A conductor of `3 m` in length is moving perpendicularly to magnetic field of `10^(-4)` tesla with the speed of `10^(2)m//s`, then the e.m.f. produced across the ends of conductor will be

Options

A. 0.3N
B. 0.9 N
C. Zero
D. `3 xx 10^(-3)N`

Answer

To find the electromotive force (e.m.f.) produced across the ends of a conductor moving in a magnetic field, we can use the formula for motional e.m.f.:

\[ \text{e.m.f.} = B \cdot L \cdot v \cdot \sin(\theta) \]

Where:
- \( B \) is the magnetic field strength,
- \( L \) is the length of the conductor,
- \( v \) is the speed of the conductor,
- \( \theta \) is the angle between the velocity vector and the magnetic field vector.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Length of the conductor, \( L = 3 \, \text{m} \)
- Magnetic field strength, \( B = 10^{-4} \, \text{T} \)
- Speed of the conductor, \( v = 10^2 \, \text{m/s} = 100 \, \text{m/s} \)
- The angle \( \theta = 90^\circ \) (since the conductor is moving perpendicularly to the magnetic field).

2. **Substitute the Values into the Formula:**
- Since the conductor is moving perpendicularly to the magnetic field, we have:
\[ \sin(90^\circ) = 1 \]
- Therefore, the formula simplifies to:
\[ \text{e.m.f.} = B \cdot L \cdot v \]

3. **Calculate the e.m.f.:**
\[ \text{e.m.f.} = (10^{-4} \, \text{T}) \cdot (3 \, \text{m}) \cdot (100 \, \text{m/s}) \]
\[ \text{e.m.f.} = 10^{-4} \cdot 3 \cdot 100 \]
\[ \text{e.m.f.} = 10^{-4} \cdot 300 \]
\[ \text{e.m.f.} = 3 \times 10^{-2} \, \text{V} = 0.03 \, \text{V} \]

4. **Conclusion:**
The e.m.f. produced across the ends of the conductor is \( 0.03 \, \text{V} \).

### Final Answer:
\[ \text{e.m.f.} = 0.03 \, \text{V} \]