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ExamsJEE MainPhysics

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:

  1. 108.25 mV
  2. 54.125 mV
  3. 88.37 mV
  4. 62.50 mV

Correct answer: 108.25 mV

Solution

Vertical field = 2.5e-4 * sin60 = 2.165e-4 Wb/m^2; v = 180 km/h = 50 m/s; emf = 2.165e-4 * 10 * 50 = 0.10825 V = 108.25 mV.

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