# Angular velocity of an atom in a magnetic field

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1. Jul 27, 2016

### Dazed&Confused

1. The problem statement, all variables and given/known data
The orbit of an electron ($-e$) around a nucleus ($Ze$) is a circular orbit of radius $a$ in a plane perpendicular to a uniform magnetic field $\textbf{B}$. By writing the equation of motion in a frame rotating with the electron, show that the angular velocity $\omega$ is given by one of the roots of the equation
$$m\omega^2 - eB\omega-Ze^2/4 \pi \epsilon_0 a^3 = 0$$
Verif that for small values of $B$ this is
$$\omega = \frac{eB}{2m}$$

2. Relevant equations

3. The attempt at a solution
So I've done the first part. The second part I would assume is just finding the root and making an approximation, so:
$$\omega = \frac{eB}{2m} \pm \sqrt{\left (\frac{eB}{2m}\right)^2 + \frac{Ze^2}{4 \pi \epsilon_0 m a^3}}$$
but the trouble is that the small $B$ would suggest the second term in the square root is dominant. The approximation they used to get the same result before was that the first term is much smaller than the second, in the square root.

Last edited: Jul 27, 2016
2. Jul 27, 2016

### haruspex

Quite so. It seems to me that the leading term is missing from the given answer, namely, the value that omega would have if B were zero.

3. Jul 28, 2016

### Dazed&Confused

I have to say the question wasn't exactly phrased like this. It wanted you to get the same result as the previous section, which was the one I quoted.

4. Jul 29, 2016

### Dazed&Confused

I think realised the issue, and it's down to me not paying attention. I looked at the example again and what they did was work in a rotating frame with angular velocity $-qB/2m$ and from that got the equation
$$\ddot{\textbf{r}} = -\frac{k}{mr^2}\hat{\textbf{r}}$$
where $k= qq'/4 \pi \epsilon_0$. This produces an elliptical orbit. The ellipse precesses with angular velocity $-qB/2m$. Now for a circle this just means that you add the two motions to find the angular velocity. The angular velocity of the ellipse turned circle is
$$\sqrt{\frac{k}{mr^3}}$$
which is precisely that second term in the square root.

For completeness if you take their example the other way and assume that the magnetic field is perpendicular to the position vector then you have
$$\ddot{\textbf{r}} = -\left ( \frac{qB}{2m}\right)^2 \textbf{r}$$
which is also an ellipse with angular velocity $qB/2m$, i.e. the first term.