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In the classical Langevin theory of diamagnetism, an externally applied magnetic field $\vec{B}$ either increases or decreases the speed of orbital motion of the electrons. But there is yet another possibility in which the orientation of the orbits can change. This is because an orbit is a circulating current loop carrying a magnetic moment $\vec{m}$ and in presence of $\vec{B}$ it feels a torque $\vec{m}\times\vec{B}$. But this change in orientation is not considered in the Langevin theory. Why?

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  • $\begingroup$ That's simply because a tilted magnetic moment can be viewed as a superposition of a vertical one (affected by Langevin diamagnetism) and a horizontal one (not affected at all). So since the tilted moments don't really add anything you might as well restrict attention to vertical ones. $\endgroup$ – knzhou Dec 1 '18 at 15:44
  • $\begingroup$ Of course to get a quantitative answer you would consider tilted moments too. But this is all beside the point, because the Langevin theory is completely wrong -- there is simply no way to get diamagnetism in a classical context. $\endgroup$ – knzhou Dec 1 '18 at 15:45
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In the classical theory of diamagnetism, a fundamental assumption is that the orbit of the electron is not influenced by the external magnetic field. This orbit, therefore, will only contribute to the creation of a magnetic dipole moment that, in a material, will give rise to a magnetization vector and, consequently, it allows ut to determine the expression of the magnetic susceptibility.

On the contrary, in the classical theory of paramagnetism, developed again by Langevin, we assume that each electron (for the sake of simplicity, in an hydrogen-like atom in which the nuclear magnetic moment is equal to zero) will have a certain intrinsic dipole moment that will align under the action of an external magnetic field: in this case, the orientation is explicitly taken into account.

What is astonishing, in this case, is that both the classical theories give results that are in good agreement with the experimental results.

However, it is possible to consider that, as you point out in your question, both the effects should in principle be present when considering the classical behavior of a single electron. Repeating therefore the calculations of these two models taking into account both effects, it is possible to show, as it was done by Van Vleck in 1932, that at the end we obtain a magnetic dipole moment for the diamagnetic theory and a magnetic dipole moment for the paramagnetic theory. Summing up these two contributions, we can obtain that the overall magnetic dipole moment is identically equal to zero: $$ \boldsymbol\mu_{tot} = \boldsymbol\mu_{diam} + \boldsymbol\mu_{param} = 0 $$ since their expressions cancel out.

It is possible to show that all these contributions will result identically equal to zero for any electron in every atom. This is consistent with the Bohr-van Leeuwen theorem, that states that magnetic effects cannot be described through classical physics (they should not just be there, if we do careful calculations!), thus requiring a fully quantum description.

Returning back to your question, my answer is that we more or less deliberately neglect it to obtain a classical description that works (been in agreement with experimental results), since including it we obtain a result in clear contradiction with experiments. If this is not satisfactory, it is just because to describe these effects we need quantum mechanics.

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  • $\begingroup$ Langevin theory of paramagnetism is for atomic or electronic moments? $\endgroup$ – mithusengupta123 Dec 1 '18 at 17:08
  • $\begingroup$ @mithusengupta123 my answer is fully related to electronic dipole moments, since (if I remember correctly) they were the only moments considered in classical theories. $\endgroup$ – JackI Dec 1 '18 at 17:18
  • $\begingroup$ Okay. Actually, you wrote atomic moments in your second paragraph. $\endgroup$ – mithusengupta123 Dec 1 '18 at 17:31
  • $\begingroup$ I'm sorry, it is an error because often I am considering an hydrogen atom and the nuclear magnetic moment to be equal to zero for the sake of simplicity, thus confusing the concept of electronic moment with atomic one. I'll correct it! $\endgroup$ – JackI Dec 1 '18 at 17:40

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