Proton in electric and magnetic fields

The proton is accelerated by the Lorentz force $\vec F_L = e (\vec E + \vec v \times \vec B)$ where $e$ is the elementary charge of the proton. For the time $t = 0$ we have $\begin{aligned} \vec F_L &= e (- 1000 \cdot \vec e_y + 10^5 \cdot 0.01 \cdot (\vec e_x \times \vec e_z) )\,\frac{\text{V}}{\text{m}} \\ &=e (- 1000 \cdot \vec e_y + 1000 \cdot (- \vec e_y) )\,\frac{\text{V}}{\text{m}} \\ &\approx -3.2 \cdot 10^{-16} \,\text{N} \cdot \vec e_y \end{aligned}$ The force is non-zero and points perpendicular to the initial direction of motion, so we can exclude the line as a possible solution for the trajectory. In particular, the electric and magnetic forces are exacly equal. Therefore, the circle and the parabola are to be excluded as possible trajectories, since these cases occur only for a purely magnetic field and a purely electric field. In the following, I will show that the solution for the trajectory is indeed a cycloid.

We are now looking for a solution of the differential equation $\frac{d \vec v(t)}{d t} = \left( \begin{array}{c} \dot v_x(t) \\ \dot v_y(t) \end{array} \right) = \frac{1}{m} \vec F_L = \frac{e}{m} \left( \begin{array}{c} v_y(t) B \\ - E - v_x(t) B \end{array} \right)$ where $B = 10\,\text{mT}$ and $E = 1\,\text{kV}/\text{m}$ are the field amplitudes. The force is zero in the case $\vec v = - \vec v_0 = - E/B \cdot \vec e_x$ , so that this implies an stationary solution $\vec v = \text{const}$ . By substitution $\vec w = \vec v + \vec v_0$ , the equations of motions simplify $\left( \begin{array}{c} \dot w_x \\ \dot w_y \end{array} \right) = \frac{e B}{m} \left( \begin{array}{c} w_y \\ - w_x \end{array} \right)$ This cases correspond to a movement in a pure magnetic field. The solution of this coupled differential equation with the initial condition $\vec w = 2 \vec v_0$ is a circular motion with the angular frequency $\omega = e B / m$ : $\begin{aligned} & & \left( \begin{array}{c} w_x \\ w_y \end{array} \right) &= \left( \begin{array}{c} 2 v_0 \cos (\omega t) \\ 2 v_0 \sin(\omega t) \end{array} \right) \\ \Rightarrow & & \left( \begin{array}{c} v_x \\ v_y \end{array} \right) &= \left( \begin{array}{c} -v_0 + 2 v_0 \cos (\omega t) \\ 2 v_0 \sin(\omega t) \end{array} \right) \\ \Rightarrow & & \left( \begin{array}{c} x \\ y \end{array} \right) &= \int_{0}^t \left( \begin{array}{c} v_x \\ v_y \end{array} \right) dt = \left( \begin{array}{c} - v_0 t + \frac{2 v_0}{\omega} \sin (\omega t) \\ -\frac{2 v_0}{\omega} \cos(\omega t) \end{array} \right) + \text{const} \end{aligned}$ The resulting trajectory thus represents a superposition of a circular motion and a rectilinear motion. This path corresponds to an elongated cycloid .