A semi-circular loop of radius R is placed in a uniform magnetic field as shown. It is pulled with a constant velocity. The induced emf in the loop is

Options

A. `Bv(piR) cos theta`
B. `Br(piR)sin theta`
C. `Bv(2R)cos theta`
D. `Bv(2R)sin theta`

Answer

To find the induced emf in a semi-circular loop of radius R that is being pulled with a constant velocity in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the radius of the semi-circular loop be \( R \).
- Let the uniform magnetic field be \( B \).
- Let the constant velocity with which the loop is pulled be \( V \).

2. **Understand the Geometry**:
- The semi-circular loop has a diameter \( 2R \).
- As the loop is pulled, only the segment of the loop that is moving through the magnetic field will contribute to the induced emf.

3. **Determine the Effective Length**:
- The effective length \( L \) of the segment of the loop that is moving through the magnetic field is equal to the diameter of the semi-circle, which is \( L = 2R \).

4. **Calculate the Induced EMF**:
- According to Faraday's law of electromagnetic induction, the induced emf \( \epsilon \) in a circuit is given by the formula:
\[ \epsilon = B \cdot V_{\perpendicular} \cdot L \]
- Here, \( V_{\perpendicular} \) is the component of the velocity perpendicular to the magnetic field. If the magnetic field is uniform and directed in a specific direction, we can express this as:
\[ V_{\perpendicular} = V \cdot \sin(\theta) \]
- Therefore, substituting \( V_{\perpendicular} \) into the emf equation gives:
\[ \epsilon = B \cdot (V \cdot \sin(\theta)) \cdot (2R) \]

5. **Final Expression for Induced EMF**:
- Thus, the induced emf in the semi-circular loop can be expressed as:
\[ \epsilon = B \cdot V \cdot 2R \cdot \sin(\theta) \]

### Conclusion:
The induced emf in the semi-circular loop being pulled in a uniform magnetic field is given by:
\[ \epsilon = B \cdot V \cdot 2R \cdot \sin(\theta) \]