charge density \( \lambda \) of the ring is given by:
\[ \lambda = \frac{q}{2\pi R} \]
The total force \( F \) acting on the ring due to the induced electric field is:
\[ F = \lambda \cdot E \cdot (2\pi R) = \left(\frac{q}{2\pi R}\right)(-\pi R^2 \alpha)(2\pi R) = -q R \alpha \]

**Hint:** Use the relationship between charge density, electric field, and the total force.

### Step 4: Calculate the Torque
The torque \( \tau \) acting on the ring due to the force is given by:
\[ \tau = R \cdot F = R \cdot (-q R \alpha) = -q R^2 \alpha \]

**Hint:** Torque is the product of the radius and the force acting at that radius.

### Step 5: Relate Torque to Angular Acceleration
Using Newton's second law for rotation, we have:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the ring. The moment of inertia \( I \) of a ring about its axis is:
\[ I = m R^2 \]
Setting the expressions for torque equal gives:
\[ -q R^2 \alpha = m R^2 \cdot \alpha \]

### Step 6: Solve for Angular Acceleration
We can now solve for the angular acceleration \( \alpha \):
\[ -q R^2 \alpha = m R^2 \cdot \alpha \implies \alpha = \frac{-q \alpha}{m} \]
Thus, the angular acceleration of the ring is:
\[ \alpha = \frac{q \alpha}{2m} \]

**Hint:** Make sure to keep track of the signs and understand the relationship between torque, moment of inertia, and angular acceleration.

### Final Answer
The angular acceleration of the ring is given by:
\[ \alpha = \frac{q \alpha}{2m} \]