A helicopter rises vertically with a speed of 100 m/s. If helicopter has length 10 m and horizontal component of earth's magnetic field is $$5\times 10^{-3} Wb/m^2$$, then the induced emf between the tip of nose and tail of helicopter is:
  • A
    50 V
  • B
    0.5 V
  • C
    5 V
  • D
    25 V

Solution

In case of motional emf, the motion of the conductor in the field exerts a force on the free charge in the conductor, so that one end of the conductor becomes positive, while the other negative resulting in a potential difference across its ends due to which a non-conservative electric field is set up in the conductor. In steady state the magnetic force on the free charge is balanced by the electric force due to induced field. 
$$qE = qvB$$
or $$q\left(\frac{V}{l}\right)=qvB$$
ie, $$V=Bvl$$ 
So, the induced emf between tip of nose and tail of helicopter is given by
 $$e=Bvl$$
$$=5\times 10^{-3}\times 10\times 100$$
$$=5V.$$