Question

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 ______.

पर्याय

• 108.25 mV

• 88.37 mV

• 62.50  mV

• 54.125 mV

Answer 1

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 108.25 mV.

Explanation:

Emf induced ε_ind = (B_v) LV and

B_v = ${\displaystyle {\text{B}}_{\text{Total}}}$sin 60°

= (2.5 × 10^-4) (sin 60°) × 10 × 180 × ${\displaystyle \frac{5}{18}}$

= ${\displaystyle 2.5\times {10}^{-4}\times \frac{\sqrt{3}}{2}\times 500}$

= 108.25 mV