The Ultimate problem!

This is not essentially a EM problem save the start, it's all mathematics as you will see.

First, take the Lorentz force i.e.

$\displaystyle F = q(E + v\times B)$

Taking x-component first,

$\displaystyle {F}_{x} = q({E}_{x} + {v}_{y}{B}_{z} - {v}_{z}{B}_{y})$

In this case,

$\displaystyle m\frac{{d}^{2}x}{d{t}^{2}} = q(0 + \frac{dy}{dt}B - 0)$

$\displaystyle = \frac{{d}^{2}x}{d{t}^{2}} = \frac{qB}{m}\frac{dy}{dt}$

We will just write

$\displaystyle x'' - \frac{qB}{m}y' = 0$

Now, taking y-component,

$\displaystyle {F}_{y} = q({E}_{y} + {v}_{z}{B}_{x} - {v}_{x}{B}_{z})$

In this case,

$\displaystyle \frac{{d}^{2}y}{d{t}^{2}} = \frac{q}{m}(E - \frac{dx}{dt}B)$

We will just write

$\displaystyle y'' + \frac{qB}{m}x' - \frac{qE}{m} = 0$

Hence, we have got our 2 DEs in 2 variables.

We will "mix" the 2 DEs together so that it becomes easy to find the answer.

To ease writing, let $\frac{qB}{m} = \omega$

We know

$\displaystyle x'' - \omega y' = 0$

Hence, $\displaystyle x' = \omega y$

Just substitute this in the 2nd DE,

$\displaystyle y'' + {\omega}^{2}y - \frac{qE}{m} = 0$

Now, we can write $\frac{qE}{m} = {\omega}^{2}\times \frac{mE}{q{B}^{2}}$ , hence, we will let $\frac{mE}{q{B}^{2}} = c$

Then,

$\displaystyle y'' + {\omega}^{2}y - {\omega}^{2}c = 0$ is our 2nd order ODE for y.

One can find solutions for it in many ways - Probably with Lagrangian or some other transform(I don't have much knowledge there). So, here we will just guess the solution.

We know that ${\omega}^{2}c$ is a constant and would be cancelled by its additive inverse, which must come from the $y$ term since for it to come from $y''$ term would create complications. Let's just keep it simple.

Then, it must be that $y''$ term must be cancelled by the rest $y$ term. And our first guess must be that $y$ looks something like $c - dcos(\omega t)$ since then its double derivative would come ${\omega}^{2}dcos(\omega t)$ which will cancel our rest in y term.

let's just keep it simple and

$\displaystyle y = c(1 - cos(\omega t)$

As a result $\displaystyle x' = c\omega(1-cos(\omega t)) \Rightarrow$ $\displaystyle x = c(\omega t - sin(\omega t))$

We have got equations for cycloid!!!

We can see now that for the trajectory to come back to x-axis, "circle of it" must traverse $2\pi$ in y-axis w.r.t x-axis or

$\displaystyle \text{Area} = \int_{\omega t=0}^{\omega t=2\pi}{y dx} = \int_{0}^{2\pi}{{c}^{2}{(1 - cos(\omega t))}^{2}} = 3\pi{c}^{2}$

$\displaystyle \text{Length} = \int_{\omega t = 0}^{\omega t =2\pi}{\sqrt{{\left(\frac{dy}{dt}\right)}^{2} + {\left(\frac{dx}{dt}\right)}^{2}} d\omega t} = 8c$

And it is obvious that $\omega$ is the angular velocity.

So our final answer becomes

$\displaystyle \frac{3\pi {c}^{2}}{8c(\omega)} = \frac{3\pi c}{8\omega}$

Substitute values for $\displaystyle c = \frac{mE}{q{B}^{2}}, \omega = \frac{qB}{m}$

$\displaystyle \frac{3 \pi \frac{mE}{q{B}^{2}}}{8 \frac{qB}{m}} = \frac{3 \pi {m}^{2}E}{8{q}^{2}{B}^{3}}$

$\displaystyle \frac{3 \pi {m}^{2}{E}^{1}{q}^{-2}{B}^{-3}}{8}$