A 10 m wire kept in east-west direction is falling with velocity 5 m/s perpendicular to the field `0.3xx10^(-4)Wb//m^(2)`.The magnitude of the induced emf in the wire is

Options

A. 0.15 V
B. 1.5 mV
C. 1.5 V
D. 15.0 V

Explanation

To find the magnitude of the induced electromotive force (emf) in the wire, we can use the formula for induced emf when a conductor moves through a magnetic field:

\[ E = B \cdot L \cdot V \]

Where:
- \(E\) is the induced emf (in volts),
- \(B\) is the magnetic field strength (in Weber per square meter, Wb/m²),
- \(L\) is the length of the wire (in meters),
- \(V\) is the velocity of the wire (in meters per second).

### Step-by-step Solution:

1. **Identify the given values**:
- Length of the wire, \(L = 10 \, \text{m}\)
- Velocity of the wire, \(V = 5 \, \text{m/s}\)
- Magnetic field strength, \(B = 0.3 \times 10^{-4} \, \text{Wb/m}^2\)

2. **Substitute the values into the formula**:
\[ E = B \cdot L \cdot V \]
\[ E = (0.3 \times 10^{-4}) \cdot 10 \cdot 5 \]

3. **Calculate the induced emf**:
- First, calculate \(B \cdot L\):
\[ B \cdot L = (0.3 \times 10^{-4}) \cdot 10 = 3 \times 10^{-4} \, \text{Wb} \]
- Now multiply by \(V\):
\[ E = (3 \times 10^{-4}) \cdot 5 = 15 \times 10^{-4} \, \text{V} \]
- Simplifying gives:
\[ E = 1.5 \times 10^{-3} \, \text{V} \]

4. **Convert to millivolts**:
\[ E = 1.5 \, \text{mV} \]

5. **Final answer**:
The magnitude of the induced emf in the wire is \(1.5 \, \text{mV}\).