Consider an airbus a380 of wing span 798 m flying horizontal

Consider an airbus a380 of wing span 79.8 m. flying horizontally at a cruise speed (Mach = 0.85). due north in a region where the earth s magnet a dip angle of 72.5 degree. (a) Find the emf induced across wing. (b) Which end of the wing is positive? Explain.

Solution

Here ,

length , L = 79.8 m

speed of plane , v = 1009.8 km/hr = 280.5 m/s

theta = 72.5 degree

B = 5.5 * 10^-5 T

a) induced emf across the wing = B * v * L * sin(72.5)

induced emf across the wing = 79.8 * 5.5 *10^-5 * 280.5 * sin(72.5)

induced emf across the wing = 1.174 V

b)

as the magnetic force on the charges is given as

F = q * V X B

the direction of magnetic force on positive charge is to the east

hence, the east end of wing is positive