A Little Charge

Hello everyone! First of all, I would like to apologize because I have really forgotten about this blog for a really long time, mainly because of academic reasons and maybe a lack of inspiration. I know it is a bit unusual for me to be writing this in English, but recently I’ve got some new friends that were interested in my blog but don’t actually speak Spanish.

Well, this time I wanted to analyse a beautiful and simple problem in physics. So… Let’s get to work!

First we consider a point charge ${q}$ moving in a straight line. We will ignore all the radiation phenomena for this problem. Let’s say we turn on a magnetic field ${\vec{B}}$ perpendicular to the trajectory of the particle. We know that the particle will describe a circular trajectory in the plane defined by the vectors ${\vec{v_0}}$ and ${\vec{v_0}\times \vec{B}}$, where ${\vec{v_0}}$ is the velocity vector in the instant we turned on the magnetic field. For the sake of clarity, we define a coordinate system ${S}$ such that the center of the circular trajectory is at the origin. ${y}$-axis is parallel to ${\vec{v_0}}$ and ${\vec{B}}$ is pointing in the ${x}$ direction. From this it follows that the circle lies on the ${yz}$-plane. It can be easily derived that the corresponding motion is describe by the following equation: $\displaystyle \vec{r}(t)=R\sin \omega t \hat{\jmath}+R\cos \omega t \hat{k} \ \ \ \ \ (1)$

where ${R=\frac{m v_0}{qB}}$ and ${\omega=\frac{qB}{m}}$. This is because the centripetal force will be only given by the force exerted by the magnetic field on the particle and since this force will always be perpendicular to the trajectory, it will describe a circle.

This is a classical problem, but now, suppose that after a while we turn on an electric field ${\vec{E}}$ pointing in the ${z}$ direction in the moment that the particle is at the upper part of the circle. This field will produce a constant force in the upper direction given by ${q E \hat{k}}$. Depending on the trajectory of the particle, this new force will have different effects:

• On the upper part: Here the particle is moving in the ${y}$ direction and feels a force perpendicular to it’s movement, but this force has a smaller magnitude than the original case because the forced exerted by the electric field is opposed to the force caused by the magnetic field. The net effect of this is an increase on the radius of curvature, given a “wider movement” to the right.
• On the right: Here the particle is moving down towards the ${-z}$ direction and fells a force to the left given by the magnetic field and a force in the ${z}$ direction given by the electric field. The latter will decrease the speed of the particle and, from the expression obtained for the radius, we know that the curvature radius will decrease. Thus, the movement from the upper part to the right can be expected to be like a circle that is closing on itself (like a spiral)
• On the lower part: In this positions both forces are pointing in the same direction and the particle is moving to the left. This will cause a stronger centripetal force, decreasing even more the radius of curvature and, being this force completely perpendicular to the velocity, this movement will be like the one of a small circle (small compared to the original movement). The net effect of this will be a shorter movement to the left.
• On the left: Here the force caused by the electric field will have the same direction of the velocity, increasing the speed. From a similar argument on the one we used on the right, this will produce an increase on the radius of curvature. This will look like a circle opening, something like a spiral as well.
The combined effect of the wider movement on the upper part and the shorter step in the lower, will cause the particle to move in the ${y}$ direction making loops on the way. Something like the next picture: Of course this is just a qualitative analysis and that isn’t enough. The trajectory describes in the picture above looks like a prolate cycloid. This curve can be thought of as the path followed by a point sticked to a rolling circle. For more information you can read a little about this curve in Wolfram. Being this figure the result of a rotating circle, it would be nice to… follow this circle!

From this last idea we can start developing the math. From the construction we know that ${\vec{B}=B\hat{\imath}}$, ${\vec{E}=E\hat{k}}$. We can consider a moving frame of reference ${S'}$ with respect to ${S}$, moving on the ${y}$ direction with a constant velocity ${\vec{V}=V\hat{\jmath}}$. Let ${\vec{v}}$ and ${\vec{v'}}$ be the velocities of the particle seen from ${S}$ and ${S'}$ respectively. From this we know that. $\displaystyle \vec{v}=\vec{V}+\vec{v'}=V\hat{\jmath}+\vec{v'} \ \ \ \ \ (2)$

And also, from the assumption that ${\vec{V}}$ is constant we have the following condition: $\displaystyle \vec{a}=\vec{a'} \ \ \ \ \ (3)$

where ${\vec{a}}$ is the acceleration measured in ${S}$, while ${\vec{a'}}$ is measured in ${S'}$

Once we do this we can procede with the equation of motion. From Newton’s second law and the expression for the Lorentz force: $\displaystyle m\vec{a}=q(\vec{E}+\vec{v}\times\vec{B}) \ \ \ \ \ (4)$

But we already defined the specific direction of some of those vectors, this together with the equation ~(2) gives us: $\displaystyle m\vec{a}=q\left( E\hat{k}+(V\hat{\jmath}+\vec{v'})\times(B\hat{\imath})\right) \ \ \ \ \ (4)$

Doing the expansion of the parenthesis and the corresponding products we have: $\displaystyle m\vec{a}=q\left( E\hat{k}-VB\hat{k}+\vec{v'}\times(B\hat{\imath})\right) \ \ \ \ \ (5)$

And now, here comes the magic, let’s consider that the velocity of the moving frame of reference is given by: $\displaystyle V=\frac{E}{B} \ \ \ \ \ (6)$

From this, the equation ~(5) becomes: $\displaystyle m\vec{a}=q\left( E\hat{k}-\frac{E}{B}B\hat{k}+\vec{v'}\times(B\hat{\imath})\right)=q\left(\vec{v'}\times\vec{B}\right) \ \ \ \ \ (7)$

This last equation can be rewritten using the equation ~(3) $\displaystyle m\vec{a'}=q\left(\vec{v'}\times\vec{B}\right) \ \ \ \ \ (8)$

And there we have it! This is the same equation that we had before we turned on the electric field, but now in the second frame reference. We now that the solution is given by: $\displaystyle \vec{r'}(t)=R\sin \omega t \hat{\jmath}+R\cos \omega t \hat{k} \ \ \ \ \ (9)$

where ${R=\frac{m |v'|}{qB}=\frac{m (v_0-V)}{qB}}$ and ${\omega=\frac{qB}{m}}$. This, together with the definition of the second frame of reference we have the following: $\displaystyle \vec{r}(t)=(R\sin \omega t + Vt)\hat{\jmath}+R\cos \omega t \hat{k} \ \ \ \ \ (10)$

And this is the equation for a cycloid! And the ratio ${\phi=\frac{R\omega}{V}=frac{v_0-V}{V}}$ determines what type of cycloid we are dealing with:

• ${\phi \geq 1 }$: Prolate cycloid
• ${\phi = 1 }$: Cycloid
• ${\phi \leq 1 }$: Curtate Cycloid

And there you have it! A beautiful solution for a beautiful problem.

Last, but not least: two weeks ago I went to the fist World Science Conference in Israel. This is a beautiful project where I saw many conferences given by Nobel laureates and met many interesting people. If you want to know more about this event, check the Official Web Site or send me a message.

Well, that’s all, I’ll try to keep writing stuff for this blog. See you later and thank you for reading.