# Do magnetic dipoles align with $\mathbf B$ or $\mathbf H$, and why are the two field so closely associated?

As the title says, do magnetic dipoles align with $$\mathbf B$$ or $$\mathbf H$$? If one and not the other, why is that?

In addition, what force is responsible for causing the magnetic dipoles to align themselves or align antiparallel with the field? As far as I'm concerned, things don't move in the direction of $$\mathbf B$$ or $$\mathbf H$$.

Also, I am confused with the general differences between $$\mathbf B$$ and $$\mathbf H$$ other than the fact that one represents the curl of the total current in a volume and the other represents the curl of the total bound current in a volume. Why then, are they used interchangeably (as I've heard)? All I can concretely say is:

$$\mathbf H \equiv \frac{1}{\mu_0} \mathbf B - \mathbf M$$

Yet Wikipedia says that $$\mathbf B$$ and $$\mathbf H$$ are "closely related" and that $$\mathbf H$$ refers to "magnetic field strength". Where is this all coming from?

So the questions I'd like addressed are:

• The titular question, and if one or not the other an explanation.
• What force is causing the dipoles to align
• Despite the definition $$\mathbf H$$ why it is often so closely associated with $$\mathbf B$$ while I don't see as much comparison with $$\mathbf E$$ and $$\mathbf D$$, for instance.
• A similar question applies to electric dipoles. Do they align with $\mathbf{E}$ or $\mathbf{D}$? – G. Smith Mar 31 '19 at 17:46
• @G.Smith Electric dipoles align with $\mathbf P$. They align themselves so as to cancel the electric field applied to it. – sangstar Mar 31 '19 at 17:54
• In vacuum there is no $\mathbf{P}$ to align with. – G. Smith Mar 31 '19 at 19:12
• @sangstar Electric dipoles align with ${\bf E}$ to form the polarization ${\bf P}$ (which is just defined to be the dipole moment per unit volume). ${\bf P}$ has no direct effect on a dipole; it only affects dipoles through the ${\bf E}$ it generates. – Buzz Apr 1 '19 at 2:46

$${\bf B}$$ is the fundamental field. At the microscopic level, everything is governed by $${\bf B}$$. It determines the forces on moving charges, the torques on magnetic dipoles (thus, it is what aligns the spins), and the induced electric field when there is time dependence.
However, $${\bf H}$$ has its usefulness as well. Before microscopic electrodynamics was understood, $${\bf H}$$ was considered the more elementary quantity for two reasons: $${\bf H}$$ behaves more like the fundamental electric field $${\bf E}$$ than $${\bf B}$$ does (both being curl-free in the absence of free currents); and $${\bf H}$$ is typically easier to determine than $${\bf B}$$, because its curl is the free current (which is externally controlled), rather than the total current (which includes induced currents in a material, which are much harder to quantify).
With (isotropic) linear magnetic materials, it does not particularly matter whether you use $${\bf B}$$ or $${\bf H}$$, because they can be easily interchanged via $${\bf B}=\mu{\bf H}$$. However, while nonlinear materials are extremely uncommon in electrostatics, they are quite common in magnetic problems. Ferromagnets like iron will carry a net magnetization even in the absence of applied fields, and their actual responses to applied external fields are complicated and hysteretic (i.e. depend on the history of the material). It is in these situations where $${\bf H}$$ really tends to be easier to work with. For example, when studying ferromagnetic systems without any free current involved, $${\bf H}$$ can be written as the gradient of the magnetic potential, making the whole mathematical apparatus of scalar potential theory available for finding solutions.
While $${\bf H}$$ is ultimately an auxiliary, it is thus much more useful than the other auxiliary field $${\bf D}$$. $${\bf D}$$ is only useful in some very particular situations, typically involving solving Gauss's Law problems in the presence of dielectrics. While $${\bf H}$$ is determined by the (externally controlled) free current, $${\bf D}$$ is determined by the free charge, which is not controlled by the experimenter. In an electrostatics experiment, what is typically set externally is a voltage, and the voltage is related to the total (free plus bound) charge and the electric field $${\bf E}$$, rather than $${\bf D}$$.
• In a nonisotropic material where $\mathbf{B}$ and $\mathbf{H}$ aren’t parallel, why would a dipole align with $\mathbf{B}$ rather than with $\mathbf{H}$? Saying that it aligns with $\mathbf{B}$ seems to me to be saying that it wouldn’t feel the magnetization $\mathbf{M}$, which I find hard to believe. – G. Smith Apr 1 '19 at 0:45
• @G.Smith See my second sentence. Classically, everything is determined by ${\bf B}$, and the only affect of a magnetization (magnetic dipole moment per unit volume) ${\bf M}$ is to produce a ${\bf B}$. In quantum systems, things get a little trickier, because there are also exchange effects between spins that are mediated by the Pauli Exclusion Principle. – Buzz Apr 1 '19 at 2:48