A 10 m long horizontal wire extends from North east to South East. It is falling with a speed of `5.0 ms^(-1)`, at right angles to the horizontal component of the earth's magnetic field, of `0.3xx10^(-4)Wb//m^(2)`. The value of the induced emf in wire is :

Options

A. `1.5xx10^(-3)V`
B. `1.1xx10^(-3)V`
C. `2.5xx10^(-3)V`
D. `0.3xx10^(-3)V`

Explanation

To find the induced electromotive force (emf) in the wire, we can use the formula for induced emf due to motion in a magnetic field:

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

Where:
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wire,
- \( V \) is the velocity of the wire,
- \( \theta \) is the angle between the direction of the magnetic field and the direction of the velocity of the wire.

### Step 1: Identify the given values
- Length of the wire, \( L = 10 \, \text{m} \)
- Speed of the wire, \( V = 5.0 \, \text{m/s} \)
- Magnetic field strength, \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \)
- The angle \( \theta = 90^\circ \) (since the wire is falling perpendicular to the magnetic field)

### Step 2: Calculate \( \sin(\theta) \)
Since \( \theta = 90^\circ \):
\[ \sin(90^\circ) = 1 \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the induced emf formula:
\[ \text{emf} = B \cdot L \cdot V \cdot \sin(\theta) \]
\[ \text{emf} = (0.3 \times 10^{-4}) \cdot (10) \cdot (5) \cdot (1) \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \text{emf} = 0.3 \times 10^{-4} \cdot 10 \cdot 5 \]
\[ \text{emf} = 0.3 \times 10^{-4} \cdot 50 \]
\[ \text{emf} = 15 \times 10^{-4} \, \text{V} \]
\[ \text{emf} = 1.5 \times 10^{-3} \, \text{V} \]

### Final Answer
The induced emf in the wire is:
\[ \text{emf} = 1.5 \times 10^{-3} \, \text{V} \]