JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Electromagnetic Induction

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 1 0^{- 4} Wb / m^{2}

and the angle of dip is

6 0^{\circ}

. The emf induced between the tips of the plane wings will be ______

Answer: (D)

Solution

The wings are horizontal and sweep a horizontal area, so the field component that drives the motional emf across the span is the vertical one:

B_{V} = B sin δ = 2.5 \times 1 0^{- 4} \times sin 6 0^{\circ} = 2.165 \times 1 0^{- 4} Wb m^{- 2}

With the span

l = 10

m and the speed

v = 180 km h^{- 1} = 50 m s^{- 1}

ε = B_{V} l v = 2.165 \times 1 0^{- 4} \times 10 \times 50 = 1.0825 \times 1 0^{- 1} V

The emf between the wing tips is

108.25

mV.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Single correct option