An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 × 10 - 4 Wb/m2 and the angle of dip is 60 °. The emf induced between the tips of the plane wings will be __________. (a) 88.37 mV (b) 62.50 mV (c) 54.125 mV (d) 108.25 mV

Answer: (d). Horizontal wings cut the vertical component: BV = Bsin δ = 2.5×10⁻⁴ × sin 60° = 2.165×10⁻⁴ T. v = 180 km/h = 50 m/s. ε = BV l v = 2.165×10⁻⁴ × 10 × 50 = 0.10825 V = 108.25 mV. So the answer is (d).

An aeroplane, with its wings spread 10 m, is flying at a speed of 180… | JEE Main 2021 (26 Feb, Shift 2)

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Motional EMF

An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 $\times$ 10 $-$ 4 Wb/m2 and the angle of dip is 60 $^{\circ}$ . The emf induced between the tips of the plane wings will be __________.

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Answer: (d)

Horizontal wings cut the vertical component: $B_V = B\sin\delta = 2.5\times10^{-4} \times \sin60^\circ = 2.165\times10^{-4}$ T.
$v = 180$ km/h $= 50$ m/s.
$\varepsilon = B_V l v = 2.165\times10^{-4} \times 10 \times 50 = 0.10825$ V
$= 108.25$ mV. So the answer is (d).

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