# What is the motion of two charges constrained by a bar in a magnetic field?

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: $$$$m\ddot{\textbf{r}}_1=\textbf{F}_1=q\dot{\textbf{r}}_1\times \textbf{B}+\frac{\mu_0q^2}{4\pi}\frac{1}{\lvert\textbf{r}_1-\textbf{r}_2\rvert^2}\dot{\textbf{r}}_1\times(\dot{\textbf{r}}_2\times\textbf{r}_1)$$$$ For the second mass we can write a similar equation, with the subscripts $$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 $$\theta$$,

$$$$\ddot{\theta}=\frac{q}{mR}\dot{\theta}\cos{\theta}$$$$

So this should be how $$\theta$$, namely the angle between bar and $$B$$, change with time.

What do you think?