Explanation

(a) As the block is of metal, the charge carriers are electrons, so for current along positive $x$-axis, the electrons are moving along negative $x$-axis, $i.e.\ \overrightarrow{v} = - v\widehat{i}$ and as the magnetic field is along the $y$-axis, $i.e.\ \overrightarrow{B} = B\widehat{j}$. So $\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})$ for this case yield $\overrightarrow{F} = ( - e)\left\lbrack - v\widehat{i} \times B\widehat{j} \right\rbrack$
$i.e.,\ \overrightarrow{F} = evB\widehat{k}\ $[As $\widehat{i} \times \widehat{j} = \widehat{k}$]
As force on electrons is towards the face $ABCD$, the electrons will accumulate on it an hence it will acquire lower potential