A straight conductor 0.1 m long moves in a uniform magnetic field 0.1 T. The velocity of the conductor is `15ms^(-1)` and is directed perpendicular to the field. The emf induced between the two ends of the conductor is

Options

A. 0.10 V
B. 0.15 V
C. 1.50 V
D. 15 V

Explanation

To find the induced electromotive force (emf) in a straight conductor moving in a magnetic field, we can use the formula:

\[ \text{emf} (E) = L \cdot v \cdot B \cdot \sin(\theta) \]

Where:
- \( E \) is the induced emf,
- \( L \) is the length of the conductor,
- \( v \) is the velocity of the conductor,
- \( B \) is the magnetic field strength,
- \( \theta \) 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, \( L = 0.1 \, \text{m} \)
- Magnetic field strength, \( B = 0.1 \, \text{T} \)
- Velocity of the conductor, \( v = 15 \, \text{m/s} \)
- Angle \( \theta = 90^\circ \) (since the conductor moves perpendicular to the magnetic field)

2. **Calculate \( \sin(\theta) \):**
- Since \( \theta = 90^\circ \), we have:
\[ \sin(90^\circ) = 1 \]

3. **Substitute the values into the emf formula:**
\[ E = L \cdot v \cdot B \cdot \sin(\theta) \]
\[ E = 0.1 \, \text{m} \cdot 15 \, \text{m/s} \cdot 0.1 \, \text{T} \cdot 1 \]

4. **Perform the multiplication:**
\[ E = 0.1 \cdot 15 \cdot 0.1 = 0.15 \, \text{V} \]

5. **Conclusion:**
The induced emf between the two ends of the conductor is \( 0.15 \, \text{V} \).

### Final Answer:
The emf induced between the two ends of the conductor is **0.15 volts**.