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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
Options:
\( 400 \mathrm{~ms}^{-1} \). 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
Solution:
1474 Upvotes
Verified Answer
The correct answer is:
\( 1.6 \mathrm{~V} \)
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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