A metal conductor of length 1m rotates vertically about one of its ends at angular velocity 5 radians per second. If the horizontal component of earth's magnetic field is `0.2xx10^(-4)T`, then the emf developed between the two ends of hte conductor is

Options

A. 5mV
B. `50 (mu)V`
C. `5 mu V`
D. `50 mV`

Explanation

To find the induced electromotive force (emf) developed between the two ends of a rotating conductor, we can use the formula for induced emf in a rotating conductor in a magnetic field:

\[ \text{emf} = \frac{1}{2} B L^2 \omega \]

Where:
- \( B \) = magnetic field strength (in Tesla)
- \( L \) = length of the conductor (in meters)
- \( \omega \) = angular velocity (in radians per second)

### Step-by-Step Solution:

1. **Identify the given values**:
- Length of the conductor, \( L = 1 \, \text{m} \)
- Angular velocity, \( \omega = 5 \, \text{rad/s} \)
- Horizontal component of Earth's magnetic field, \( B = 0.2 \times 10^{-4} \, \text{T} \)

2. **Substitute the values into the formula**:
\[ \text{emf} = \frac{1}{2} \times (0.2 \times 10^{-4}) \times (1)^2 \times (5) \]

3. **Calculate the emf**:
- First, calculate \( (1)^2 = 1 \).
- Then, substitute back into the equation:
\[ \text{emf} = \frac{1}{2} \times (0.2 \times 10^{-4}) \times 1 \times 5 \]
\[ = \frac{1}{2} \times 0.2 \times 5 \times 10^{-4} \]
\[ = 0.5 \times 10^{-4} \, \text{V} \]
\[ = 0.5 \times 10^{-4} \, \text{V} = 50 \, \mu\text{V} \]

4. **Conclusion**:
The induced emf developed between the two ends of the conductor is \( 50 \, \mu\text{V} \).

### Final Answer:
The emf developed between the two ends of the conductor is \( 50 \, \mu\text{V} \).