In an atom there are $n$ electrons and the $n$th electron creates a magnetic field, as it has velocity and the other $n-1$ electrons will be affected by this force. But still, why aren't we considering this?
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2$\begingroup$ The electron is not in an orbit, but in an orbital, en.wikipedia.org/wiki/Atomic_orbital . no continuous velocity describes the probable location of the electron(s) $\endgroup$– anna vCommented Jul 22, 2022 at 16:17
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3$\begingroup$ Nevertheless the p, d, and higher orbitals to have non-zero angular momentum and therefore an associated magnetic dipole moment, so I think this is a fair question. $\endgroup$– John RennieCommented Jul 22, 2022 at 16:36
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Actually in quantum mechanics one of the first corrections we do to the hydrogen atom can be thought as the effect of what you are describing at the quantum level.
A key example of this phenomenon is the spin–orbit interaction leading to shifts in an electron's atomic energy levels, due to electromagnetic interaction between the electron's magnetic dipole, its orbital motion, and the electrostatic field of the positively charged nucleus. This phenomenon is detectable as a splitting of spectral lines, which can be thought of as a Zeeman effect product of two relativistic effects: the apparent magnetic field seen from the electron perspective and the magnetic moment of the electron associated with its intrinsic spin.
A similar effect, due to the relationship between angular momentum and the strong nuclear force, occurs for protons and neutrons moving inside the nucleus, leading to a shift in their energy levels in the nucleus shell model.
Also in the field of spintronics, spin–orbit effects for electrons in semiconductors and other materials are explored for technological applications. The spin–orbit interaction is one cause of magnetocrystalline anisotropy and the spin Hall effect.
It is pretty cool to play around when you see this for the first time, changing reference frames and trying to understand it for every frame, try it yourself!
For more insight check what the spin-orbit interaction is: https://en.wikipedia.org/wiki/Spin%E2%80%93orbit_interaction
answered Jul 22, 2022 at 16:50