Q 12-06-128JEE MainJEE Main 2019 (10 Jan, Shift 1)Easy
A solid metal cube of edge length $2\ \text{cm}$ is moving in the positive $y$-direction, at a constant speed of $6\ \text{m s}^{-1}$. There is a uniform magnetic field of $0.1\ \text{T}$ in the positive $z$-direction. The potential difference between the two faces of the cube, perpendicular to the $x$-axis, is
Answer: (A) $12\ \text{mV}$
The force on free charges is along $\vec v\times\vec B \propto \hat j\times\hat k = \hat i$, so charges separate between the faces perpendicular to $x$.
$$V = Bvl = 0.1\times6\times0.02 = 0.012\ \text{V} = 12\ \text{mV}$$
Solution by Sreeraj P, M.Sc Physics