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 find the induced EMF between the tip of the nose and the tail of the helicopter, we can use the formula for induced EMF in a conductor moving through a magnetic field:

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

Where:
- \( v \) = velocity of the helicopter (in m/s)
- \( B_H \) = horizontal component of the Earth's magnetic field (in Wb/m²)
- \( L \) = length of the helicopter (in m)

### Step 1: Identify the given values
- Velocity of the helicopter, \( v = 100 \, \text{m/s} \)
- Horizontal component of Earth's magnetic field, \( B_H = 5 \times 10^{-3} \, \text{Wb/m}^2 \)
- Length of the helicopter, \( L = 10 \, \text{m} \)

### Step 2: Substitute the values into the formula
Now, we can substitute the values into the induced EMF formula:

\[ E = 100 \, \text{m/s} \cdot (5 \times 10^{-3} \, \text{Wb/m}^2) \cdot 10 \, \text{m} \]

### Step 3: Perform the multiplication
Calculating the multiplication step-by-step:

1. Calculate \( 100 \times 5 \):
\[ 100 \times 5 = 500 \]

2. Now multiply by \( 10^{-3} \):
\[ 500 \times 10^{-3} = 0.5 \]

3. Finally, multiply by \( 10 \):
\[ 0.5 \times 10 = 5 \]

### Step 4: Conclusion
Thus, the induced EMF between the tip of the nose and the tail of the helicopter is:

\[ E = 5 \, \text{V} \]

### Final Answer
The induced EMF is \( 5 \, \text{V} \).

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