A conducting ring of radius ris placed perpendicularly inside a time varying magnetic field given by `B=B_(0) + alphal`. `B_(0)` and a are positive constants. E.m.f. induced in the ring is

Options

A. `-pi alpha r`
B. `-pi alpha r^(2)`
C. `-pi alpha^(2) r^(2)`
D. `-pi alpha^(2) r`

Answer

To find the induced electromotive force (e.m.f.) in a conducting ring placed in a time-varying magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Magnetic Field**:
The magnetic field is given by \( B = B_0 + \alpha t \), where \( B_0 \) and \( \alpha \) are positive constants, and \( t \) is time.

2. **Identify the Area of the Ring**:
The area \( A \) of the conducting ring with radius \( R \) is given by:
\[ A = \pi R^2 \]

3. **Calculate the Magnetic Flux**:
The magnetic flux \( \Phi \) through the ring is given by:
\[ \Phi = B \cdot A = (B_0 + \alpha t) \cdot (\pi R^2) \]
Since the magnetic field is perpendicular to the area vector of the ring, we can simplify this to:
\[ \Phi = (B_0 + \alpha t) \pi R^2 \]

4. **Apply Faraday's Law of Electromagnetic Induction**:
According to Faraday's law, the induced e.m.f. \( \mathcal{E} \) is given by the negative rate of change of magnetic flux:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]

5. **Differentiate the Magnetic Flux**:
We need to differentiate \( \Phi \) with respect to time \( t \):
\[ \frac{d\Phi}{dt} = \frac{d}{dt} \left( (B_0 + \alpha t) \pi R^2 \right) \]
Since \( B_0 \) is a constant, its derivative is zero, and the derivative of \( \alpha t \) is \( \alpha \):
\[ \frac{d\Phi}{dt} = \pi R^2 \cdot \alpha \]

6. **Calculate the Induced E.M.F.**:
Now substituting back into the equation for e.m.f.:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\pi R^2 \alpha \]
Therefore, the induced e.m.f. in the ring is:
\[ \mathcal{E} = -\pi R^2 \alpha \]

### Final Answer:
The induced e.m.f. in the conducting ring is:
\[ \mathcal{E} = -\pi R^2 \alpha \]