Suppose that you have a rod in a uniform magnetic field $B$. The field points in the $z$ direction, while the rod is initially perpendicular to it. At the two ends of the rod, two point-like charges with the same mass $m$ and charge $q$ are fixed. Suppose that you give to the charges an instantaneous velocity, along the $z$ axis. The velocities have the same magnitude, say $v$, but they point in opposite direction. What is the subsequent motion of the bar?
I would say that the bar cannot translate. Rather, it rotates in some complicated way in a plane with increasingly high angular speed.
Let us write down the equation of motion for the masses. For the first point: \begin{equation} m\ddot{\textbf{r}}_1=\textbf{F}_1=q\dot{\textbf{r}}_1\times \textbf{B}+\frac{\mu_0q^2}{4\pi}\frac{1}{\lvert\textbf{r}_1-\textbf{r}_2\rvert^2}\dot{\textbf{r}}_1\times(\dot{\textbf{r}}_2\times\textbf{r}_1) \end{equation} For the second mass we can write a similar equation, with the subscripts $1$ and $2$ switched.
To derive that, we take into account the fact that when one of the charges moves, it generates a magnetic field (Biot-Savart law). So the other charge feels a Lorentz force due to the initial field $B$ and the new field.
Analyzing the forces, it seems that they are equal and opposite. So there is a torque that makes the rod rotate in a circumference, and there is an angular acceleration.
Hence, we can set $\lvert \textbf{r}_1 \rvert=\lvert \textbf{r}_2 \rvert=R$, $\lvert\textbf{r}_1-\textbf{r}_2\rvert=2R\hspace{3mm}$, as well as $\hspace{3 mm}\dot{\textbf{r}}_1=-\dot{\textbf{r}}_2=R\dot{\theta}\hat{e}_{\theta}$, $\hspace{3 mm}\dot{\textbf{r}}_1 \cdot \textbf{r}_1=0 $.
Then we use the fundamental law of rotations, $\tau=I\ddot{\theta}$, where $\tau$ is the torque, $I=2mR^2$ the moment of inertia around the origin, which we can put in the middle of the rod. Since $\tau=\textbf{r}_1\times \textbf{F}_1+\textbf{r}_2\times\textbf{F}_2$, we have to take the cross product of the right -hand side of the equations of motion (which are the forces) with $\textbf{r}_1$ and $\textbf{r}_2$ respectively to get $\tau$. If we do this, after some simplification, we get a non linear differential equation for $\theta$,
\begin{equation} \ddot{\theta}=\frac{q}{mR}\dot{\theta}\cos{\theta} \end{equation}
So this should be how $\theta$, namely the angle between bar and $B$, change with time.
What do you think?
