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An air plane, with 20 m wingspread is flying at 250 ms^-1 parallel to the earth’s surface at a place where the horizontal component of earth’s magnetic field is $2 \times 10^{- 5} T$ and angle of dip is 60⁰. The magnitude of the induced emf between the tips of the wings is

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$\frac{\sqrt{2}}{10} V$

$\frac{1}{10} V$

$\frac{1}{5} V$

$\frac{\sqrt{3}}{10} V$

answer is C.

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Detailed Solution

Given data:

Length of the wing span l=20m

Earth’s magnetic field $B = 2 \times 10^{- 5} T$

$B_{v} = B_{H} \tan 60^{0} = 2 \sqrt{3} \times 10^{- 5} T$

Velocity $v = 250 m / s$

$E = Blv$

$E = 2 \sqrt{3} \times 10^{- 5} \times 250 \times 20 = \frac{\sqrt{3}}{10} V$

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