the uniform magnetic field perpendicular to the plane of a conducting ring of radius a change at the rate of `alpha`, then

Options

A. all the points ont the ring are the same potential
B. the emf induced in the ring is `pia^(2)alpha`
C. electric field intensity E at anny point on the ring is zero
D. `E=(aalpha)/(2))`

Answer

To solve the problem, we need to determine the induced EMF in a conducting ring when the uniform magnetic field perpendicular to its plane changes at a rate of `alpha`.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Radius of the conducting ring: \( a \)
- Rate of change of magnetic field: \( \frac{dB}{dt} = \alpha \)

2. **Calculate the Area of the Ring:**
- The area \( A \) of the conducting ring is given by the formula:
\[ A = \pi a^2 \]

3. **Use Faraday's Law of Electromagnetic Induction:**
- According to Faraday's law, the induced EMF (\( \mathcal{E} \)) in a closed loop is given by:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} \]
- Where \( \Phi_B \) is the magnetic flux. The magnetic flux \( \Phi_B \) through the ring is given by:
\[ \Phi_B = B \cdot A = B \cdot \pi a^2 \]

4. **Calculate the Rate of Change of Magnetic Flux:**
- Since the magnetic field \( B \) changes at a rate of \( \alpha \), we have:
\[ \frac{d\Phi_B}{dt} = \frac{d}{dt}(B \cdot \pi a^2) = \pi a^2 \cdot \frac{dB}{dt} = \pi a^2 \cdot \alpha \]

5. **Substitute into the Induced EMF Formula:**
- Therefore, the induced EMF becomes:
\[ \mathcal{E} = -\frac{d\Phi_B}{dt} = -\pi a^2 \cdot \alpha \]
- The magnitude of the induced EMF is:
\[ |\mathcal{E}| = \pi a^2 \cdot \alpha \]

6. **Determine the Electric Field in the Ring:**
- The induced EMF can also be related to the electric field \( E \) around the ring:
\[ \mathcal{E} = E \cdot L \]
- Where \( L \) is the circumference of the ring, given by \( L = 2\pi a \).
- Thus, we have:
\[ E \cdot 2\pi a = \pi a^2 \cdot \alpha \]
- Solving for \( E \):
\[ E = \frac{\pi a^2 \cdot \alpha}{2\pi a} = \frac{a \cdot \alpha}{2} \]

### Conclusion:
The induced EMF in the conducting ring is \( |\mathcal{E}| = \pi a^2 \cdot \alpha \) and the electric field intensity \( E \) at any point on the ring is \( E = \frac{a \cdot \alpha}{2} \).