Q 12-06-099JEE MainJEE Main 2022 (27 Jun, Shift 2)Medium
A metallic rod of length $20$ cm is placed in North-South direction and is moved at a constant speed of $20\ \text{m s}^{-1}$ towards East. The horizontal component of the Earth's magnetic field at that place is $4\times10^{-3}$ T and the angle of dip is $45^\circ$. The emf induced in the rod is ______ mV.
Numerical value type. Enter your answer.
Answer: 16
A N-S rod moving east cuts only the vertical component of the field:
$$B_V = B_H\tan45^\circ = 4\times10^{-3}\ \text{T}$$
$$\varepsilon = B_V v l = 4\times10^{-3}\times20\times0.2 = 16\times10^{-3}\ \text{V} = 16\ \text{mV}$$
Solution by Sreeraj P, M.Sc Physics