# Motion of charged particle in uniform magnetic field and a radially symmetric electric field

This question posted by me on MSE talks about a physics problem. This is what it was: (I hope someone can help me with this)

Consider a region of 2-dimensional space with a uniform magnetic field of magnitude $$B$$ directed perpendicular to the plane. With respect to some fixed point (origin), at a point with position vector $$\vec {r}$$, there exists an electric field $$\vec{E} = E_0\vec {r}$$ where $$E_0$$ is some positive constant. Both the magnetic and electric fields are constant with respect to time and exist independently of each other. We have a particle of mass $$m$$ that has charge $$q$$ (where $$q > 0$$) which we release in this region (from rest) at a distance $$a$$ from the origin.

The aim is to realise (and prove) that the subsequent path described by the particle is an epicycloid

However, this is true only if $$E_0 < \frac {qB^2}{4m}$$ Apparently, for larger magnitudes of electric field the motion of particle becomes unbounded (its distance from the origin keeps increasing with time).

Is this a known result? I can't seem to find this anywhere on the internet. How do I prove this?

I used polar coordinates $$(r, \theta)$$ with the same origin as above. I write $$\omega$$ for $$\frac {d \theta}{d t}$$. The equations would be: (assuming $$\vec{B}$$ is along $$\vec{\theta} \times \vec {r}$$) $$m (\ddot{r} - r \omega^2) =qE_0 r - qr\omega B \dots(1)$$ $$m(2 \dot{r} \omega + r \dot{\omega}) = q\dot{r} B \dots (2)$$ From $$(2)$$, multiply both sides with $$r$$ and integrate w.r.t. time: (this is equivalent to writing torque equation about origin) $$r^2 \omega = \frac {qB}{2m} (r^2 - a^2) \dots (3)$$ Substitute $$\omega$$ from $$(3)$$ into $$(1)$$ to get: $$\ddot {r} = \left( \frac {qE_0}{m} - \left( \frac {qB}{2m} \right)^2 \right) r + \left( \frac {qB}{2m} \right)^2 \frac {a^4}{r^3}$$ Integrating both sides with respect to $$r$$: $$(\dot {r})^2 = \left( \frac {qE_0}{m} - \left( \frac {qB}{2m} \right)^2 \right) (r^2 - a^2) + \left( \frac {qB}{2m} \right)^2 \frac {a^2(r^2 - a^2)}{r^2}$$ In an attempt to eliminate the variable time from these equations, I tried to compute $$\frac{dr}{rd\theta}$$ as $$\frac {\dot {r}}{r \omega}$$ : $$\left(\frac {dr}{r d \theta} \right)^2 = \frac {a^2 - r^2 \left(1- \frac {4mE_0}{qB^2} \right)}{r^2 - a^2} = \frac {a^2}{b^2} \frac {(b^2 - r^2)}{(r^2 - a^2)}$$ where $$1- \frac {4mE_0}{qB^2} = \frac {a^2}{b^2}$$ for some $$b$$. ($$b$$ happens to be the maximum distance from the origin during motion)
Now, although this does let me describe $$\theta$$ as a function of $$r$$, I am having trouble concluding from there that the path is an epicycloid. As an example, this should represent a part of the epicycloid: $$2 \theta = \frac {b}{a} \cos^{-1} \left(\frac {a^2 + b^2 - 2 r^2}{b^2 - a^2} \right) - \cos ^{-1} \left(\frac {\frac {2a^2b^2}{r^2} - a^2 -b^2}{b^2 - a^2} \right)$$ This is where I am stuck

Edit: Thanks to Hint II by Frobenius, specifically the substitution $$\xi = x + iy$$, I am now able to conclude that $$\xi$$ is of the form: $$\frac {b+a}{b-a} e^{i \phi} - e^{i \phi \frac {b + a}{b - a}}$$ Where $$\phi$$ is a real parameter (happens to be proportional to time).
This I am able to realise and prove as an epicycloid. All that remains to figure out is how do my equations lead to an epicycloid as well..

• It is not a good practice the "homework-and-exercises" be untagged by the OP. Commented Jun 7, 2020 at 19:20

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For the equation of motion we have \begin{align} & q\left(\mathbf E\boldsymbol{+}\mathbf v\boldsymbol{\times}\mathbf B\right)\boldsymbol{=}m\mathbf a \quad \boldsymbol{\Longrightarrow} \quad q E_0\,\mathbf r\boldsymbol{+}q\,\mathbf{\dot{r}}\boldsymbol{\times}\mathbf B\boldsymbol{=}m\mathbf{\ddot{r}} \quad \stackrel{\mathbf B\boldsymbol{=}B_0 \mathbf e_z}{\boldsymbol{=\!=\!=\!\Longrightarrow}} \nonumber\\ & q E_0\left(x,y\right)\boldsymbol{+}q B_0 \left(\dot{y},\boldsymbol{-}\dot{x}\right)\boldsymbol{=}m\left(\ddot{x},\ddot{y}\right) \quad \boldsymbol{\Longrightarrow} \quad \tag{01}\label{01} \end{align} so the following system of two scalar linear differential equations of 2nd order \begin{align} \ddot{x}\boldsymbol{-}b\dot{y}\boldsymbol{-}\epsilon x & \boldsymbol{=}0 \tag{02a}\label{02a}\\ \ddot{y}\boldsymbol{+}b\dot{x}\boldsymbol{-}\epsilon y & \boldsymbol{=}0 \tag{02b}\label{02b}\\ \epsilon \boldsymbol{=}\dfrac{qE_0}{m}\,,\quad b&\boldsymbol{=}\dfrac{qB_0}{m} \tag{02c}\label{02c} \end{align} Multiplying \eqref{02b} by the imaginary unit $$\,i\,$$ and adding/subtracting the result to/from \eqref{02a} we have respectively \begin{align} \left(\ddot{x}\boldsymbol{+}i\ddot{y}\right)\boldsymbol{+}i b\left(\dot{x}\boldsymbol{+}i\dot{y}\right)\boldsymbol{-}\epsilon \left(x\boldsymbol{+}iy\right) & \boldsymbol{=}0 \tag{03a}\label{03a}\\ \left(\ddot{x}\boldsymbol{-}i\ddot{y}\right)\boldsymbol{-}i b\left(\dot{x}\boldsymbol{-}i\dot{y}\right)\boldsymbol{-}\epsilon \left(x\boldsymbol{-}iy\right) & \boldsymbol{=}0 \tag{03b}\label{03b} \end{align} Defining the complex variable $$$$\xi\boldsymbol{=}x\boldsymbol{+}iy \tag{04}\label{04}$$$$ equations \eqref{03a},\eqref{03b} yield respectively \begin{align} \ddot{\xi}\boldsymbol{+}i b\,\dot{\xi}\boldsymbol{-}\epsilon \,\xi & \boldsymbol{=}0 \tag{05a}\label{05a}\\ \ddot{\overline{\xi}}\boldsymbol{-}i b\,\dot{\overline{\xi}}\boldsymbol{-}\epsilon \,\overline{\xi} & \boldsymbol{=}0 \tag{05b}\label{05b} \end{align} Note that equation \eqref{05b} is the complex conjugate of \eqref{05a}, so in order to find explicitly the orbit of the particle it's sufficient to solve one of them, let it be equation \eqref{05a}.
Beyond the possibility to use an integral transform, for example Laplace or Fourier or etc, it's convenient to make the following trick that converts the second order equation \eqref{05a} to a first order. Suppose that we split the coefficient $$\,ib\,$$ to two coefficients $$$$ib\boldsymbol{=}\rho_{1}\boldsymbol{+}\rho_{2} \tag{06}\label{06}$$$$ Then equation \eqref{05a} could be expressed as $$$$\left(\ddot{\xi}\boldsymbol{+}\rho_{1}\dot{\xi}\right)\boldsymbol{+}\rho_{2}\left(\dot{\xi}\boldsymbol{-}\dfrac{\epsilon}{\rho_{2}} \,\xi \right) \boldsymbol{=}0 \tag{07}\label{07}$$$$ If it would be possible the two coefficients $$\,\rho_{1},\rho_{2}\,$$ of equation \eqref{06} to satisfy also the condition $$$$\boldsymbol{-}\dfrac{\epsilon}{\rho_{2}} \boldsymbol{=}\rho_{1} \tag{08}\label{08}$$$$ then \eqref{06} yields $$$$\left(\ddot{\xi}\boldsymbol{+}\rho_{1}\dot{\xi}\right)\boldsymbol{+}\rho_{2}\left(\dot{\xi}\boldsymbol{+}\rho_{1} \,\xi \right) \boldsymbol{=}0 \tag{09}\label{09}$$$$ and defining the new complex variable $$$$\eta \boldsymbol{=}\dot{\xi}\boldsymbol{+}\rho_{1} \,\xi \tag{10}\label{10}$$$$ we have the first order equation $$$$\dot{\eta}\boldsymbol{+}\rho_{2} \,\eta \boldsymbol{=}0 \tag{11}\label{11}$$$$ From equations \eqref{06}, \eqref{08} the unknown coefficients $$\,\rho_{1},\rho_{2}\,$$ must be the roots of the quadratic equation $$$$\rho^2\boldsymbol{-}ib\rho-\epsilon\boldsymbol{=}0 \tag{12}\label{12}$$$$ Looking carefully in \eqref{12} we note that its discriminant is $$$$\Delta\boldsymbol{=}4\epsilon\boldsymbol{+}\left(\boldsymbol{-}ib\right)^2\boldsymbol{=}4\epsilon\boldsymbol{-}b^2\stackrel{\eqref{02c}}{\boldsymbol{=\!=\!=}}4\left(\dfrac{qE_0}{m}\right)\boldsymbol{-}\left(\dfrac{qB_0}{m}\right)^2 \tag{13}\label{13}$$$$ that is $$$$\Delta\boldsymbol{=}\dfrac{4q}{m}\left(E_0\boldsymbol{-}\dfrac{qB^2_0}{4m}\right)\boldsymbol{\longrightarrow} \begin{cases} \boldsymbol{<}0 \quad \text{if} \quad E_0\boldsymbol{<}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \boldsymbol{=}0 \quad \text{if} \quad E_0\boldsymbol{=}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \boldsymbol{>}0 \quad \text{if} \quad E_0\boldsymbol{>}\dfrac{qB^2_0}{4m}\vphantom{\dfrac{a}{\dfrac{a}{b}}} \end{cases} \tag{14}\label{14}$$$$
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