A non - conducting ring of mass m = 4 kg and radius R = 10 cm has charge Q = 2 C uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that the plane of the ring is parallel to the surface. A vertical magnetic field `B=4t^(3)T` is switched on at t = 0. At t = 5 s ring starts to rotate about the vertical axis through the centre. The coefficient of friction between the ring and the surface is found to be `(k)/(24)`. Then the value of k is

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a non-conducting ring with mass \( m = 4 \, \text{kg} \), radius \( R = 10 \, \text{cm} = 0.1 \, \text{m} \), and charge \( Q = 2 \, \text{C} \). A vertical magnetic field \( B = 4t^3 \, \text{T} \) is switched on at \( t = 0 \). The ring starts to rotate about its vertical axis at \( t = 5 \, \text{s} \). We need to find the coefficient of friction \( \mu \) expressed as \( \frac{k}{24} \) and determine the value of \( k \).

### Step 2: Calculate the Induced Electric Field
Using Faraday's law of electromagnetic induction, we know that the induced electric field \( E \) in the ring can be calculated from the rate of change of magnetic flux.

1. **Magnetic Flux (\( \Phi \))**:
\[ \Phi = B \cdot A = B \cdot \pi R^2 \]
where \( A = \pi R^2 \) is the area of the ring.

2. **Rate of Change of Magnetic Flux**:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(B \cdot \pi R^2) = \pi R^2 \frac{dB}{dt} \]
Given \( B = 4t^3 \):
\[ \frac{dB}{dt} = 12t^2 \]
Therefore,
\[ \frac{d\Phi}{dt} = \pi R^2 \cdot 12t^2 \]

3. **Induced Electric Field**:
By Faraday's law:
\[ E \cdot 2\pi R = \frac{d\Phi}{dt} \]
Thus,
\[ E = \frac{12\pi R^2 t^2}{2\pi R} = 6Rt^2 \]

### Step 3: Calculate the Torque Due to Electric Field
The force \( F \) on the ring due to the electric field is:
\[ F = Q \cdot E = Q \cdot 6Rt^2 \]
The torque \( \tau \) about the center of the ring is:
\[ \tau = F \cdot R = Q \cdot 6Rt^2 \cdot R = 6QR^2t^2 \]

### Step 4: Calculate the Maximum Static Friction Torque
The maximum static friction force \( F_s \) is given by:
\[ F_s = \mu mg \]
Thus, the torque due to static friction is:
\[ \tau_{fs} = F_s \cdot R = \mu mgR \]

### Step 5: Set the Torques Equal
At the moment the ring starts to rotate, the torques are equal:
\[ 6QR^2t^2 = \mu mgR \]
Cancelling \( R \) from both sides:
\[ 6QRt^2 = \mu mg \]
Solving for \( \mu \):
\[ \mu = \frac{6QRt^2}{mg} \]

### Step 6: Substitute Values
Substituting \( Q = 2 \, \text{C} \), \( R = 0.1 \, \text{m} \), \( t = 5 \, \text{s} \), \( m = 4 \, \text{kg} \), and \( g = 10 \, \text{m/s}^2 \):
\[ \mu = \frac{6 \cdot 2 \cdot 0.1 \cdot 5^2}{4 \cdot 10} \]
Calculating:
\[ \mu = \frac{6 \cdot 2 \cdot 0.1 \cdot 25}{40} = \frac{30}{40} = 0.75 \]

### Step 7: Relate \( \mu \) to \( k \)
Given \( \mu = \frac{k}{24} \):
\[ 0.75 = \frac{k}{24} \]
Thus,
\[ k = 0.75 \cdot 24 = 18 \]

### Final Answer
The value of \( k \) is \( \boxed{18} \).