An airplane with a 20 wingspread is flying at 250m/s straight south parallel to the earth's surface. The earths magnetic field has a horizontal component of `2xx10^(-5)Wb//m^(2)` and the dip angle is `60^(@)`. Calculate the induced e.m.f. between the plane tips.

Options

A. 0.174 V
B. 0.173 V
C. 1.173 V
D. 0.163 V

Explanation

To calculate the induced e.m.f. between the tips of the airplane wings, we will use the formula for motional electromotive force (e.m.f.), which is given by:

\[ \text{Induced e.m.f.} = B \cdot L \cdot V \]

where:
- \( B \) is the magnetic field strength,
- \( L \) is the length of the wingspan,
- \( V \) is the velocity of the airplane.

### Step 1: Identify the given values
- Wingspan \( L = 20 \, \text{m} \)
- Velocity \( V = 250 \, \text{m/s} \)
- Horizontal component of the Earth's magnetic field \( B_H = 2 \times 10^{-5} \, \text{Wb/m}^2 \)
- Dip angle \( \theta = 60^\circ \)

### Step 2: Calculate the vertical component of the magnetic field
The vertical component \( B_V \) can be calculated using the relationship between the horizontal and vertical components of the magnetic field and the dip angle:

\[ B_V = B_H \cdot \tan(\theta) \]

Substituting the values:

\[ B_V = (2 \times 10^{-5}) \cdot \tan(60^\circ) \]

Since \( \tan(60^\circ) = \sqrt{3} \):

\[ B_V = (2 \times 10^{-5}) \cdot \sqrt{3} \approx 3.464 \times 10^{-5} \, \text{Wb/m}^2 \]

### Step 3: Calculate the total magnetic field \( B \)
The total magnetic field \( B \) can be calculated using the Pythagorean theorem:

\[ B = \sqrt{B_H^2 + B_V^2} \]

Calculating \( B \):

\[ B = \sqrt{(2 \times 10^{-5})^2 + (3.464 \times 10^{-5})^2} \]

Calculating each term:

\[ B_H^2 = 4 \times 10^{-10} \]
\[ B_V^2 = 12 \times 10^{-10} \]

Adding them:

\[ B^2 = 4 \times 10^{-10} + 12 \times 10^{-10} = 16 \times 10^{-10} \]

Taking the square root:

\[ B = 4 \times 10^{-5} \, \text{Wb/m}^2 \]

### Step 4: Calculate the induced e.m.f.
Now we can substitute the values into the induced e.m.f. formula:

\[ \text{Induced e.m.f.} = B \cdot L \cdot V \]

Substituting the values:

\[ \text{Induced e.m.f.} = (4 \times 10^{-5}) \cdot (20) \cdot (250) \]

Calculating:

\[ \text{Induced e.m.f.} = 4 \times 20 \times 250 \times 10^{-5} \]
\[ = 20000 \times 10^{-5} \]
\[ = 0.2 \, \text{V} \]

### Final Answer
The induced e.m.f. between the tips of the airplane wings is \( 0.2 \, \text{V} \).