Question

Airplane emf A Boeing KC-135A airplane has a wingspan of 39.9 m and flies at constant altitude in a northerly direction with a speed of 910 km/h . You may want to review (Pages 819 - 821) . If the vertical component of the Earth's magnetic field is 4.8×10−6 T , and its horizontal component is 1.5×10−6 T , what is the induced emf between the wing tips?

Answer 1

Each of the wing has length of 39.9 m. So the tip-to-tip distance between the two wings will be 39.9 x 2 m = 79.8 m. So this metallic length swipes a distance of 910 km per hour or

910*1000/3600 m per sec = 252.78 m/s. So the overall area swept will be 252.78 * 79.8 m²/s.

We know magnetic flux            

Here area S is horizontal and so dS vector will point veritically. So only the vertical component of B will contribute here as the horizontal part will give zero.

So,

But from Lenz's law we know that induced emf

So we obtain the induced emf to be 0.097 Joule/Coulomb.