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If the vertical component of earth's magnetic field is $0.5 \times$ $10^{-4} \mathrm{~T}$ at a point. When an aeroplane of wing span $4 \mathrm{~m}$ is moving horizontally at this place at $360 \mathrm{kmh}^{-1}$, then the motional emf formed across the ends of the wings is
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The correct answer is:
$20 \times 10^{-3} \mathrm{~V}$
Vertical component of earth Magnetic field, $B_v=0.5 \times 10^{-4} \mathrm{~T}$ Aeroplane wing, $\ell=4 \mathrm{~m}$.
Velocity, $\mathrm{v}=360 \mathrm{~km} / \mathrm{h}=\frac{360 \times 1000}{3600}=100 \mathrm{~m} / \mathrm{s}$ Emf is given by
$\begin{aligned}
& \varepsilon=\mathrm{Bv} \ell \\
& =0.5 \times 10^{-4} \times 100 \times 4 \\
& =2 \times 10^{-2} \\
& \varepsilon=20 \times 10^{-3} \mathrm{~V}
\end{aligned}$
Velocity, $\mathrm{v}=360 \mathrm{~km} / \mathrm{h}=\frac{360 \times 1000}{3600}=100 \mathrm{~m} / \mathrm{s}$ Emf is given by
$\begin{aligned}
& \varepsilon=\mathrm{Bv} \ell \\
& =0.5 \times 10^{-4} \times 100 \times 4 \\
& =2 \times 10^{-2} \\
& \varepsilon=20 \times 10^{-3} \mathrm{~V}
\end{aligned}$
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