A rod of length l rotates with a 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 two ends of the rod is

Options

A. zero
B. `(1)/(2)Blomega^2`
C. `Blomega^2`
D. `2Blomega^2`

Answer

To solve the problem of finding the potential difference between the two ends of a rotating rod in a 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 \( \omega \).
- A uniform magnetic field \( B \) exists parallel to the axis of rotation.

2. **Identify the Linear Velocity**:
- The linear velocity \( v \) of a point at a distance \( x \) from the center of the rod is given by:
\[ v = \omega x \]

3. **Consider a Differential Element**:
- Take a small strip of the rod at a distance \( x \) from the center with a width \( dx \).
- The induced electromotive force (emf) \( dE \) across this strip can be expressed as:
\[ dE = B \cdot v \cdot dx = B \cdot (\omega x) \cdot dx \]

4. **Integrate to Find Total EMF**:
- To find the total potential difference \( E \) between the two ends of the rod, integrate \( dE \) from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ E = \int_{-\frac{L}{2}}^{\frac{L}{2}} B \omega x \, dx \]

5. **Perform the Integration**:
- Factor out constants \( B \) and \( \omega \):
\[ E = B \omega \int_{-\frac{L}{2}}^{\frac{L}{2}} x \, dx \]
- The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
- Evaluating the integral from \( -\frac{L}{2} \) to \( \frac{L}{2} \):
\[ E = B \omega \left[ \frac{x^2}{2} \right]_{-\frac{L}{2}}^{\frac{L}{2}} = B \omega \left( \frac{(\frac{L}{2})^2}{2} - \frac{(-\frac{L}{2})^2}{2} \right) \]
- Both terms yield the same value, thus:
\[ E = B \omega \left( \frac{L^2}{8} - \frac{L^2}{8} \right) = 0 \]

6. **Conclusion**:
- The potential difference \( E \) between the two ends of the rod is:
\[ E = 0 \]