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:

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

Where:
- \(E\) is the induced emf,
- \(B\) is the magnetic field strength,
- \(L\) is the length of the wire,
- \(v\) is the velocity 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:**
\[ E = B \cdot L \cdot v \]
\[ E = (0.3 \times 10^{-4}) \cdot (10) \cdot (5) \]

3. **Calculate the Product:**
- First, calculate \(L \cdot v\):
\[ L \cdot v = 10 \cdot 5 = 50 \]
- Now multiply by \(B\):
\[ E = 0.3 \times 10^{-4} \cdot 50 \]

4. **Perform the Multiplication:**
\[ E = 15 \times 10^{-4} \, \text{V} \]

5. **Convert to Standard Form:**
\[ E = 1.5 \times 10^{-3} \, \text{V} \]

### Final Answer:
The induced emf in the wire is \(1.5 \times 10^{-3} \, \text{V}\) or \(1.5 \, \text{mV}\).