# Is magnetic dipole moment the same as electric dipole moment?

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?

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 $$\rho$$.