An air plane , with 20 m wingspread is flying at `250 ms^(-1)` parallel to the earth's surface at a place whre the horizontal component of earth's magnetic field is `2 xx 10^(-5)` T and angle of dip is `60^@`. The magnitude of the induced emf between the tips of the wings is

Options

A. `1/(10) V`
B. `sqrt2/(10) V`
C. `sqrt3/(10) V`
D. `1/5 V`

Explanation

To find the magnitude of the induced electromotive force (emf) between the tips of the wings of the airplane, we can follow these steps:

### Step 1: Identify the given values
- Wingspan (L) = 20 m
- Speed of the airplane (V) = 250 m/s
- Horizontal component of the Earth's magnetic field (B_h) = 2 × 10^(-5) T
- Angle of dip (δ) = 60°

### Step 2: Calculate the vertical component of the magnetic field (B_v)
Using the relationship between the horizontal and vertical components of the Earth's magnetic field and the angle of dip:
\[ B_v = B_h \cdot \tan(\delta) \]
Where:
- \( \tan(60°) = \sqrt{3} \)

Substituting the known values:
\[ B_v = (2 \times 10^{-5} \, \text{T}) \cdot \tan(60°) \]
\[ B_v = (2 \times 10^{-5} \, \text{T}) \cdot \sqrt{3} \]

### Step 3: Calculate the induced emf (ε)
The formula for induced emf in a moving conductor in a magnetic field is given by:
\[ \epsilon = B_v \cdot L \cdot V \]
Substituting the values we have:
\[ \epsilon = (2 \times 10^{-5} \cdot \sqrt{3}) \cdot (20) \cdot (250) \]

### Step 4: Simplify the expression
Calculating the terms:
1. Calculate \( 2 \times 20 = 40 \)
2. Then, \( 40 \times 250 = 10000 \)
3. Now, multiply by \( \sqrt{3} \):
\[ \epsilon = (10000 \cdot \sqrt{3}) \times 10^{-5} \]
\[ \epsilon = \sqrt{3} \times 10^{-1} \]
\[ \epsilon = \frac{\sqrt{3}}{10} \, \text{V} \]

### Final Answer
The magnitude of the induced emf between the tips of the wings is:
\[ \epsilon = \frac{\sqrt{3}}{10} \, \text{V} \]

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