KCET · Physics · Mathematics in Physics
A jet plane of wing span \( 20 \mathrm{~m} \) is travelling towards west at a speed of
\( 400 \mathrm{~ms}^{-1} \). If the earth's total magnetic field is \( 4 \times 10^{-4} \mathrm{~T} \) and the dip angle is \( 30^{\circ} \), at that
place, the voltage difference developed across the ends of the wing is
- A \( 1.6 \mathrm{~V} \)
- B \( 3.2 \mathrm{~V} \)
- C \( 0.8 \mathrm{~V} \)
- D \( 6.4 \mathrm{~V} \)
Answer & Solution
Correct Answer
(A) \( 1.6 \mathrm{~V} \)
Step-by-step Solution
Detailed explanation
Given, wing span, \(1=20 \mathrm{~m}\); speed of jet plane, \(v=400 \mathrm{~ms}^{-1}\), Earth's magnetic field, \(B=4 \times 10^{-4} \mathrm{~T}\); angle of dip, \(\theta=\)
\(30^{\circ}\).
Now, emf induced
\(e=B_{v} l v\)
Here,
\(B_{v}=B \sin \theta=4 \times 10^{-4} \times \sin 30^{\circ}=4 \times 10^{-4} \times \frac{1}{2}=2 \times 10^{-4} \mathrm{~T}\)
Therefore,
\(e=2 \times 10^{-4} \times 400 \times 20=16 \times 10^{-1} \mathrm{~V}=1.6 \mathrm{~V}\)
Thus, voltage difference developed across the ends of the wing is \(1.6 \mathrm{~V}\)
\(30^{\circ}\).
Now, emf induced
\(e=B_{v} l v\)
Here,
\(B_{v}=B \sin \theta=4 \times 10^{-4} \times \sin 30^{\circ}=4 \times 10^{-4} \times \frac{1}{2}=2 \times 10^{-4} \mathrm{~T}\)
Therefore,
\(e=2 \times 10^{-4} \times 400 \times 20=16 \times 10^{-1} \mathrm{~V}=1.6 \mathrm{~V}\)
Thus, voltage difference developed across the ends of the wing is \(1.6 \mathrm{~V}\)
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