The magnetic field perpendicular to the plane of a conducting ring of radius r changes at the rate.`(dB)/(dt)` . Then

Options

A. the emf induced in the ring is `pi r^(2) (dB)/(dt)`
B. the emf induced in the ring is `2pi r (dB)/(dt)`
C. diametrically opposite points on the ring are at the same potential
D. all points on the ring are at the same potential

Answer

To solve the problem, we need to analyze the situation where the magnetic field perpendicular to the plane of a conducting ring is changing over time. We will use Faraday's law of electromagnetic induction to find the induced electromotive force (emf) in the ring.

### Step-by-Step Solution:

1. **Identify the Magnetic Flux (Φ):**
The magnetic flux (Φ) through the conducting ring can be expressed as:
\[ \Phi = B \cdot A \]
where \(B\) is the magnetic field and \(A\) is the area of the ring. For a ring of radius \(r\), the area \(A\) is given by:
\[ A = \pi r^2 \]
Thus, the magnetic flux becomes:
\[ \Phi = B \cdot \pi r^2 \]

2. **Differentiate the Magnetic Flux with Respect to Time:**
To find the induced emf (E), we need to differentiate the magnetic flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt}(\pi r^2 B) \]
Since the radius \(r\) is constant, we can simplify this to:
\[ \frac{d\Phi}{dt} = \pi r^2 \frac{dB}{dt} \]

3. **Apply Faraday's Law of Electromagnetic Induction:**
According to Faraday's law, the induced emf (E) in the ring is equal to the negative rate of change of magnetic flux:
\[ E = -\frac{d\Phi}{dt} \]
Substituting our expression for \(\frac{d\Phi}{dt}\):
\[ E = -\pi r^2 \frac{dB}{dt} \]

4. **Determine the Sign of the Induced emf:**
The negative sign indicates the direction of the induced emf according to Lenz's law, but for the magnitude of the emf, we can write:
\[ E = \pi r^2 \frac{dB}{dt} \]

5. **Analyze the Options:**
Now we can analyze the options provided in the question:
- The first option states that the induced emf is \(E = \pi r^2 \frac{dB}{dt}\), which matches our derived expression.
- The second option states \(E = 2\pi r \frac{dB}{dt}\), which is incorrect.
- The third option claims that diametrically opposite points on the ring are at the same potential, which is incorrect as there will be a potential difference due to the induced emf.
- The fourth option states that all points on the ring are at the same potential, which is correct because the induced emf distributes evenly around the ring.

### Conclusion:
The correct answers are:
- The induced emf in the ring is given by \(E = \pi r^2 \frac{dB}{dt}\) (Option 1).
- All points on the ring are at the same potential (Option 4).