A non-conducting ring of mass `m` and radius `R` has a charge `Q` uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that plane of the ring is parallel to the surface. A vertical magnetic field `B = B_0t^2` tesla is switched on. After 2 a from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre.
(a) Find friction coefficient `mu` between the ring and the surface.
(b) If magnetic field is switched off after `4 s`, then find the angle rotated by the ring before coming to stop after switching off the magnetic field.

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Finding the Coefficient of Friction (μ)

1. **Understanding the System**:
- We have a non-conducting ring of mass `m` and radius `R` with a uniform charge `Q` distributed over its circumference.
- A vertical magnetic field `B = B_0 t^2` is applied, which changes with time.

2. **Calculating Magnetic Flux (Φ)**:
- The magnetic flux through the ring is given by:
\[ \Phi = B \cdot A = B_0 t^2 \cdot \pi R^2 \]
- Here, \( A = \pi R^2 \) is the area of the ring.

3. **Finding Induced EMF (ε)**:
- The induced electromotive force (EMF) is given by Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi}{dt} \]
- Calculating the derivative:
\[ \epsilon = -\frac{d}{dt}(B_0 t^2 \cdot \pi R^2) = -\pi R^2 \cdot 2B_0 t = 2\pi R^2 B_0 t \]

4. **Induced Electric Field (E)**:
- The induced electric field \( E \) around the ring can be expressed as:
\[ E = \frac{\epsilon}{2\pi R} = \frac{2B_0 t R}{2} = B_0 t R \]

5. **Force on the Ring**:
- The electrostatic force \( F \) acting on the ring due to the induced electric field is:
\[ F = Q \cdot E = Q \cdot (B_0 t R) = Q B_0 t R \]

6. **Torque (τ) Acting on the Ring**:
- The torque due to this force about the center of the ring is:
\[ \tau = F \cdot R = Q B_0 t R^2 \]

7. **Torque Due to Friction**:
- The torque due to friction is given by:
\[ \tau_{\text{friction}} = \mu m g R \]

8. **Setting Torques Equal**:
- At \( t = 2 \) seconds, the ring is about to rotate, so we set the torques equal:
\[ Q B_0 (2) R^2 = \mu m g R \]

9. **Solving for μ**:
- Rearranging gives:
\[ \mu = \frac{Q B_0 (2) R}{m g} \]

### Part (b): Finding the Angle Rotated (θ) After Switching Off the Magnetic Field

1. **Angular Acceleration (α)**:
- After the magnetic field is switched off at \( t = 4 \) seconds, the only torque acting on the ring is due to friction:
\[ \tau = \mu m g R \]
- The moment of inertia \( I \) of the ring is \( I = m R^2 \).

2. **Relating Torque and Angular Acceleration**:
- Using \( \tau = I \alpha \):
\[ \mu m g R = m R^2 \alpha \implies \alpha = \frac{\mu g}{R} \]

3. **Finding Angular Velocity (ω) at \( t = 4 \) seconds**:
- The angular velocity \( \omega \) at \( t = 4 \) seconds can be found from the previous calculations:
\[ \omega = \frac{2 Q B_0}{m} \]

4. **Using Kinematic Equation**:
- To find the angle \( \theta \) rotated before coming to a stop:
\[ \omega^2 = 2 \alpha \theta \]
- Rearranging gives:
\[ \theta = \frac{\omega^2}{2 \alpha} \]

5. **Substituting Values**:
- Substitute \( \omega \) and \( \alpha \):
\[ \theta = \frac{\left(\frac{2 Q B_0}{m}\right)^2}{2 \cdot \frac{\mu g}{R}} = \frac{2 Q^2 B_0^2 R}{m^2 \mu g} \]

### Final Answers:
- (a) The coefficient of friction \( \mu = \frac{Q B_0 (2) R}{m g} \)
- (b) The angle rotated \( \theta = \frac{2 Q^2 B_0^2 R}{m^2 \mu g} \)