Q 12-06-095JEE MainJEE Main 2022 (26 Jun, Shift 2)Medium
A metallic conductor of length $1\ \text{m}$ rotates in a vertical plane parallel to the east-west direction about one of its ends with angular velocity $5\ \text{rad s}^{-1}$. If the horizontal component of earth's magnetic field is $0.2\times10^{-4}\ \text{T}$, then the emf induced between the two ends of the conductor is
Answer: (B) $50\ \mu\text{V}$
The rod rotates in the east–west vertical plane, so the horizontal (north–south) component of the earth's field is perpendicular to its plane of rotation.
$$\varepsilon = \frac12B_H\omega l^2 = \frac12\times0.2\times10^{-4}\times5\times1^2 = 5\times10^{-5}\ \text{V} = 50\ \mu\text{V}$$
Solution by Sreeraj P, M.Sc Physics