A 10 m long horizontal wire extends from North east ro 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 in a conductor moving through a magnetic field:

\[ \text{Induced EMF} (V) = B \cdot L \cdot v \]

Where:
- \( V \) is the induced emf,
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wire,
- \( v \) is the speed of the wire.

### Step-by-step Solution:

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, \( B = 0.3 \times 10^{-4} \, \text{Wb/m}^2 \)

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

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

4. **Convert to standard form:**
\[ V = 1.5 \times 10^{-3} \, \text{V} \]

5. **Final answer:**
The value of the induced emf in the wire is:
\[ V = 1.5 \times 10^{-3} \, \text{V} \text{ or } 1.5 \, \text{mV} \]