Condition
(i)For directions of $\overline{\mathit{E}},\overline{\mathit{B}},\overrightarrow{\mathit{v}}$
(ii)For magnitudes of $\overrightarrow{\mathit{E}},\overrightarrow{\mathit{B}},\overrightarrow{\mathit{v}}$
(i)The velocity $\overrightarrow{\mathit{v}}$ , of the charged particles, and the $\overrightarrow{\mathit{E}}$ and $\overrightarrow{\mathit{B}}$ vectors, should be mutually perpendicular.
Also the forces on $\mathit{q}$ , due to $\overrightarrow{\mathit{E}}$ and $\overrightarrow{\mathit{B}}$, must be oppositely directed.
(Also accept if the student draws a diagram to show the directions.)

(ii) $\mathit{q}\mathit{E}=\mathit{q}\mathit{v}\mathit{B}$

[Alternatively, The student may write : Force due to electric field $=\mathit{q}\overrightarrow{\mathit{E}}$
Force due to magnetic field $=\mathit{q}\left(\overrightarrow{\mathit{v}}\times \overrightarrow{\mathit{B}}\right)$
The required condition is
$\mathit{q}\overrightarrow{\mathit{E}}=-\mathit{q}\left(\overrightarrow{\mathit{v}}\times \overrightarrow{\mathit{B}}\right)$