I heart physics

The equation of motion for the electron in the polar coordinate around the wire is:

$m(2\dot{r}\dot{\theta} + r \ddot{\theta}) = m\frac{d(r^2\dot{\theta})}{rdt}= eB\dot{r}=eB\frac{dr}{dt} \Rightarrow m\frac{d(r^2\dot{\theta})}{dt}= eB\dot{r}=eB\frac{d(r^2)}{2dt}$

$\Rightarrow mr^2\dot{\theta} =\frac{eBr^2}{2} \Rightarrow \dot{\theta}=\frac{eB}{2m}=\omega$

$m(\ddot{r}-r\dot{\theta}^2)=eE-eBr\dot{\theta} \Rightarrow \ddot{r}+\frac{e^2B^2}{4m^2}r=\ddot{r}+\omega^2 r=eE \Rightarrow r(t)=\frac{4m^2E}{eB^2}-R cos{(\omega t + \phi)}$

Using the boundary condition $r(t=0)=0$ and $\dot{r}(t=0)=v\approx 0$ we get $\phi=0$ and $R=\frac{4m^2E}{eB^2}$ :

$r=\frac{4m^2E}{eB^2}(1-\cos{(\omega t)})=\frac{4m^2E}{eB^2}(1-\cos{\theta})$

The orbit looks like a heart!

The total length is pretty easy to calculate:

$s=\int_0^{2\pi} (dr^2 + r^2 d\theta^2)^{1/2} = \int_0^{2\pi} \frac{4m^2E}{eB^2}\sqrt{2+2\cos{\theta}} d\theta = \frac{32 m^2E}{eB}=32(m)$

Here's how the graph looks like: