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With over 15 years of related experience at the fore, IKEA is dedicated to offering customers outstanding service and product selection.Our multi-channel information and communication solutions are designed to help you adapt to the modern world.Explore the possibilities, see the possibilities, make the possibilities!Q:
How is the relationship between a macroscopic object’s rotational and orbital angular momentum related to a microscopic object’s spin?
A system of two charged particles with charge $q$ and mass $m$ produces a magnetic dipole moment given by
$$\mu = \frac{q\hbar}{4\pi\epsilon_0}\left(3\frac{\cos\theta -1}{r^3}+\frac{1-3\cos^2\theta}{r^5}\right)$$
when acted on by an external magnetic field. Thus, there is no centrifugal force acting on the system since the angular momentum is zero, despite the fact that the system has nonzero angular momentum. This is analogous to the fact that there is no ordinary magnetic force acting on the system. It would seem that the magnetic field due to a magnet exerts only its orbital angular momentum on a particle in it, but not its angular momentum. How can this be?
A:
While the two particles do have nonzero spin, the magnetic dipole moment of the whole system, including both the particles’ magnetic dipole moments, is given by
\begin{align*}
\mu &= \frac{q\hbar}{4\pi\epsilon_0}\left(\mathbf{r}_1\cdot\frac{q\hbar\mathbf{r}_2}{4\pi\epsilon_0 r^3}+\mathbf{r}_2\cdot\frac{q\hbar\mathbf{r}_1}{4\pi\epsilon_0 r^3}\right)\\
&= \frac{q\hbar}{4\pi\epsilon_0}\frac{(\mathbf{r}_1\cdot\mathbf{r}_2)\mathbf{r}_1}{
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