Explain: Velocity selector.

(N/A) charge $q$ moving with velocity $\vec{v}$ in the presence of both electric and magnetic fields experiences a Lorentz force given by:
$\vec{F} = \vec{F}_{E} + \vec{F}_{B} = q\vec{E} + q(\vec{v} \times \vec{B}) \dots (1)$
Consider a configuration where the electric field $\vec{E}$ and magnetic field $\vec{B}$ are perpendicular to each other and both are perpendicular to the velocity $\vec{v}$ of the particle,as shown in the figure.
Let $\vec{E} = E\hat{j}$ and $\vec{B} = B\hat{k}$,with the particle moving along the $x$-axis $(\vec{v} = v\hat{i})$.
The electric force is $\vec{F}_{E} = q\vec{E} = qE\hat{j}$.
The magnetic force is $\vec{F}_{B} = q(\vec{v} \times \vec{B}) = q(v\hat{i} \times B\hat{k}) = -qvB\hat{j} \dots (2)$ (since $\hat{i} \times \hat{k} = -\hat{j}$).
Thus,$\vec{F}_{E}$ and $\vec{F}_{B}$ act in opposite directions.
If we adjust the magnitudes of $\vec{E}$ and $\vec{B}$ such that the magnitudes of the two forces are equal,the total force on the charge becomes zero,and the particle moves undeflected.
Setting $|\vec{F}_{E}| = |\vec{F}_{B}|$,we get $qE = qvB$.
Therefore,$v = \frac{E}{B} \dots (3)$.
This condition allows us to select charged particles of a specific velocity $v = \frac{E}{B}$ from a beam containing particles with different speeds,regardless of their charge or mass. The crossed $\vec{E}$ and $\vec{B}$ fields act as a velocity selector,allowing only particles with speed $\frac{E}{B}$ to pass through undeflected.