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 __________.

Solution

Physics sir · 3 steps

A wing moving horizontally can only cut the vertical part of earth's field, so use B sin δ.

1. Which component. The wings lie horizontal and the plane moves horizontally. The horizontal component of the field is parallel to the plane of motion, so it gives no emf. Only the vertical component B_V = B sin δ is cut by the wings.
2. Convert the speed. v = 180 km/h = 180 × 5/18 = 50 m s⁻¹. The vertical field is B_V = 2.5 × 10⁻⁴ × sin 60° = 2.165 × 10⁻⁴ Wb m⁻².

   ε = B_V l v = 2.165 × 10⁻⁴ × 10 × 50 = 0.10825 V

3. So. The emf between the wing tips is 108.25 mV, option (d).

The idea to keep. For a rod moving horizontally, only the field component perpendicular to both the rod and the velocity counts. For a plane in flight that is the vertical component, B sin δ.

Answer from the NTA final answer key. Solution by PrepWiser.