An aeroplane with its wings spread `10 m` is flying with speed `180 km/h` in horizontal direction. The total intensity of earth's field is `2.5 X 10^-4` Tesla and angle of dip is `60°`. Then find emf induced between the tips of the plane wings.

Options

A. `108.25mV`
B. `54mV`
C. `216mV`
D. `140mV`

Answer

To solve the problem of finding the induced EMF between the tips of the airplane wings, we can follow these steps:

### Step 1: Understand the given data
- Wingspan of the airplane \( L = 10 \, \text{m} \)
- Speed of the airplane \( v = 180 \, \text{km/h} \)
- Total intensity of Earth's magnetic field \( B = 2.5 \times 10^{-4} \, \text{T} \)
- Angle of dip \( \theta = 60^\circ \)

### Step 2: Convert the speed from km/h to m/s
To convert the speed from kilometers per hour to meters per second, we use the conversion factor \( \frac{5}{18} \):
\[ v = 180 \, \text{km/h} \times \frac{5}{18} = 50 \, \text{m/s} \]

### Step 3: Determine the vertical component of the magnetic field
The angle of dip indicates that the magnetic field has both horizontal and vertical components. The vertical component \( B_v \) can be calculated using:
\[ B_v = B \sin(\theta) \]
Substituting the values:
\[ B_v = 2.5 \times 10^{-4} \, \text{T} \cdot \sin(60^\circ) = 2.5 \times 10^{-4} \, \text{T} \cdot \frac{\sqrt{3}}{2} \]
Calculating \( B_v \):
\[ B_v = 2.5 \times 10^{-4} \cdot 0.866 = 2.165 \times 10^{-4} \, \text{T} \]

### Step 4: Calculate the induced EMF
The induced EMF (\( \mathcal{E} \)) can be calculated using the formula:
\[ \mathcal{E} = B_v \cdot L \cdot v \]
Substituting the values:
\[ \mathcal{E} = (2.165 \times 10^{-4} \, \text{T}) \cdot (10 \, \text{m}) \cdot (50 \, \text{m/s}) \]
Calculating \( \mathcal{E} \):
\[ \mathcal{E} = 2.165 \times 10^{-4} \cdot 10 \cdot 50 = 1.0825 \times 10^{-2} \, \text{V} = 10.825 \, \text{mV} \]

### Step 5: Final answer
The induced EMF between the tips of the airplane wings is:
\[ \mathcal{E} = 108.25 \, \text{mV} \]