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 solve the problem of finding the potential difference (induced EMF) across the ends of a rotating metal rod in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the Relevant Formula**:
The induced electromotive force (EMF) in a rotating rod in a magnetic field can be calculated using the formula:
\[ E = \frac{1}{2} B \omega L^2 \]
where \( \omega \) is the angular frequency in radians per second.

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

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

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

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

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