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An aircraft with a wingspan of \( 40 \mathrm{~m} \) flies with a speed of \( 1080 \mathrm{~km} \mathrm{~h}^{-1} \) in the eastward
direction at a constant altitude in the northern hemisphere, where the vertical component of the
earth's magnetic field \( 1.75 \times 10^{-5} \mathrm{~T} \). Then the emf developed between the tips of the wings is
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direction at a constant altitude in the northern hemisphere, where the vertical component of the
earth's magnetic field \( 1.75 \times 10^{-5} \mathrm{~T} \). Then the emf developed between the tips of the wings is
Solution:
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Verified Answer
The correct answer is:
\( 0.21 \mathrm{~V} \)
Given, length of wing, $l=40 \mathrm{~m}$; speed, $v=1080 \mathrm{kmh}^{-1}$
$=1080 \times \frac{5}{18} \mathrm{~ms}^{-1}$
Earth's magnetic field,$B=1.75 \times 10^{-5} \mathrm{~T}$
Then emf
$E=B l v=\left(1.75 \times 10^{-5}\right) \times 40 \times 1080 \times \frac{5}{18}$
$=21000 \times 10^{-5}=0.21 \mathrm{~V}$
Therefore, the emf developed between the tips of the wing is $0.21 \mathrm{~V}$
$=1080 \times \frac{5}{18} \mathrm{~ms}^{-1}$
Earth's magnetic field,$B=1.75 \times 10^{-5} \mathrm{~T}$
Then emf
$E=B l v=\left(1.75 \times 10^{-5}\right) \times 40 \times 1080 \times \frac{5}{18}$
$=21000 \times 10^{-5}=0.21 \mathrm{~V}$
Therefore, the emf developed between the tips of the wing is $0.21 \mathrm{~V}$
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