A charge q is uniformly distributed along the periphery of a non conducting uniform circular disc pivoted at its centre in a vertical plane and can freely rotate about a horizontal axis passing through its centre. A cylindrical region with its axis passing through the centre of the disc has a varying magnetic field B = kt as shown in the adjacent figure. wrapped over the disc as shown and remains stationary, then

Options

A. `m=(kqR)/(2g)`
B. `m=(2kqR)/(2g)`
C. Induced electric field at the centre of the disc, E=kR/2
D. Induced electric field at all points on the periphery of the disc , E=2kR

Answer

To solve the problem, we need to analyze the situation step by step. The charge \( q \) is uniformly distributed along the periphery of a non-conducting circular disc that can rotate about a horizontal axis through its center. A cylindrical region with a varying magnetic field \( B = kt \) is present. We will find the induced electric field and the forces acting on the disc.

### Step 1: Understanding the Magnetic Field
The magnetic field is given by \( B = kt \), where \( k \) is a constant and \( t \) is time. This indicates that the magnetic field is increasing linearly with time.

**Hint:** Remember that the magnetic field is varying with time, which will lead to an induced electric field.

### Step 2: Calculate the Induced EMF
The induced electromotive force (EMF) \( \epsilon \) in a closed loop can be calculated using Faraday's law of electromagnetic induction:
\[ \epsilon = -\frac{d\Phi_B}{dt} \]
where \( \Phi_B \) is the magnetic flux through the area enclosed by the loop.

For a circular loop of radius \( R \):
\[ \Phi_B = B \cdot A = B \cdot \pi R^2 \]
Substituting \( B = kt \):
\[ \Phi_B = kt \cdot \pi R^2 \]

Now, differentiate this with respect to time \( t \):
\[ \frac{d\Phi_B}{dt} = \pi R^2 k \]
Thus, the induced EMF becomes:
\[ \epsilon = -\pi R^2 k \]

**Hint:** The negative sign indicates the direction of the induced EMF according to Lenz's law, but we will focus on its magnitude for calculating the induced electric field.

### Step 3: Calculate the Induced Electric Field
The induced electric field \( E \) around the circular path can be related to the induced EMF:
\[ \epsilon = E \cdot 2\pi R \]
Substituting for \( \epsilon \):
\[ -\pi R^2 k = E \cdot 2\pi R \]
Solving for \( E \):
\[ E = -\frac{kR}{2} \]

**Hint:** The induced electric field is directed in a way to oppose the change in magnetic flux.

### Step 4: Calculate the Force on the Charge
The force \( F \) acting on the charge \( q \) due to the induced electric field is given by:
\[ F = qE \]
Substituting for \( E \):
\[ F = q \left(-\frac{kR}{2}\right) = -\frac{qkR}{2} \]

**Hint:** The negative sign indicates the direction of the force, which is opposite to the direction of the electric field.

### Step 5: Considering the Weight of the Disc
The weight of the disc is given by \( mg \). For the disc to be in equilibrium, the upward force due to the electric field must balance the downward gravitational force:
\[ mg = \frac{qkR}{2} \]

**Hint:** This equation relates the mass of the disc to the charge and the induced electric field.

### Step 6: Solve for Mass \( m \)
Rearranging the equation gives:
\[ m = \frac{qkR}{2g} \]

**Hint:** This equation shows how the mass of the disc is influenced by the charge distribution, the rate of change of the magnetic field, and the gravitational acceleration.

### Final Answer
The mass \( m \) of the disc is given by:
\[ m = \frac{qkR}{2g} \]