📂Electrodynamics

Changes in Electron Orbits Due to External Magnetic Fields and Diamagnetism

Explanation¹

Let’s imagine an electron orbiting around the nucleus with a radius of $R$. Although the moving point charge does not become a steady current, it appears to do so because of its high velocity. The period is calculated by dividing the distance traveled by the speed, thus

$$ T=\dfrac{2\pi R}{v} $$

The current $I$ is the amount of charge passing per unit time, and since the electron passes through a point on its orbit once per period,

$$ I=\dfrac{-e}{T}=-\dfrac{ev}{2 \pi R} $$

Therefore, the magnetic dipole moment of this electron’s orbit is

$$ \mathbf{m}=I\pi R^{2}=-\dfrac{1}{2}evR \hat{\mathbf{z}} $$

When there is an external magnetic field $\mathbf{B}$, it experiences a torque but it is very hard to tilt the orbit, so the magnetic dipole due to the electron’s orbital motion does not significantly contribute to paramagnetism. The important effect here is that the velocity of the electron changes depending on the direction of the external magnetic field.

Let’s consider the force acting on the electron at point $P$. Since the electric force at all points except $Q$ cancels each other out, we only need to consider the electric force at $Q$. And the centripetal force that a body in circular motion experiences is mass*velocity$^{2}$*radius$^{-1}$. Since both forces are directed towards the center of the circle, if we simply express the magnitude

$$ \dfrac{1}{4 \pi \epsilon_{0}}\dfrac{e^{2}}{R^{2}}=m_{e}\dfrac{v^{2}}{R} $$

Here $m_{e}$ is the mass of the electron, not to be confused with the dipole moment $m$. If there is an external magnetic field, the term $-e(\mathbf{v} \times \mathbf{B} )$ for the magnetic force is added to the left side. To align this force’s direction towards the center of the circle as well, let’s assume as shown in the image below that the external magnetic field is perpendicular to the plane of the electron’s motion. The equation of motion, including the term for the magnetic field, is as follows.

$$ \dfrac{1}{4 \pi \epsilon_{0}} \dfrac{e^{2}}{R^{2}}+e\bar{v}B=m_{e}\dfrac{{\bar{v}}^{2}}{R} $$

By substituting the equation of motion that we formulated without the term for the magnetic force into the above, we get the following.

$$ m_{e}\dfrac{v^{2}}{R}+e\bar{v}B=m_{e}\dfrac{{\bar{v}}^{2}}{R} \\[1em] \implies e\bar{v}B=\dfrac{m_{e}}R({\bar{v}}^{2}-v^{2})=\dfrac{m_{e}}{R}(\bar{v}+v)(\bar{v}-v) $$

Let the change in speed be $\Delta v=\bar{v}-v$. If the change is small, then $\bar{v} \approx v$, therefore the equation simplifies to

$$ evB = \dfrac{2vm_{e}}{R}\Delta v \\ \implies \Delta v = \dfrac{eRB}{2m_{e}} $$

At the beginning of the text, we expressed the dipole moment of the electron’s orbit $\mathbf{m}$ in terms of velocity. Thus, if the velocity changes, the dipole moment also changes, and the change is as follows.

$$ \Delta \mathbf{m} = -\dfrac{1}{2}e ( \Delta v)R\hat{\mathbf{z}}=-\dfrac{e^{2}R^{2}}{4m_{e}}\mathbf{B} $$

We can see the important fact that the change in the dipole moment $\mathbf{m}$ is in the opposite direction of the external magnetic field $\mathbf{B}$. This results in diamagnetism. While this is a general phenomenon occurring in all atoms, its effect is much weaker compared to paramagnetism, hence it is observed only in atoms where electron count is even and paramagnetism does not manifest.