A copper rod of length l is rotated about one end perpendicular to the uniform magnetic field B with constant angular velocity `omega` . The induced e.m.f. between its two ends is

Options

A. `Bomegal^(2)`
B. `(3)/(2)Bomegal^(2)`
C. `(1)/(2)Bomegal^(2)`
D. `2 Bomegal^(2)`

Answer

To find the induced e.m.f. (electromotive force) between the two ends of a copper rod of length \( l \) that is rotated about one end perpendicular to a uniform magnetic field \( B \) with a constant angular velocity \( \omega \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a copper rod of length \( l \) rotating about one end in a uniform magnetic field \( B \).
- The rod is perpendicular to the magnetic field, and it rotates with a constant angular velocity \( \omega \).

2. **Induced EMF Formula**:
- The induced e.m.f. \( \mathcal{E} \) can be calculated using Faraday's law of electromagnetic induction, which states:
\[ \mathcal{E} = -\frac{d\Phi}{dt} \]
where \( \Phi \) is the magnetic flux.

3. **Calculating Magnetic Flux**:
- The magnetic flux \( \Phi \) through the area swept by the rod can be expressed as:
\[ \Phi = B \cdot A \]
where \( A \) is the area swept by the rod as it rotates.

4. **Area Swept by the Rod**:
- As the rod rotates, it sweeps out a circular sector. The angle \( \theta \) that the rod sweeps in time \( t \) is given by:
\[ \theta = \omega t \]
- The area \( A \) swept by the rod at time \( t \) is:
\[ A = \frac{1}{2} L^2 \theta = \frac{1}{2} L^2 (\omega t) \]
- Therefore, the magnetic flux becomes:
\[ \Phi = B \cdot \frac{1}{2} L^2 (\omega t) \]

5. **Finding the Rate of Change of Flux**:
- To find the induced e.m.f., we need to differentiate the magnetic flux with respect to time:
\[ \frac{d\Phi}{dt} = \frac{d}{dt} \left( B \cdot \frac{1}{2} L^2 (\omega t) \right) = B \cdot \frac{1}{2} L^2 \omega \]

6. **Calculating Induced EMF**:
- Substituting this into the e.m.f. formula gives:
\[ \mathcal{E} = -\frac{d\Phi}{dt} = -\left( B \cdot \frac{1}{2} L^2 \omega \right) \]
- The negative sign indicates the direction of the induced e.m.f. according to Lenz's law, but the magnitude of the induced e.m.f. is:
\[ \mathcal{E} = \frac{1}{2} B L^2 \omega \]

### Final Answer:
The induced e.m.f. between the two ends of the copper rod is:
\[ \mathcal{E} = \frac{1}{2} B L^2 \omega \]