Q 12-06-110JEE MainJEE Main 2021 (26 Feb, Shift 2)Medium
An aeroplane, with its wings spread $10$ m, is flying at a speed of $180$ km h$^{-1}$ in a horizontal direction. The total intensity of earth's field at that part is $2.5\times10^{-4}$ Wb m$^{-2}$ and the angle of dip is $60^\circ$. The emf induced between the tips of the plane wings will be
Answer: (A) $108.25$ mV
A horizontally moving wing cuts the vertical component of the earth's field:
$$B_V = B\sin60^\circ = 2.5\times10^{-4}\times0.866 = 2.165\times10^{-4}\ \text{T}$$
$v = 180$ km h$^{-1}$ $= 50$ m s$^{-1}$.
$$\varepsilon = B_Vlv = 2.165\times10^{-4}\times10\times50 = 0.10825\ \text{V} = 108.25\ \text{mV}$$
Solution by Sreeraj P, M.Sc Physics