A helicopter rises vertically upwards with a speed of 100 `m//s`. If the helicopter has a length of 10m and horizontal component of earth's magnetic field is `5 xx 10^(-3)Wb//m^(2)`, then the induced emf between the tip of the nose and the tail of the helicopter is

Options

A. 50 V
B. 0.5 V
C. 5 V
D. 25 V

Answer

To solve the problem of calculating the induced EMF (electromotive force) between the tip of the nose and the tail of a helicopter rising vertically, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the helicopter, \( V = 100 \, \text{m/s} \)
- Length of the helicopter, \( l = 10 \, \text{m} \)
- Horizontal component of Earth's magnetic field, \( B = 5 \times 10^{-3} \, \text{Wb/m}^2 \)

2. **Understand the Induced EMF Formula:**
The induced EMF (ε) in a conductor moving through a magnetic field can be calculated using the formula:
\[ \epsilon = B \cdot V \cdot l \]
where:
- \( \epsilon \) is the induced EMF,
- \( B \) is the magnetic field strength,
- \( V \) is the velocity of the conductor,
- \( l \) is the length of the conductor.

3. **Substitute the Values into the Formula:**
Plugging in the values we have:
\[ \epsilon = (5 \times 10^{-3} \, \text{Wb/m}^2) \cdot (100 \, \text{m/s}) \cdot (10 \, \text{m}) \]

4. **Calculate the Induced EMF:**
Performing the multiplication:
\[ \epsilon = 5 \times 10^{-3} \times 100 \times 10 \]
\[ \epsilon = 5 \times 10^{-3} \times 1000 \]
\[ \epsilon = 5 \, \text{V} \]

5. **Conclusion:**
The induced EMF between the tip of the nose and the tail of the helicopter is \( 5 \, \text{V} \).