Metal ring of radius R is placed perpendicular to uniform magnetic field B. Magnetic field starts changing at a rate `alpha`

Options

A. Electric field at any point on the ring is zero.
B. Electric field at any point on the ring is `alpha (R)//(2)`
C. Emf induced in the ring is zero
D. Emf induced in the ring is `piaR^(2)`

Explanation

To solve the problem, we need to determine the induced electromotive force (EMF) and the electric field at any point on the metal ring when the magnetic field is changing at a rate of \( \alpha \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a metal ring of radius \( R \) placed perpendicular to a uniform magnetic field \( B \).
- The magnetic field \( B \) is changing at a rate \( \alpha \), which can be mathematically represented as \( \frac{dB}{dt} = \alpha \).

2. **Calculating Magnetic Flux**:
- The magnetic flux \( \Phi \) through the ring is given by the formula:
\[ \Phi = B \cdot A \]
where \( A \) is the area of the ring. Since the ring is perpendicular to the magnetic field, we can simplify this to:
\[ \Phi = B \cdot \pi R^2 \]
Here, \( \pi R^2 \) is the area of the circle formed by the ring.

3. **Finding the Induced EMF**:
- According to Faraday's law of electromagnetic induction, the induced EMF (\( \mathcal{E} \)) is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
- Since \( \Phi = B \cdot \pi R^2 \), we differentiate it with respect to time:
\[ \frac{d\Phi}{dt} = \pi R^2 \frac{dB}{dt} \]
- Substituting \( \frac{dB}{dt} = \alpha \):
\[ \frac{d\Phi}{dt} = \pi R^2 \alpha \]
- Thus, the induced EMF is:
\[ \mathcal{E} = -\pi R^2 \alpha \]

4. **Calculating the Electric Field**:
- The induced EMF is also related to the electric field \( E \) around the ring and the circumference of the ring:
\[ \mathcal{E} = E \cdot L \]
where \( L \) is the circumference of the ring, given by \( L = 2\pi R \).
- Therefore, we can express the electric field as:
\[ E = \frac{\mathcal{E}}{L} = \frac{-\pi R^2 \alpha}{2\pi R} \]
- Simplifying this gives:
\[ E = -\frac{R \alpha}{2} \]

5. **Final Results**:
- The induced EMF is:
\[ \mathcal{E} = \pi R^2 \alpha \]
- The electric field at any point on the ring is:
\[ E = \frac{R \alpha}{2} \]

### Summary of Answers:
- Induced EMF: \( \mathcal{E} = \pi R^2 \alpha \)
- Electric Field: \( E = \frac{R \alpha}{2} \)