A charge $q$ moving with velocity $\vec{v}$ in presence of both electric and magnetic fields experiences a force given by,

$\overrightarrow{\mathrm{F}}=\overrightarrow{\mathrm{F}}_{\mathrm{E}}+\overrightarrow{\mathrm{F}}_{\mathrm{B}}$

$=\overrightarrow{\mathrm{E}} q+q(\vec{v}+\overrightarrow{\mathrm{B}}) \ldots \text { (1) }$

Electric and magnetic field are perpendicular to each other and also perpendicular to the velocity of the particle as shown in figure.

$\overrightarrow{\mathrm{F}}_{\mathrm{E}}= q \overrightarrow{\mathrm{E}}=q \mathrm{E} \hat{\mathrm{j}}$

$\text { and } \overrightarrow{\mathrm{F}}_{\mathrm{B}} =q(\vec{v} \times \overrightarrow{\mathrm{B}})$ $…(2)$ 

$=q(v \hat{i} \times \mathrm{B} \hat{k})$

$\overrightarrow{\mathrm{F}}_{\mathrm{B}}=-q v \mathrm{~B}(\hat{j}) \ldots \text { (3) }(\because \hat{i} \times \hat{k}=-\hat{j})$

Thus, $\vec{F}_{E}$ and $\vec{F}_{B}$ both are opposite to each other.

Suppose, we adjust the value of $\overrightarrow{\mathrm{E}}$ and $\overrightarrow{\mathrm{B}}$ such that magnitudes of the two forces are equal. Then, total force on the charge is zero and the charge will move in the fields undeflected.

Thus, $\mathrm{E} q=q v \mathrm{~B}$

$\therefore v=\frac{\mathrm{E}}{\mathrm{B}}$ $…(4)$ 

This condition can be used to select charged particles of a particular velocity out of a beam containing charges moving with different speeds (irrespective of their charge and mass). The crossed $\vec{E}$ and $\vec{B}$ fields therefore, serve as a velocity selector. Only particles with speed $\frac{E}{B}$ pass undeflected through the region of crossed fields.