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
$e=Bvl$
$=5\times 10^{-3}\times 10\times 100$
$=5V.$