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In the problem below, why does the second method used to calculate magnetic dipole moment $\vec {\mu}$ essentially use the definition of the electric dipole moment $\vec {p}$? Is magnetic dipole moment the same as electric dipole moment? Can this method always be used? If yes, what is the need to differentiate between the two?

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Electric dipole moment and magnetic dipole moment are not the same quantity, because (of course) an electric dipole moment creates an electric field, while a magnetic dipole moment creates a magnetic field.

However, in the presence of magnetized matter and in the absence of free currents, it is possible to define a (fictitious) "magnetic charge" $\rho_m = -\vec{\nabla} \cdot \vec{M}$. (On the boundary between media, we can similarly define $\sigma_m = \vec{M} \cdot \hat{n}$.) The auxiliary field $\vec{H}$ then satisfies $\vec{\nabla} \cdot \vec{H} = \rho_m$ and $\vec{\nabla} \times \vec{H} = 0$, so we can use the same mathematical techniques to find $\vec{H}$ as we do to find the electric field outside a known charge distribution.

In particular, this allows us to define a "Coulomb's Law" for $\vec{H}$, in analogy to that for the electric field: $$ \vec{H}(\vec{r}) = \frac{1}{4 \pi} \iiint \rho_m(\vec{r}')\frac{\vec{r} - \vec{r}'}{|\vec{r} - \vec{r}'|^3} \, d^3\vec{r}'. $$ One can then perform a multipole expansion for this $\vec{H}$ in powers of $r^{-1}$; and the "dipole term" for this expansion is defined in terms of $\rho_m$ in exactly the same way that $\vec{p}$ is related to $\rho$.

For more information on this technique, I recommend Zangwill's Modern Electrodynamics, as well as my answers here and here. Or, perhaps, "Slide 5" from your instructor's notes.

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  • $\begingroup$ Thank you! Slide 5 didn't help but your links did. $\endgroup$
    – dimes
    Oct 16 '20 at 19:49
  • $\begingroup$ @dimes Can you please share the link to your instructor's notes slides? $\endgroup$ Oct 2 '21 at 8:02

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