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An aeroplane is travelling horizontally towards west with a speed of $540 \mathrm{kmh}^{-1}$. The wing span of the plane is 20 $\mathrm{m}$. If the horizontal component of the earth's magnetic field at the location is $2.5 \sqrt{3} \times 10^{-4} \mathrm{~T}$ and the dip angle is $30^{\circ}$, the potential difference developed between the ends of the wing is
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Verified Answer
The correct answer is:
$0.75 \mathrm{~V}$
Speed of aeroplane, $v=540 \mathrm{~km} / \mathrm{h}$
$=540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}$
Distance between the ends of the wings, $1=20 \mathrm{~m}$
Magnetic field, $B=2.5 \sqrt{3} \times 10^{-4} \mathrm{~T}$
Dip angle, $\theta=30^{\circ}$
The potential differences developed between the ends of the wings
$\begin{aligned}
& \mathrm{E}=\mathrm{B}_{1+} \mathrm{lv} \sin 30 \quad \because \mathrm{B}_{1+}=\mathrm{B} \tan 30 \\
& =2.5 \sqrt{3} \times 10^{-4} \times 20 \times 540 \times \frac{5}{18} \times \tan 30 \\
& =0.75 \mathrm{~V}
\end{aligned}$
$=540 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}$
Distance between the ends of the wings, $1=20 \mathrm{~m}$
Magnetic field, $B=2.5 \sqrt{3} \times 10^{-4} \mathrm{~T}$
Dip angle, $\theta=30^{\circ}$
The potential differences developed between the ends of the wings
$\begin{aligned}
& \mathrm{E}=\mathrm{B}_{1+} \mathrm{lv} \sin 30 \quad \because \mathrm{B}_{1+}=\mathrm{B} \tan 30 \\
& =2.5 \sqrt{3} \times 10^{-4} \times 20 \times 540 \times \frac{5}{18} \times \tan 30 \\
& =0.75 \mathrm{~V}
\end{aligned}$
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