Question

The aircraft with a wing span of $ 40m $ flies with a speed of $ 1080km{h^{ - 1}} $ in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth’s magnetic field is $ 1.75 \times {10^{ - 5}}T $ . Then the e.m.f. developed between the tips of the wings is:
(A) $ 0.5V $
(B) $ 0.34V $
(C) $ 0.21V $
(D) $ 2.1V $

Answer 1

Hint :Here we have to find the e.m.f. induced between the tips of the wings, as we have been given the magnetic field , speed and length of wing span so, we have to use the concept of e.m.f. developed due to the magnetic field and applying that formula. Convert the kilometer unit to meters for better calculation.

Complete Step By Step Answer:
According to the question the given data is as follows:
 $ l = 40m $….(length of wing span)
 $ v = 1080km{h^{ - 1}} $….(speed of aircraft)
 $ B = 1.75 \times {10^{ - 5}}T $…..(magnetic field)
Now let us convert speed in $ m{s^{ - 1}} $ such that,
 $ v = 1080km{h^{ - 1}} = \dfrac{{1080 \times 1000m}}{{60 \times 60s}} $ $ = 300m{s^{ - 1}} $
Now the formula for induced e.m.f. is given by
 $ e = Blv $……(definition)
Put all the required values we have in above equation, we get
 $ e = 1.75 \times {10^{ - 5}} \times 40 \times 300 $
On calculating, $ e = 0.21volt $
Thus, the e.m.f. developed between the tips of the wings is $ 0.21volt $ or $ 0.21V $
The correct option is C.

Note :
One must read the question carefully and understand what concept is to be used here because there are lots of terms given in the question like hemisphere and direction of the aircraft with respect to the earth’s magnetic field. If e.m.f. is asked then it is directly proportional to the speed of the object, magnetic field and length of the object.