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

Options

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

Explanation

To solve the problem, we need to analyze the situation where a uniform magnetic field perpendicular to the plane of a conducting ring of radius \( a \) is changing at the rate of \( \alpha \). We will derive the induced electromotive force (emf) and the electric field in the ring step by step.

### Step 1: Calculate the Magnetic Flux
The magnetic flux \( \Phi \) through the conducting ring is given by the formula:
\[ \Phi = B \cdot A \]
where \( B \) is the magnetic field and \( A \) is the area of the ring. The area \( A \) of the ring with radius \( a \) is:
\[ A = \pi a^2 \]
Thus, the magnetic flux becomes:
\[ \Phi = B \cdot \pi a^2 \]

### Step 2: Relate the Change in Magnetic Field to Time
Given that the magnetic field \( B \) is changing at a rate \( \alpha \), we can express this as:
\[ \frac{dB}{dt} = \alpha \]

### Step 3: Apply Faraday's Law of Electromagnetic Induction
According to Faraday's law, the induced emf \( \mathcal{E} \) in the ring is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
Substituting the expression for magnetic flux, we have:
\[ \mathcal{E} = -\frac{d}{dt}(B \cdot \pi a^2) \]
Since \( A \) is constant, we can write:
\[ \mathcal{E} = -\pi a^2 \frac{dB}{dt} \]
Substituting \( \frac{dB}{dt} = \alpha \):
\[ \mathcal{E} = -\pi a^2 \alpha \]
The magnitude of the induced emf is:
\[ |\mathcal{E}| = \pi a^2 \alpha \]

### Step 4: Calculate the Induced Electric Field
The induced electric field \( E \) around the ring can be related to the induced emf by the formula:
\[ \mathcal{E} = \oint E \cdot dl \]
For a circular path of radius \( a \), the length \( dl \) is the circumference:
\[ dl = 2\pi a \]
Thus, we have:
\[ \mathcal{E} = E \cdot 2\pi a \]
Substituting the expression for induced emf:
\[ \pi a^2 \alpha = E \cdot 2\pi a \]
Solving for \( E \):
\[ E = \frac{\pi a^2 \alpha}{2\pi a} = \frac{a \alpha}{2} \]

### Step 5: Check the Potential Difference
To check if all points on the ring are at the same potential, we consider a small segment \( dl \) of the ring. The potential difference \( dV \) across this segment can be expressed as:
\[ dV = E \cdot dl \]
If we assume a current \( I \) flows through the ring, the potential drop across the resistance \( R \) of the ring can be expressed as:
\[ dV = -I \cdot dR \]
Since \( dR \) is proportional to \( dl \), we find that the potential difference across any segment is zero if the induced electric field is uniform. Thus, all points on the ring are at the same potential.

### Final Conclusion
The induced emf in the ring is \( \pi a^2 \alpha \), the induced electric field is \( \frac{a \alpha}{2} \), and all points on the ring are at the same potential.