A metal disc of radius R rotates with an angular velcoity `omega` about an axis perpendiclar to its plane passing through its centre in a magnetic field B acting perpendicular to the plane of the disc. Calculate the induced emf between the rim and the axis of the disc.

Options

A. `pwBR^(2)`
B. `wBR^(2)`
C. `pwB(R^(2))/2`
D. `wB (R^(2))/2`

Answer

To solve the problem of calculating the induced EMF between the rim and the axis of a rotating metal disc in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a metal disc of radius \( R \) rotating with an angular velocity \( \omega \) about an axis perpendicular to its plane and passing through its center.
- A magnetic field \( B \) is acting perpendicular to the plane of the disc.

2. **Induced EMF Formula**:
- The induced EMF (\( E \)) in a circuit is given by the formula:
\[ E = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux.

3. **Calculating Magnetic Flux**:
- The magnetic flux \( \Phi \) through the disc is given by:
\[ \Phi = B \cdot A \]
where \( A \) is the area swept by the rotating disc. Initially, when the disc has not rotated, the area is \( 0 \).

4. **Area Swept During Rotation**:
- After one complete rotation, the area swept by the radius \( R \) is:
\[ A = \pi R^2 \]
Therefore, the final magnetic flux after one complete rotation is:
\[ \Phi_{\text{final}} = B \cdot \pi R^2 \]

5. **Time Period of Rotation**:
- The time period \( T \) for one complete rotation is given by:
\[ T = \frac{2\pi}{\omega} \]

6. **Change in Magnetic Flux**:
- The change in magnetic flux \( d\Phi \) over the time period \( T \) is:
\[ d\Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} = B \cdot \pi R^2 - 0 = B \cdot \pi R^2 \]

7. **Calculating Induced EMF**:
- Now substituting \( d\Phi \) and \( dt \) into the EMF formula:
\[ E = -\frac{d\Phi}{dt} = -\frac{B \cdot \pi R^2}{T} \]
Substituting \( T = \frac{2\pi}{\omega} \):
\[ E = -\frac{B \cdot \pi R^2}{\frac{2\pi}{\omega}} = -\frac{B \cdot \pi R^2 \cdot \omega}{2\pi} \]
Simplifying this gives:
\[ E = \frac{B R^2 \omega}{2} \]

8. **Final Result**:
- The induced EMF between the rim and the axis of the disc is:
\[ E = \frac{B R^2 \omega}{2} \]