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)`

Answer

To solve the problem, we need to analyze the situation involving a non-conducting ring placed on a rough horizontal surface in a time-varying magnetic field. We will derive the expression for the coefficient of friction (μ) that allows the ring to start rotating after 2 seconds.

### Step-by-Step Solution:

1. **Identify the Magnetic Field and its Rate of Change**:
The magnetic field is given by \( B = 4t^2 \). To find the induced electromotive force (EMF), we first need to calculate the rate of change of the magnetic field:
\[ \frac{dB}{dt} = \frac{d}{dt}(4t^2) = 8t \]

2. **Calculate the Induced EMF**:
The induced EMF (ε) in the ring can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux. The area \( A \) of the ring is \( \pi R^2 \), thus:
\[ \Phi = B \cdot A = B \cdot \pi R^2 = 4t^2 \cdot \pi R^2 \]
Therefore, the induced EMF is:
\[ \epsilon = -\frac{d}{dt}(4t^2 \cdot \pi R^2) = -8t \cdot \pi R^2 \]

3. **Determine the Induced Electric Field**:
The induced electric field \( E \) can be related to the induced EMF by the formula:
\[ \epsilon = E \cdot (2\pi R) \]
Thus, we can express the electric field as:
\[ E = \frac{\epsilon}{2\pi R} = \frac{-8t \cdot \pi R^2}{2\pi R} = -4tR \]

4. **Calculate the Torque due to Electric Force**:
The electric force \( F \) acting on the charge \( q \) distributed uniformly on the ring due to the electric field is:
\[ F = qE = q(-4tR) = -4qtR \]
The torque \( \tau \) caused by this force about the center of the ring is:
\[ \tau = F \cdot R = -4qtR^2 \]

5. **Calculate the Torque due to Friction**:
The torque due to friction \( \tau_f \) that opposes the motion is given by:
\[ \tau_f = \mu mgR \]

6. **Set the Torques Equal at t = 2 seconds**:
The ring starts rotating at \( t = 2 \) seconds, so we set the magnitudes of the torques equal:
\[ 4q(2)R^2 = \mu mgR \]

7. **Solve for the Coefficient of Friction (μ)**:
Rearranging the equation gives:
\[ \mu = \frac{4q(2)R^2}{mgR} = \frac{8qR}{mg} \]

### Final Expression for Coefficient of Friction:
\[ \mu = \frac{8qR}{mg} \]