Introduction
An eletric field is a vector function of space $\bm{E}=\bm{E}(\bm{r})$ (and maybe of time). The electric force on a charge $q$ is $\bm{F}_E=q\bm{E}$. Newton’s law is
$$ m\frac{d\bm{v}}{dt} = q\bm{E}_E. $$
[$m$ is the mass of the charged particle.]
Remark. The particle is accelerating in the direction of the field.
Charge in magnetic field
Assume a magnetic field $\bm{B}$. The force on the charged particle is
$$ \bm{F}_B = q\,\bm{\upsilon}\times\bm{B}. $$
Assume a uniform field $\bm{B}=B\bm{\hat{z}}$. Newton’s law gives
$$ m\frac{d\bm{v}}{dt} = q\,\bm{v}\times\bm{B} \Rightarrow \frac{d\bm{v}}{dt} = \frac{qB}{m}\,\bm{v}\times\bm{\hat{z}}. $$
Let us assume motion on the xy-plane, thus the velocity components are given by
$$ \begin{aligned} \dot{v}_x & = \omega_c v_y \\ \dot{v}_y & = -\omega_c v_x\end{aligned}\qquad \qquad\omega_c=\frac{qB}{m}. $$
Remark. The speed (magnitude of velocity) of the particle does not charge during the motion. We can show this as following.
$$ \frac{d |\bm{v}|^2}{dt} = \frac{d (\bm{v}\cdot\bm{v})}{dt} = 2\bm{v}\cdot \frac{d\bm{v}}{dt} = 2\bm{v}\cdot \left( \frac{q}{m}\bm{v}\times\bm{B} \right) = 0.\;\square $$
The equations of motion take a compact form is we define the complex variable
$$ \tilde{v} = v_x + i v_y. $$
We have