Question

A jet plane is flying west at the speed of 1,200 kmh"¹ parallel to the surface of earth. What is the voltage difference developed between the ends of its wing 24 m long, if the earth's magnetic field at the location has a magnitude of 4.92 x 10"⁴T and angle of dip is 30°?​

Answer 1

Given :

Speed of jet plane =v= 1200 km/h

= 1200x5/18 m/s

l=Distance= 24m

earth's magnetic field=B= 4.92 x 10 ⁻⁴T

angle of dip=δ=30°

Formula to be used :

e= BVl=Bsinδvl

e=[4.92 x 10 ⁻⁴sin30°x1200x5x24 ]/18

=4.92 x 10 ⁻⁴ x1200x5x24x1/18 x 2

=19680x 10 ⁻⁴  V

=1.968V

∴voltage difference developed between the ends of its wing  is 1.968V

Answer 2

e = B v l cosФ

v = 500/3 m/s ; B = 4.92 X 10⁻⁴ T ; l = 24 m ; cosФ = cos30 = √3 /2

e = 4.92 X 10⁻⁴ X 500 X 24 X √3 / 6

e = 5.904 V