Q.

An aeroplane, with its wings spread 10m, is flying at a speed of 180km/h in a horizontal direction. The total intensity of earth’s field at that part is 2.5×10−4Wb/m2 and the angle of dip is 60o . The emf induced between the tips of the plane wings will be ______

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a

108.25 mV

b

54.125 mV

c

88.37 mV

d

62.50mV

answer is A.

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Detailed Solution

Vertical component of magnetic field, Bv=Besin⁡δ=2.5×10−4×32TEmf induced =BvLv=2.5×10−4×32×10×50=108.25mV
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An aeroplane, with its wings spread 10m, is flying at a speed of 180km/h in a horizontal direction. The total intensity of earth’s field at that part is 2.5×10−4Wb/m2 and the angle of dip is 60o . The emf induced between the tips of the plane wings will be ______