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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~ $9.10*10^{-31}$.

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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 $$B = \frac{\mu_0 e v}{4 \pi r^2}.$$

But, there are a few important problems with this model, it ignores the fact that the proton and the electron acts like miniature magnets in their own right because of spin, have a look at the following reference, H.C. Ohanian, "What is spin?", Am. J. Phys. 54 (1986) 500–505.

However, much more importantly! is that the model of an electron orbiting a proton is wrong. Because of its wave nature, the electron in its ground state is actually smeared symmetrically about the proton (ignoring spin-spin effects) and the magnetic field turns out to be zero (this may be expected if the electron is smeared spherically symmetric about the proton, there is no special direction that you'd expect any magnetic field to point in, as distinct from the planetary model where you might expect a field perpendicular to the plane of the orbit).

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  • $\begingroup$ I think the answer would be 12.5 tesla. Right. $\endgroup$ – Sai Saandeep Jun 22 '16 at 17:47
  • $\begingroup$ 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.) $\endgroup$ – jim Jun 22 '16 at 19:45
  • $\begingroup$ a bit old but for future visitors: I also calculated about B=12.5T however I used an orbital radius of around 0.0529nm when clculating $\endgroup$ – Downgoat Apr 19 at 7:45

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