A non-conducting ring having q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field `B = 4t^(2)` is switched on at time t = 0. Mass of the ring is m and radius is R.
The ring starts rotating after 2 s, the coefficient of friction between the ring and the table is

Options

A. `(4qmR)/(g)`
B. `(2qmR)/(g)`
C. `(8qR)/(mg)`
D. `(qR)/(2mg)`

Explanation

To solve the problem, we need to find the coefficient of friction (μ) between the non-conducting ring and the rough horizontal surface. The ring starts rotating after 2 seconds when a vertical time-varying magnetic field is applied.

### Step-by-step Solution:

1. **Identify the given parameters:**
- Charge on the ring, \( q \)
- Mass of the ring, \( m \)
- Radius of the ring, \( R \)
- Magnetic field, \( B(t) = 4t^2 \)
- Time when the ring starts rotating, \( t = 2 \, \text{s} \)

2. **Calculate the induced electric field (E):**
- The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A = B \cdot \pi R^2 \]
- The induced electromotive force (emf) is given by Faraday's law:
\[ \text{emf} = -\frac{d\Phi}{dt} \]
- Since \( B(t) = 4t^2 \):
\[ \Phi = 4t^2 \cdot \pi R^2 \]
- Differentiate \( \Phi \) with respect to time \( t \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(4t^2 \cdot \pi R^2) = 8t \cdot \pi R^2 \]
- Therefore, the induced emf (which is equal to the electric field times the circumference of the ring) is:
\[ \text{emf} = -8t \cdot \pi R^2 \]
- The electric field \( E \) around the ring is:
\[ E = \frac{\text{emf}}{2\pi R} = -\frac{8t \cdot \pi R^2}{2\pi R} = -4tR \]

3. **Calculate the electric field at \( t = 2 \) seconds:**
- Substitute \( t = 2 \):
\[ E = -4(2)R = -8R \]
- We can neglect the negative sign for magnitude:
\[ E = 8R \]

4. **Calculate the torque due to the electric field:**
- The force \( F \) acting on the ring due to the electric field is:
\[ F = qE = q(8R) = 8qR \]
- The torque \( \tau_1 \) due to this force is:
\[ \tau_1 = F \cdot R = (8qR) \cdot R = 8qR^2 \]

5. **Calculate the torque due to friction:**
- The frictional force \( F_f \) is given by:
\[ F_f = \mu mg \]
- The torque \( \tau_2 \) due to friction is:
\[ \tau_2 = F_f \cdot R = (\mu mg) \cdot R \]

6. **Set the torques equal for equilibrium:**
- At the point when the ring starts rotating, the torques are equal:
\[ \tau_1 = \tau_2 \]
\[ 8qR^2 = \mu mgR \]

7. **Solve for the coefficient of friction \( \mu \):**
- Cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ 8qR = \mu mg \]
- Rearranging gives:
\[ \mu = \frac{8qR}{mg} \]

### Final Answer:
The coefficient of friction \( \mu \) between the ring and the table is:
\[ \mu = \frac{8qR}{mg} \]