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: Calculate the Magnetic Flux (Φ)
The magnetic field is given as \( B = 4t^3 \, \text{T} \). The area \( A \) of the ring is given by:
\[ A = \pi R^2 \]
where \( R = 0.1 \, \text{m} \) (10 cm converted to meters). Thus,
\[ A = \pi (0.1)^2 = \pi (0.01) = 0.01\pi \, \text{m}^2 \]

The magnetic flux \( \Phi \) through the ring is:
\[ \Phi = B \cdot A = (4t^3) \cdot (0.01\pi) = 0.04\pi t^3 \]

### Step 2: Calculate the EMF (Electromotive Force)
Using Faraday's law of electromagnetic induction, the EMF \( \mathcal{E} \) induced in the ring is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Calculating the derivative:
\[ \mathcal{E} = -\frac{d}{dt}(0.04\pi t^3) = -0.12\pi t^2 \]

### Step 3: Calculate the EMF at \( t = 5 \, \text{s} \)
Substituting \( t = 5 \):
\[ \mathcal{E} = -0.12\pi (5^2) = -0.12\pi (25) = -3\pi \, \text{V} \]

### Step 4: Calculate the Force on the Ring
The force \( F \) acting on the ring due to the electric field generated by the EMF is:
\[ F = Q \cdot \mathcal{E} \]
where \( Q = 2 \, \text{C} \):
\[ F = 2 \cdot (3\pi) = 6\pi \, \text{N} \]

### Step 5: Relate the Force to Friction
The frictional force \( F_f \) that prevents the ring from sliding is given by:
\[ F_f = \mu mg \]
where \( m = 4 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \):
\[ F_f = \mu (4)(10) = 40\mu \]

### Step 6: Set the Forces Equal
At the moment the ring starts to rotate, the force due to the EMF equals the frictional force:
\[ 6\pi = 40\mu \]
Thus,
\[ \mu = \frac{6\pi}{40} = \frac{3\pi}{20} \]

### Step 7: Relate Coefficient of Friction to \( k \)
We know from the problem statement that:
\[ \mu = \frac{k}{24} \]
Setting the two expressions for \( \mu \) equal gives:
\[ \frac{3\pi}{20} = \frac{k}{24} \]

### Step 8: Solve for \( k \)
Cross-multiplying to solve for \( k \):
\[ 3\pi \cdot 24 = 20k \]
\[ k = \frac{72\pi}{20} = \frac{18\pi}{5} \]

### Step 9: Approximate \( k \)
Using \( \pi \approx 3.14 \):
\[ k \approx \frac{18 \cdot 3.14}{5} \approx \frac{56.52}{5} \approx 11.30 \]

### Conclusion
The value of \( k \) is approximately \( 11.30 \).