A conducting ring of radius r and resistance R is placed in region of uniform time varying magnetic field B which is perpendicular to the plane of the ring. It the magnetic field is changing at a rate `alpha`, then the current induced in the ring is

Options

A. `(pir^2 alpha)/(2R)`
B. `(2pi r alpha)/(R)`
C. `(pi r alpha)/(R)`
D. `(pir^(2) alpha)/R`

Answer

To find the induced current in a conducting ring placed in a time-varying magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Radius of the ring: \( r \)
- Resistance of the ring: \( R \)
- Rate of change of magnetic field: \( \alpha \) (i.e., \( \frac{dB}{dt} = \alpha \))

2. **Understand the Concept of Induced EMF**:
- According to Faraday's law of electromagnetic induction, the induced electromotive force (EMF) in a closed loop is equal to the negative rate of change of magnetic flux through the loop.
- The formula for induced EMF (\( \mathcal{E} \)) is given by:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
- Where \( \Phi \) is the magnetic flux given by \( \Phi = B \cdot A \cdot \cos(\theta) \). Here, \( A \) is the area of the ring and \( \theta \) is the angle between the magnetic field and the normal to the surface of the ring.

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

4. **Determine the Magnetic Flux**:
- Since the magnetic field \( B \) is perpendicular to the plane of the ring, \( \theta = 0 \) and \( \cos(0) = 1 \).
- Thus, the magnetic flux becomes:
\[ \Phi = B \cdot A = B \cdot \pi r^2 \]

5. **Calculate the Rate of Change of Magnetic Flux**:
- The rate of change of magnetic flux is:
\[ \frac{d\Phi}{dt} = \frac{d(B \cdot \pi r^2)}{dt} = \pi r^2 \frac{dB}{dt} = \pi r^2 \alpha \]

6. **Find the Induced EMF**:
- Substituting into the induced EMF formula:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\pi r^2 \alpha \]
- The magnitude of the induced EMF is:
\[ |\mathcal{E}| = \pi r^2 \alpha \]

7. **Calculate the Induced Current**:
- Using Ohm's law, the induced current \( I \) can be calculated as:
\[ I = \frac{\mathcal{E}}{R} = \frac{\pi r^2 \alpha}{R} \]

### Final Answer:
The induced current in the ring is:
\[ I = \frac{\pi r^2 \alpha}{R} \]