A metal conductor of length `1 m` 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. `5muV`
B. `5mV`
C. `50muV`
D. `50mV`

Explanation

To solve the problem of finding the EMF developed between the two ends of a rotating conductor in a magnetic field, we can follow these steps:

### 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. **Understand the Formula for Induced EMF:**
The formula to calculate the induced EMF (\( \epsilon \)) in a rotating conductor in a magnetic field is given by:
\[ \epsilon = \frac{1}{2} B \omega L^2 \]

3. **Substitute the Given Values into the Formula:**
Plugging in the values we have:
\[ \epsilon = \frac{1}{2} \times (0.2 \times 10^{-4}) \times 5 \times (1)^2 \]

4. **Calculate the EMF:**
Simplifying the expression:
\[ \epsilon = \frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1 \]
\[ = \frac{1}{2} \times 1 \times 10^{-4} \]
\[ = 0.5 \times 10^{-4} \, \text{V} \]
\[ = 5 \times 10^{-5} \, \text{V} \]

5. **Convert to Microvolts:**
Since \( 1 \, \text{V} = 10^6 \, \mu\text{V} \):
\[ 5 \times 10^{-5} \, \text{V} = 50 \, \mu\text{V} \]

6. **Final Answer:**
The induced EMF developed between the two ends of the conductor is:
\[ \epsilon = 50 \, \mu\text{V} \]