A conductor of length 5 cm is moved paralllel to itself with a speed of `2m//s`, perpendicular to a uniform magnetic field of `10^(-3) Wb//m^(2).` the induced emf generated is

Options

A. `2xx10^(-3)V`
B. `1xx10^(-3)V`
C. `1xx10^(-4)V`
D. `2xx10^(-4)V`

Explanation

To solve the problem of finding the induced EMF generated in a conductor moving in a magnetic field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the conductor (L) = 5 cm = 0.05 m (convert cm to m)
- Speed of the conductor (v) = 2 m/s
- Magnetic field (B) = \(10^{-3}\) Wb/m²

2. **Understand the Formula for Induced EMF:**
The induced EMF (\( \mathcal{E} \)) in a conductor moving in a magnetic field is given by the formula:
\[ \mathcal{E} = L \cdot v \cdot B \cdot \sin(\theta) \]
where:
- \(L\) = length of the conductor
- \(v\) = speed of the conductor
- \(B\) = magnetic field strength
- \(\theta\) = angle between the direction of motion and the magnetic field

3. **Determine the Angle:**
Since the conductor is moving perpendicular to the magnetic field, the angle \( \theta = 90^\circ \). Therefore, \( \sin(90^\circ) = 1 \).

4. **Substitute the Values into the Formula:**
\[ \mathcal{E} = L \cdot v \cdot B \cdot \sin(90^\circ) \]
\[ \mathcal{E} = 0.05 \, \text{m} \cdot 2 \, \text{m/s} \cdot 10^{-3} \, \text{Wb/m}^2 \cdot 1 \]

5. **Calculate the Induced EMF:**
\[ \mathcal{E} = 0.05 \cdot 2 \cdot 10^{-3} \]
\[ \mathcal{E} = 0.1 \cdot 10^{-3} = 1 \cdot 10^{-4} \, \text{V} \]

6. **Final Result:**
The induced EMF generated is \( \mathcal{E} = 1 \times 10^{-4} \, \text{V} \).

### Conclusion:
The induced EMF generated in the conductor is \( 1 \times 10^{-4} \, \text{V} \).