Find magnetic field of parabaloid-spiral particle path Given the components of the particle's path (tcos(at), tsin(at), R(x^2 + y^2)), where a and R are constants, how can I find the magnetic force causing this? 
I would like to understand if this can be done with a constant electric field and changing magnetic field and how this can be solved. Is there a way to solve for both B and E?

My attempt would be to take the Lorentz Force equation as a differential and remove E from the equation. From what I understand, the path of the point charge is determined by the velocity-cross-magnetic field only (not sure), so just finding B without E might suffice. Is this accurate, and can it lead to a solution for E?
 A: Great Question!
First of all, you can solve this problem for the Magnetic Field and the Electromagnetic Field. You won't have to do any work with the Lorentz Force Equation. Instead you will use the solutions to Maxwell's Equations.
$$\pmb{\mathcal{E}}=-\nabla \phi-\frac{\partial{\pmb{A}}}{\partial{t}}$$
$$\pmb{\mathcal{B}}=\nabla \times\pmb{A}$$
where $\phi$ is the scalar potential, $\pmb{A}$ is the vector potential,$\nabla$ is the del vector operator. There are many ways to calculate these potentials, but the formulas that I will use are called the Liénard–Wiechert potentials. The Liénard-Wiechert potentials calculate $\phi$ and $\pmb{A}$ of a point charge moving with arbitrary velocity. The potentials are:
$$\phi (\pmb{r},t)=\frac{q}{4 \pi  \epsilon_0  (1-\pmb{n}\cdot
  \pmb{ \beta_s} ) \left\| \pmb{r}-\pmb{r_s}\right\| } $$
$$A(\pmb{r},t)=\frac{\pmb{\beta _s} \phi (\pmb{r},t)}{c}$$
where
$\pmb{\beta _s}=\frac{1}{c}\frac{\partial{\pmb{r_s}}}{\partial{t_s}}$
$\pmb{n}=\frac{r-r_s}{\left\| r-r_s\right\| }$
$t_s=t-\frac{\left\| r-r_s\right\|
   }{c}$
$\pmb{r_s}$ is the position of the point charge in retarded time as experienced by the point charge (due to relativity),$t_s$ is retarded time, $\pmb{r}$ is any point in space, $t$ is time as experience by an observer at $\pmb{r}$, $\epsilon_0$ is the permittivity of a vacuum, $c$ is the speed of light, and $q$ is the charge of the point charge. Now that we have our equations we can start to solve. The vector that describes the movement of the point charge is 
$$\pmb{r_s}=\left(
\begin{array}{c}
 t_s \cos (a t_s) \\
 t_s \sin (a t_s) \\
 R \left(t_s^2 \cos ^2(a t_s)+t_s^2 \sin ^2(a
   t_s)\right) \\
\end{array}
\right)$$
So,
$$\pmb{\beta _s}=\left(
\begin{array}{c}
 \frac{\cos \left(a t_s\right)-a \sin \left(a
   t_s\right) t_s}{c} \\
 \frac{\sin \left(a t_s\right)+a \cos \left(a
   t_s\right) t_s}{c} \\
 \frac{R \left(2 t_s \cos ^2\left(a t_s\right)+2
   \sin ^2\left(a t_s\right) t_s\right)}{c} \\
\end{array}
\right)$$
$$\pmb{n}=\left\{\frac{x}{\sqrt{\left(z-R \left(t_s^2 \sin
   ^2\left(a t_s\right)+t_s^2 \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a
   t_s\right)\right){}^2}}-\frac{t_s \cos
   \left(a t_s\right)}{\sqrt{\left(z-R
   \left(t_s^2 \sin ^2\left(a t_s\right)+t_s^2
   \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a
   t_s\right)\right){}^2}},\frac{y}{\sqrt{\left(
   z-R \left(t_s^2 \sin ^2\left(a
   t_s\right)+t_s^2 \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a
   t_s\right)\right){}^2}}-\frac{t_s \sin
   \left(a t_s\right)}{\sqrt{\left(z-R
   \left(t_s^2 \sin ^2\left(a t_s\right)+t_s^2
   \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a t_s\right)\right){}^2}},-\frac{R
   t_s^2 \cos ^2\left(a
   t_s\right)}{\sqrt{\left(z-R \left(t_s^2 \sin
   ^2\left(a t_s\right)+t_s^2 \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a t_s\right)\right){}^2}}-\frac{R
   t_s^2 \sin ^2\left(a
   t_s\right)}{\sqrt{\left(z-R \left(t_s^2 \sin
   ^2\left(a t_s\right)+t_s^2 \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a
   t_s\right)\right){}^2}}+\frac{z}{\sqrt{\left(
   z-R \left(t_s^2 \sin ^2\left(a
   t_s\right)+t_s^2 \cos ^2\left(a
   t_s\right)\right)\right){}^2+\left(x-t_s \cos
   \left(a t_s\right)\right){}^2+\left(y-t_s
   \sin \left(a t_s\right)\right){}^2}}\right\}$$
(Sorry this vector looks like a list) Then, we can get our potentials, but I can't put the potentials into this answer because when I entered the potentials into the box, it said I exceeded 30,000 characters. This also means I can't put the solutions, but I can put some images of the resulting fields!

A Stream plot of the Electric Field section where y=0. The black dot is the point charge.

Same graphics, but this animation depicts the Magnetic Field.
I hope this helped!
