A metal rod of length L is placed normal to a magnetic field and rotated through one end of rod in circular path with frequency f. The potential difference between it ends will be-

Options

A. `piL^(2)Bf`
B. `BL//f`
C. `piL^(2)B//f`
D. fBL

Explanation

To find the potential difference between the ends of a metal rod of length \( L \) that is rotated in a magnetic field with frequency \( f \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- A metal rod of length \( L \) is placed perpendicular (normal) to a magnetic field \( B \).
- The rod is rotated about one of its ends in a circular path with a frequency \( f \).

2. **Identify the Formula for EMF**:
- The electromotive force (emf) induced in a rotating rod in a magnetic field is given by the formula:
\[ E = \frac{1}{2} B \omega L^2 \]
where \( E \) is the induced emf, \( B \) is the magnetic field strength, \( \omega \) is the angular velocity, and \( L \) is the length of the rod.

3. **Convert Frequency to Angular Velocity**:
- The angular velocity \( \omega \) can be related to the frequency \( f \) by the formula:
\[ \omega = 2 \pi f \]

4. **Substitute \( \omega \) into the EMF Formula**:
- Replace \( \omega \) in the emf formula:
\[ E = \frac{1}{2} B (2 \pi f) L^2 \]

5. **Simplify the Expression**:
- Simplifying the equation gives:
\[ E = B \pi f L^2 \]

6. **Final Result**:
- Therefore, the potential difference (emf) between the ends of the rod is:
\[ E = \pi L^2 B f \]

### Conclusion:
The potential difference between the ends of the rod is given by:
\[ E = \pi L^2 B f \]