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, we will break it down into two parts as specified in the question.

### Part (a): Finding the friction coefficient \( \mu \)

1. **Understanding the Induced Electric Field**:
The magnetic field is given by \( B = B_0 t^2 \). The induced electric field \( E \) in the ring can be calculated using Faraday's law of electromagnetic induction, which states:
\[ \oint E \cdot dl = -\frac{d\Phi_B}{dt} \]
where \( \Phi_B \) is the magnetic flux through the ring.

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

3. **Finding the Rate of Change of Flux**:
Now, we differentiate \( \Phi_B \) with respect to time \( t \):
\[ \frac{d\Phi_B}{dt} = \frac{d}{dt}(B_0 t^2 \cdot \pi R^2) = 2B_0 t \cdot \pi R^2 \]

4. **Using Faraday's Law**:
Substituting this into Faraday's law gives:
\[ E \cdot 2\pi R = -2B_0 t \cdot \pi R^2 \]
Simplifying, we find:
\[ E = -\frac{B_0 t R}{1} \]

5. **Calculating the Force and Torque**:
The force \( F \) due to the electric field on the charge \( Q \) is:
\[ F = E \cdot Q = -\frac{B_0 t R}{1} \cdot Q \]
The torque \( \tau \) exerted by this force about the center of the ring is:
\[ \tau = F \cdot R = -\frac{B_0 t R^2 Q}{1} \]

6. **Setting Up the Friction Condition**:
At \( t = 2 \) seconds, the ring is just about to rotate, meaning the torque due to friction \( \tau_f \) must equal the torque due to the electric field:
\[ \tau_f = \mu mg R \]
Therefore, we have:
\[ \frac{B_0 (2) R^2 Q}{1} = \mu mg R \]

7. **Solving for \( \mu \)**:
Rearranging gives:
\[ \mu = \frac{B_0 (2) R Q}{mg} \]

### Part (b): Finding the angle rotated before stopping

1. **Finding Angular Velocity**:
After \( t = 2 \) seconds, the net torque \( \tau_{net} \) acting on the ring is:
\[ \tau_{net} = \tau_{electric} - \tau_f = \frac{B_0 t R^2 Q}{1} - \mu mg R \]
Substituting \( \mu \):
\[ \tau_{net} = \frac{B_0 t R^2 Q}{1} - \frac{B_0 (2) R Q}{mg} mg R \]
This simplifies to:
\[ \tau_{net} = \frac{B_0 t R^2 Q}{1} - 2 B_0 R^2 Q \]

2. **Finding Angular Acceleration**:
Using \( \tau_{net} = I \alpha \) where \( I = mR^2 \):
\[ \alpha = \frac{\tau_{net}}{I} = \frac{\frac{B_0 t R^2 Q}{1} - 2 B_0 R^2 Q}{mR^2} \]

3. **Integrating to Find Angular Velocity**:
From \( t = 2 \) to \( t = 4 \), we integrate to find the angular velocity \( \omega \):
\[ d\omega = \alpha dt \]

4. **Using Kinematics**:
The final angular velocity \( \omega \) at \( t = 4 \) seconds can be calculated, and then we can use the kinematic equation:
\[ \omega^2 = \omega_0^2 - 2\alpha \theta \]
where \( \omega_0 \) is the initial angular velocity at \( t = 4 \) seconds.

5. **Finding the Angle \( \theta \)**:
Rearranging gives:
\[ \theta = \frac{\omega^2}{2\alpha} \]

### Summary of Solutions:
1. **Friction Coefficient \( \mu \)**:
\[ \mu = \frac{B_0 (2) R Q}{mg} \]

2. **Angle Rotated \( \theta \)**:
\[ \theta = \frac{\omega^2}{2\alpha} \]