A particle with mass $m$ and charge $q$ moves in a constant magnetic field $B$. Show that, if the initial velocity is perpendicular to $B$, the path is circular and the angular velocity is

\begin{equation*} \omega = \frac{q}{mv}B \end{equation*}


Charged particles in a magnetic field feel a force perpendicular to their velocity. The Lorentz force $F$ in a constant magnetic field $B$ on a particle of charge $q$ is given by

$$F = qvB$$

where $v$ is the the particle's velocity. The velocity $v$ will always be perpendicular to the magnetic field thus resulting in a circular orbit. This gives rise to the second force acting on $q$ from the centripetal acceleration, thus the angular velocity is

$$F = qvB$$ $$ma_{c} = qvB$$ $$m\frac{v^{2}}{r} = qvB$$ $$m \omega = qB$$ $$\omega = \frac{q}{m}B $$

There must be a typo in the literature b/c there is a $\frac{1}{v}$ factor in the answer that I did not get.


closed as off-topic by ACuriousMind, user12029, Kyle Kanos, John Rennie, Martin Aug 31 '15 at 12:11

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  • $\begingroup$ I don't think you've shown sufficient research effort in this problem! Two tips: 1. the question likely intends for you to set $\vec{E}=0$. 2. restrict $\vec{v}=\vec{p}/m$ to a plane, and write out the resulting system of ordinary differential equations to the two by two matrix equation $\dot{\bf f}=A\cdot \bf f$ where $f$ is your two row column vector, and $A$ is a matrix. $\endgroup$ – user12029 Aug 29 '15 at 20:04
  • $\begingroup$ What would the matrix A represent? $\endgroup$ – Joshua Burrow Aug 29 '15 at 20:48
  • $\begingroup$ That's just the form of the equation. $\bf f$ could be the momentum or the velocity (doesn't matter), and $A$ should have units of inverse time. $\endgroup$ – user12029 Aug 29 '15 at 20:51
  • $\begingroup$ Isn't $\omega=v/r$? $\endgroup$ – Kyle Kanos Aug 30 '15 at 2:02

Your work is fine (depending on your units) and what you were asked to show is wrong. Though I do object to saying you have a force equal to $ma_c,$ I would just say that a net force orthogonal to the velocity makes it go in a circle of radius $r$ where $F=mv^2/r.$

And the problem is famous. The fact that the frequency doesn't depend on the velocity or the radius is the heart of how a cyclotron works and that frequency is the cyclotron frequency.

That said there is the issue of units. Do magnetic fields and electric fields have the same units? It depends which system you use. And then the Lorentz Force law could look like $\vec F=q\left(\vec E +\frac{1}{c}\vec v \times \vec B\right).$ And that carries all the way through. So it is possible that someone meant to have a speed there ... the speed of light.

You could always go learn about different unit systems for electromagnetism and see which formulas loom different in different units. If you are interacting with someone that is forgetting which system they are using from time to time it be helpful to learn them yourself. Then you will be prepared or if it is too hard you will be compassionate.

Different units systems are popular in different fields and regions so you might have to learn different ones anyway eventually.

  • $\begingroup$ This makes sense to me b/c I know the amplitude of the electric field $\vec{E}$ is larger than the amplitude of the magnetic induction field $\vec{B}$ by a factor of $\frac{1}{c}$ I define the units as such $[\vec{E}] = \frac{V}{m}$ and $[\vec{B}] = V \frac{s}{m^{2}$. I will do more research on the theory of a cyclotron. $\endgroup$ – Joshua Burrow Aug 29 '15 at 22:31

What happens if the particle is on the circular path?

Hint. There must be something compensating the Lorentz force.

EDIT: I think your identity for the angular velocity is wrong anyways. It should be

$$\omega = \frac{qB}{m}$$

  • $\begingroup$ Newtons 2nd Law applied to a circular path. Hence $\endgroup$ – Joshua Burrow Aug 29 '15 at 20:27
  • $\begingroup$ Which ones are the two forces involved? $\endgroup$ – root Aug 29 '15 at 20:31
  • $\begingroup$ The force of the B field acting on the charged particle and the force due to the centripetal acceleration. $\endgroup$ – Joshua Burrow Aug 29 '15 at 20:41
  • $\begingroup$ Yep, that's right. And your answer is right as well. $\endgroup$ – root Aug 29 '15 at 20:44
  • $\begingroup$ Thanks for your guidance! I'll be sure to question the author about the typo in the answer. $\endgroup$ – Joshua Burrow Aug 29 '15 at 20:49

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