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`

Answer

To find the induced electromotive force (emf) generated when a conductor is moved in a magnetic field, we can use the formula for motional emf:

\[ \text{emf} = V \cdot B \cdot L \]

where:
- \( V \) is the velocity of the conductor,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the conductor.

### Step-by-Step Solution:

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

2. **Substitute the Values into the Formula:**
\[ \text{emf} = V \cdot B \cdot L \]
\[ \text{emf} = 2 \, \text{m/s} \cdot 10^{-3} \, \text{T} \cdot 0.05 \, \text{m} \]

3. **Calculate the Induced emf:**
\[ \text{emf} = 2 \cdot 10^{-3} \cdot 0.05 \]
\[ \text{emf} = 10^{-2} \cdot 10^{-3} = 10^{-4} \, \text{V} \]

4. **Final Result:**
The induced emf generated is \( 10^{-4} \, \text{V} \) or \( 0.0001 \, \text{V} \).

### Conclusion:
The induced emf generated is \( 10^{-4} \, \text{V} \).