An aeroplane in which the distance between the tips of wings is `50 m` is flying horizontal with a speed of `360km//hr` over a place where the verticle components of earth magnetic field is `2.0xx10^(-4)webr//m^(2)`. The potential different between the tips of wings would be

Options

A. 0.1 V
B. 0.01 V
C. 0.2 V
D. 1.0 V

Explanation

To solve the problem, we need to calculate the potential difference (or electromotive force, EMF) between the tips of the wings of the airplane using the formula for EMF induced in a moving conductor in a magnetic field.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance between the tips of the wings (L) = 50 m
- Speed of the airplane (V) = 360 km/hr
- Vertical component of Earth's magnetic field (B) = 2.0 × 10^(-4) Wb/m²

2. **Convert the Speed from km/hr to m/s:**
\[ V = 360 \, \text{km/hr} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{hr}}{3600 \, \text{s}} = \frac{360000 \, \text{m}}{3600 \, \text{s}} = 100 \, \text{m/s} \]

3. **Use the Formula for EMF:**
The formula for the electromotive force (EMF) induced in a conductor moving through a magnetic field is given by:
\[ \text{EMF} = L \times V \times B \]
where:
- L = length of the conductor (distance between the tips of the wings)
- V = velocity of the conductor (speed of the airplane)
- B = magnetic field strength

4. **Substitute the Values into the Formula:**
\[ \text{EMF} = 50 \, \text{m} \times 100 \, \text{m/s} \times (2.0 \times 10^{-4} \, \text{Wb/m}^2) \]

5. **Calculate the EMF:**
\[ \text{EMF} = 50 \times 100 \times 2.0 \times 10^{-4} = 50 \times 0.02 = 1 \, \text{V} \]

6. **Conclusion:**
The potential difference between the tips of the wings is **1 volt**.