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 for motional emf:

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

Where:
- \(E\) is the induced emf,
- \(B\) is the magnetic field strength,
- \(l\) is the length of the conductor,
- \(v\) is the velocity of the conductor,
- \(\theta\) is the angle between the direction of motion and the magnetic field.

### 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. **Substitute the values into the formula**:
Since the angle \(\theta\) is \(90^\circ\), \(\sin(90^\circ) = 1\). Therefore, the formula simplifies to:
\[ E = B \cdot l \cdot v \]

3. **Calculate the induced emf**:
\[ E = 0.1 \, \text{T} \cdot 0.1 \, \text{m} \cdot 15 \, \text{m/s} \]
\[ E = 0.1 \cdot 0.1 \cdot 15 = 0.15 \, \text{V} \]

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

### Final Answer:
The induced emf is \(0.15 \, \text{V}\).

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