Question

an aeroplane is flying horizontally from west to east with a velocity of 900 km/hr calculate the potential difference developed between the ends of its wings having a span of 20 m the horizontal component of the earths magnetic field is 5×10⁻⁴T and the angle of dip is 30°

Answer 1

First of all we will convert the velocity into m/s as it is provided in km/hr.

It will be 250 m/s

We know that the length of wing is 20 m

and horizontal component is 5×10⁻⁴

and the angle of dip is 30°

E = Blvsin(θ)

E = Blv Sin 30

E = 5*10⁻⁴ * 20*250* sin 30

= 1.25v

Answer 2

_____?

Solution!!

Potential difference developed between the ends of the wings e = Blv

Given velocity ;

v = 900 km/hr

v = 250 m/s

The horizontal component of Earth's magnetic field = 5* 10^-4 T

wing spam (I) = 20m

Now..,

The vertical component of Earth's Magnetic field

Bv = BH tanδ

= 5* 10^-4 (tan 30°) T

∴The potential difference,

e = 5*10^-4 (tan 30°) * 2 * 250

e = 5 * 10^-4 * 20 * 250

√3

e = 1.44 V✅