JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Electromagnetic Induction

As the metal cube drifts through the field its free charges feel $q\vec{v}\times \vec{B}$ and pile up until the built-up electric field cancels the magnetic push, leaving a motional emf $\mathcal{E}=Bvd$. Here $\vec{v}$ is along $+y$ and $\vec{B}$ along $+z$, so $\vec{v}\times \vec{B}$ points along $x$, and the relevant edge is the one perpendicular to $x$, $d=0.02$ m. Thus $\mathcal{E}=0.1\times 6\times 0.02=0.012$ V, that is $12$ mV, appearing across the two faces perpendicular to the $x$-axis.