Show the motion of a charge in a uniform magnetic field is a circle

Question

Show that the locus of the charge in a uniform magnetic field is a circle (in the plane perpendicular to the magnetic field). Assume the charge has initial velocity vector perpendicular to the magnetic field.

By Lorentz Force Law, we have $\vec{F}=q\;\vec{v}\times\vec{B}$, where $\vec{F}$ is the magnetic force on the charge, $q$ is the charge, $\vec{v}$ is the velocity vector of the charge, and $\vec{B}$ is the magnetic field vector. First, I chose the plane that is perpendicular to magnetic field to be the $xy$-plane, and $z$-axis is chosen such that the magnetic field is in the negative $z$ direction. Rewriting the relation: $$m\vec{a}= qB<v_y,-v_x,0>, $$where $v_x$ and $ v_y$ are the velocity components in $x$ and $y$ directions. Then, we get three differential equations. Let $\vec{r}$ be the position vector of the charge, and $k=\frac{qB}{m}$: \begin{align*} \frac{d^2{r_x}}{dt^2}&=kv_y,\\ \frac{d^2{r_y}}{dt^2}&=-kv_x,\\ \frac{d^2{r_z}}{dt^2}&=0. \end{align*} I chose the initial position of the charge to be $\vec{r}(0)=\vec{0}$. But I am stuck here, because I have no idea how to relate $v_x$, and $v_y$.

Can I do things like: $$r_x(t)=r(t)cos(\phi(t))?$$

Could someone lead me a help, a small hint is enough? Thank you.

Answer 1

The issue is that you choose the coordinate origin as your initial position. If the trajectory of the charge is a circle and hence, the initial position is on the circle too, the center of the circle would then be someplace away from the origin. Contrary to that, you are likely assuming the center at the origin by your cosine formula (and presumably the corresponding sine formula for $r_y=r \sin \phi(t)$).

Better not assume anything about the initial position, and just make the sine/cosine approach. You can just move your general solution to match the initial conditions in the end, because the uniform field is translationally invariant.

By the way, there is nothing wrong in specifically assuming $r(t)=r=const.$ and $\phi(t)=\omega t$. If that solves your problem, it follows from the uniqueness of the solution of the linear differential equation system that this is the only solution.

Answer 2

Usually in problems involving higher order differential equations, It is easy to take $v=\frac{dx}{dt}$ as the variable if possible. Here the differential equations can be written as :

$\frac{dv_x}{dt} = k v_y \tag{1}\label{1}$

$\frac{dv_y}{dt} = - kv_x \tag{2}\label{2}$

Now take time - derivative of \eqref{1} and replace $\frac{dv_y} {dt}$ on RHS with \eqref{2}.

We would have two second order differential equations :

$\frac{d^2v_x}{dt^2} =- k^2 v_x$

$\frac{d^2v_y}{dt^2} = - k^2v_y$

You can solve these equations and after you find $v_x$ and $v_y$ as functions of $t$, integrate them to get $x(t)$ and $y(t)$