# How to find the magnetic field due to a revolving electron of hydrogen atom in first orbit

So, I was thinking about the Bohr model of atom and I started to wonder how we could find the magnetic field due to a revolving electron (produced at the location of proton) of hydrogen atom in first orbit. Example:- How to find the magnetic field due to a revolving electron of hydrogen atom in first orbit? Given h ~$6.625*10^{-24}$;charge of electron~$1.6*10^{-19}$; $pi~ 3.141$; mass of electron~ $9.10*10^{-31}$.

If you naively use a Bohr-like model for the hydrogen atom, then the electron in its ground state is imagined as moving in a circular orbit of radius $r$ and moving with a speed $v$. In this case you could argue the electron is moving, moving charge is current, current creates a magnetic field. Following this model you might expect the magnetic field at the centre of the loop. From classical electromagnetism the magnetic field at the centre of a loop of radius $r$ carrying a current $I$ is $B = \frac{\mu_0 I}{2 r}$.
The question now becomes what do you use for the current. You're aware that the electron isn't a continuous charge distribution so that you have to use the following definition of current, namely current is the rate of change of charge passing you $I = \frac{\Delta Q}{\Delta t}$. Now, if the electron is moving fast enough in it's orbit you can imagine it to be roughly "smeared out" along its path. The electron takes an amount of time $\Delta t$ to move all the way round the orbit of length $2 \pi r$ and since its speed is $v$, this gives $\Delta t = \frac{2 \pi r}{v}$ and the appropriate current to use as $I = \frac{ev}{2 \pi r}$. Plugging this in gives $$B = \frac{\mu_0 e v}{4 \pi r^2}.$$
• Using SI units through out so the result should be in Tesla: $\frac{\mu_0}{4 \pi} = 10^{-7}, e \approx 1.6 \times 10^{-19} C, v \approx 3 \times 10^8/137 ms^{-1}, r \approx.0.1 \times 10^{-9} m$ gives $B \approx 7 T$? (thanks for the correction.) – jim Jun 22 '16 at 19:45