A copper rod of length L rotates with an angular with an angular speed `'omega' ` in a uniform magnetic field B. find the emf developed between the two ends of the rod. The field in perpendicular to the motion of the rod.

Options

A. `Bomegal^(2)`
B. `(1)/(2)Bomegal^(2)`
C. `2Bomegal^(2)`
D. `(1)/(4)Bomegal^(2)`

Answer

To find the electromotive force (emf) developed between the two ends of a rotating copper rod in a uniform magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a copper rod of length \( L \) rotating with an angular speed \( \omega \) in a uniform magnetic field \( B \).
- The magnetic field is perpendicular to the motion of the rod.

2. **Determine the Area Swept by the Rod**:
- As the rod rotates, it sweeps out a circular area. The area \( A \) swept by the rod during one complete rotation can be calculated as:
\[ A = \pi L^2 \]
- Here, \( \pi L^2 \) is the area of the circle with radius \( L \).

3. **Calculate the Change in Magnetic Flux**:
- The magnetic flux \( \Phi \) through the area \( A \) in the magnetic field \( B \) is given by:
\[ \Phi = B \cdot A = B \cdot (\pi L^2) \]

4. **Determine the Time for One Complete Rotation**:
- The time \( T \) taken for one complete rotation is related to the angular speed \( \omega \):
\[ T = \frac{2\pi}{\omega} \]

5. **Use Faraday's Law of Electromagnetic Induction**:
- The induced emf \( E \) can be calculated using Faraday's law, which states that the induced emf is equal to the rate of change of magnetic flux:
\[ E = \frac{\Delta \Phi}{\Delta t} \]
- Substituting the change in flux and the time for one complete rotation:
\[ E = \frac{B \cdot (\pi L^2)}{T} \]

6. **Substitute for Time \( T \)**:
- Now substitute \( T \) from step 4 into the equation:
\[ E = \frac{B \cdot (\pi L^2)}{\frac{2\pi}{\omega}} = B \cdot (\pi L^2) \cdot \frac{\omega}{2\pi} \]

7. **Simplify the Expression**:
- The \( \pi \) terms cancel out:
\[ E = \frac{B \cdot L^2 \cdot \omega}{2} \]

8. **Final Result**:
- Thus, the emf developed between the two ends of the rod is:
\[ E = \frac{1}{2} B L^2 \omega \]