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An aeroplane is flying horizontally with a velocity of $360 \mathrm{~km} \mathrm{~h}^{-1}$. The distance between the tips of the wings of the aeroplane is $50 \mathrm{~m}$. The vertical component of the earth's magnetic field is $4 \times 10^{-4} \mathrm{Wbm}^{-2}$. The induced emf is
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Verified Answer
The correct answer is:
$2 \mathrm{~V}$
Given, velocity of aeroplane,
$$
\begin{aligned}
v &=360 \mathrm{kmh}^{-1} \\
&=360 \times \frac{5}{18} \mathrm{~ms}^{-1}=100 \mathrm{~ms}^{-1}
\end{aligned}
$$
Distance between the tips of wings,
$$
l=50 \mathrm{~m}
$$
Vertical component of earth's magnetic field,
$$
\begin{aligned}
&B_{V}=4 \times 10^{-4} \mathrm{Wbm}^{-2} \\
&\begin{aligned}
\therefore \text { Induced emf, } e &=B_{V} v l \\
&=4 \times 10^{-4} \times 100 \times 50=2 \mathrm{~V}
\end{aligned}
\end{aligned}
$$
$$
\begin{aligned}
v &=360 \mathrm{kmh}^{-1} \\
&=360 \times \frac{5}{18} \mathrm{~ms}^{-1}=100 \mathrm{~ms}^{-1}
\end{aligned}
$$
Distance between the tips of wings,
$$
l=50 \mathrm{~m}
$$
Vertical component of earth's magnetic field,
$$
\begin{aligned}
&B_{V}=4 \times 10^{-4} \mathrm{Wbm}^{-2} \\
&\begin{aligned}
\therefore \text { Induced emf, } e &=B_{V} v l \\
&=4 \times 10^{-4} \times 100 \times 50=2 \mathrm{~V}
\end{aligned}
\end{aligned}
$$
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