Motion of a Charged Particle in Constant Electric and Magnetic Fields

The motion of a particle with charge $q$ and mass $m$ in a constant magnetic field aligned along the $z$-axis, $\vec{B}=B_z\,\hat{z}$, and an arbitrary electric field $\vec{E}$, is described by,

$y$

$x$

 $$\vec{r} =\left( x_0+\frac{E_y}{B_z}t+R\cos(\omega t + \phi)-R\cos(\phi)\right)\,\hat{x} \\ + \left( y_0 -\frac{E_x}{B_z}t+ R\sin(\omega t + \phi)-R\sin(\phi)\right)\,\hat{y} + \left(z_0 +v_{z0}t +\frac{qE_z}{2m}t^2\right)\,\hat{z},$$ $$\vec{v} =\frac{d\vec{r}}{dt}= \left(\frac{E_y}{B_z}-\omega R\sin(\omega t + \phi)\right)\,\hat{x}+ \left(-\frac{E_x}{B_z}+\omega R\cos(\omega t + \phi)\right)\,\hat{y} + \left(v_{z0}+\frac{qE_z}{m}t\right)\,\hat{z},$$ $$\vec{a} = \frac{d\vec{v}}{dt}= -\omega^2 R\cos(\omega t + \phi)\,\hat{x} - \omega^2 R\sin(\omega t + \phi)\,\hat{y} + \frac{qE_z}{m}\,\hat{z},$$ $$\vec{F} = m\vec{a} = -m\omega^2 R\cos(\omega t + \phi)\,\hat{x} - m\omega^2 R\sin(\omega t + \phi)\,\hat{y} + qE_z\,\hat{z},$$

$x_0=0$ [m], $y_0=0$ [m], $v_{x0}=0$ [m/s],
$m=1$ [kg], $q=1$ [C]

$B_{z}=$ 0 [T]

$E_x=$ 2 [V/m]

$E_y=$ 2 [V/m]

$v_{x0}=$ 4 [m/s]

$v_{y0}=$ 4 [m/s]

 $$\omega = -\frac{qB_z}{m},\qquad \tan(\phi)= \frac{\frac{E_y}{B_z}-v_{x0}}{\frac{E_x}{B_z}+v_{y0}}, \\ R=\frac{1}{|\omega|}\sqrt{\left(v_{x0}-\frac{E_y}{B_z}\right)^2+\left(v_{y0}+\frac{E_x}{B_z}\right)^2}.$$

Consider electric and magnetic fields that are constant in space and time. Assume that the magnetic field is aligned with the $z$-axis but that the electric field can be in any direction. The force on a charged particle in these fields is the Lorentz force,

$$\vec{F} =q(\vec{E}+\vec{v}\times\vec{B}).$$

The force in the $z$-direction is a constant, $qE_z$, so the motion in the $z$-direction is $z=z_0+v_{z0}t+\frac{qE_z}{2m}t^2$, where $z_0$ and $v_{z0}$ are the position and the velocity in the $z$-direction at $t=0$. Both electric and magnetic forces act in the $x$- and $y$-directions which results in a circular motion that moves with a constant velocity in a direction perpendicular to the component of the electric field in the $x-y$ plane. The position vector that describes this motion is,

$$\vec{r} =\left( x_0+\frac{E_y}{B_z}t+R\cos(\omega t + \phi)-R\cos(\phi)\right)\,\hat{x} + \left( y_0 -\frac{E_x}{B_z}t+ R\sin(\omega t + \phi)-R\sin(\phi)\right)\,\hat{y} + \left(z_0 +v_{z0}t +\frac{qE_z}{2m}t^2\right)\,\hat{z},$$

Here $R$ is the radius of the circle, $\omega$ is the angular frequency, and $\phi$ is a phase that will be specified below. Note that at $t=0$ the position of the particle is $(x_0,y_0,z_0)$. Differentiating to determine the velocity vector yields,

$$\vec{v} =\frac{d\vec{r}}{dt}= \left(\frac{E_y}{B_z}-\omega R\sin(\omega t + \phi)\right)\,\hat{x}+ \left(-\frac{E_x}{B_z}+\omega R\cos(\omega t + \phi)\right)\,\hat{y} + \left(v_{z0}+\frac{qE_z}{m}t\right)\,\hat{z}.$$

Evaluating this expression at $t=0$ results in the relations $v_{x0} = \frac{E_y}{B_z}-\omega R\sin(\phi)$ and $v_{y0} = -\frac{E_x}{B_z}+\omega R\cos(\phi)$ which can be used to determine $\phi$ and $R$,

$$\tan(\phi)= \frac{\frac{E_y}{B_z}-v_{x0}}{\frac{E_x}{B_z}+v_{y0}}, \qquad R=\frac{1}{|\omega|}\sqrt{\left(v_{x0}-\frac{E_y}{B_z}\right)^2+\left(v_{y0}+\frac{E_x}{B_z}\right)^2}.$$

Differentiating the velocity yields the acceleration,

$$\vec{a} = \frac{d\vec{v}}{dt}= -\omega^2 R\cos(\omega t + \phi)\,\hat{x} - \omega^2 R\sin(\omega t + \phi)\,\hat{y} + \frac{qE_z}{m}\,\hat{z}.$$

Multiplying by the mass gives the force vector,

$$\vec{F} = m\vec{a} = -m\omega^2 R\cos(\omega t + \phi)\,\hat{x} - m\omega^2 R\sin(\omega t + \phi)\,\hat{y} + qE_z\,\hat{z}.$$

This expression for the force can be equated with $\vec{F} =q(\vec{E}+\vec{v}\times\vec{B})$. The equations for the $x$- and $y$-components are,

$$ -m\omega^2 R\cos(\omega t + \phi)= \left(qE_x-q\frac{E_x}{B_z}B_z+q\omega R\cos(\omega t + \phi)B_z\right),$$ $$ - m\omega^2 R\sin(\omega t + \phi)=\left( qE_y -q\frac{E_y}{B_z}B_z+q\omega R\sin(\omega t + \phi)B_z\right).$$

The terms containing the electric fields cancel out and comparing the remaining terms we find that the angular frequency is,

$$\omega = -\frac{qB_z}{m}. $$

This problem can also be solved numerically with a Numerical 6th order differential equation solver.