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

Options

A. `(1)/(2) B omegal^(2)`
B. `(3)/(4) B omega l^(2)`
C. `B omega l^(2)`
D. `2 B omega l^(2)`

Answer

To find the induced e.m.f. (electromotive force) between the two ends of a rotating copper rod in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a copper rod of length \( L \) that is rotated about one end.
- The rod is perpendicular to a uniform magnetic field \( B \).
- The rod is rotated with a constant angular velocity \( \omega \).

2. **Identify the Velocity of a Point on the Rod**:
- Consider a small element of the rod at a distance \( x \) from the pivot (the end about which the rod is rotating).
- The linear velocity \( V \) of this element due to rotation is given by:
\[ V = \omega x \]

3. **Calculate the Induced Electric Field**:
- The induced electric field \( E \) at this point in the rod can be expressed using the formula:
\[ E = B V = B (\omega x) = B \omega x \]

4. **Determine the Potential Difference**:
- The potential difference \( dV \) across a small segment \( dx \) of the rod can be expressed as:
\[ dV = E \cdot dx = (B \omega x) \cdot dx \]

5. **Integrate to Find Total Potential Difference**:
- To find the total potential difference \( V \) between the two ends of the rod, we need to integrate \( dV \) from \( x = 0 \) to \( x = L \):
\[ V = \int_0^L dV = \int_0^L B \omega x \, dx \]
- Performing the integration:
\[ V = B \omega \int_0^L x \, dx = B \omega \left[ \frac{x^2}{2} \right]_0^L = B \omega \left( \frac{L^2}{2} \right) \]
- Thus, we find:
\[ V = \frac{1}{2} B \omega L^2 \]

6. **Conclusion**:
- The induced e.m.f. between the two ends of the rod is:
\[ \text{Induced e.m.f.} = \frac{1}{2} B \omega L^2 \]

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