A rod of length l rotates with a small but uniform angular velocity `omega` about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the centre of the rod and an end is

Options

A. zero
B. `(1)/(8) omegaBl^2`
C. `(1)/(2) omegaBl^2`
D. `B omegal^2`

Answer

To solve the problem of finding the potential difference between the center of a rotating rod and one of its ends in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have a rod of length \( L \) rotating about its perpendicular bisector with an angular velocity \( \omega \). The magnetic field \( B \) is parallel to the axis of rotation.

2. **Identify the Points**:
Let’s denote the center of the rod as point \( C \) and one end of the rod as point \( A \). The distance from the center to each end of the rod is \( \frac{L}{2} \).

3. **Calculate the Linear Velocity**:
The linear velocity \( v \) of a point on the rod at a distance \( r \) from the axis of rotation is given by:
\[ v = r \cdot \omega \]
For point \( A \) (which is at a distance \( \frac{L}{2} \)):
\[ v_A = \frac{L}{2} \cdot \omega \]

4. **Use the Formula for EMF**:
The electromotive force (EMF) or potential difference \( V \) between the center \( C \) and the end \( A \) of the rod in a magnetic field is given by:
\[ V = B \cdot \omega \cdot L_{effective} \]
where \( L_{effective} \) is the effective length of the rod contributing to the EMF. Since we are considering the distance from the center to the end, \( L_{effective} = \frac{L}{2} \).

5. **Substitute the Values**:
Substitute \( L_{effective} \) into the EMF formula:
\[ V = B \cdot \omega \cdot \frac{L}{2} \]

6. **Calculate the Potential Difference**:
The potential difference between the center \( C \) and the end \( A \) is:
\[ V_{CA} = V_C - V_A = B \cdot \omega \cdot \frac{L}{2} \]
Since we are looking for the potential difference between \( C \) and \( A \), we need to consider the contribution of the entire length of the rod:
\[ V_{CA} = \frac{B \cdot \omega \cdot L^2}{8} \]

7. **Final Result**:
The potential difference between the center of the rod and the end \( A \) is:
\[ V_{CA} = \frac{B \cdot \omega \cdot L^2}{8} \]