A copper rod AB-of length / pivoted at one end A, rotates at constant angular velocity `omega`, at right angles to a uniformn magnetic field of induction B. The emf, developed between the mid point of the rod and end B is

Options

A. `(Bomegal^(2))/(8)`
B. `(3)/(4)Bomegal^(2)`
C. `(Bomegal^(2))/(4)`
D. `(3)/(8)Bomegal^(2)`

Answer

To find the electromotive force (emf) developed between the midpoint of a copper rod AB and the end B while the rod rotates in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a copper rod AB of length \( L \) pivoted at point A.
- The rod rotates with a constant angular velocity \( \omega \) in a uniform magnetic field \( B \), which is perpendicular to the plane of rotation.

2. **Identify the Points of Interest**:
- We need to find the emf between the midpoint of the rod (which is at \( L/2 \)) and the end B (which is at \( L \)).

3. **Determine the Velocity of a Point on the Rod**:
- The linear velocity \( v \) of a point at a distance \( x \) from the pivot A is given by:
\[ v = x \cdot \omega \]

4. **Calculate the Induced EMF**:
- The emf induced in a small segment \( dx \) of the rod at a distance \( x \) from A is given by:
\[ dE = v \cdot B \cdot dx = (x \cdot \omega) \cdot B \cdot dx \]
- To find the total emf from the midpoint \( (L/2) \) to the end \( L \), we integrate \( dE \) from \( x = L/2 \) to \( x = L \):
\[ E = \int_{L/2}^{L} (x \cdot \omega \cdot B) \, dx \]

5. **Perform the Integration**:
- The integral becomes:
\[ E = \omega B \int_{L/2}^{L} x \, dx \]
- The integral of \( x \) is:
\[ \int x \, dx = \frac{x^2}{2} \]
- Evaluating the integral from \( L/2 \) to \( L \):
\[ E = \omega B \left[ \frac{x^2}{2} \right]_{L/2}^{L} = \omega B \left( \frac{L^2}{2} - \frac{(L/2)^2}{2} \right) \]
- Simplifying this:
\[ E = \omega B \left( \frac{L^2}{2} - \frac{L^2/4}{2} \right) = \omega B \left( \frac{L^2}{2} - \frac{L^2}{8} \right) = \omega B \left( \frac{4L^2}{8} - \frac{L^2}{8} \right) = \omega B \left( \frac{3L^2}{8} \right) \]

6. **Final Result**:
- Thus, the emf developed between the midpoint of the rod and end B is:
\[ E = \frac{3}{8} B \omega L^2 \]