What does the $B$-field refer to?

Question

I'm just looking for some guidance, it would be nice if someone could explain this. So the $B$ field in $B=\mu H$, as I understand is the induced magnetic field in some material due to this material’s presence in some $H$ field. So the $B$ field here doesn't say anything about the flux density of the magnetic field source; it only tells us about the induced magnetic field in some material affected by this source of the magnetic field. If this is the case then why do we calculate the density of the magnetic field at some distance $r$ from a wire using a formula $B=\mu_{0}H$ when this would calculate the induced magnetic field density in the vacuum, it has nothing to do with the $B$ field of the source. One example where this is done is when calculating the force between two wires $F_1=B_2\times I_1\times L_1$ it is said that instead of $B_2$ we can substitute $\frac{\mu_{0}\times I}{l}$ but this wouldn't address the $B$ field produced by the first wire, but the induced field in the vacuum which doesn't have much effect on the second wire right?

Answer 1

$B$ is the magnetic field. In the Lorentz force law $\vec{F}=q(\vec{E} + \vec{v}\times \vec{B})$, which tells you how charges respond to electric and magnetic fields, the field that appears is $B$, not $H$. So the value of the magnetic field, $B$, at a given point in space, tells you the magnetic force that will be experienced by a current element at that point.

Having said that, the auxiliary field $H$ is a useful quantity when dealing with magnetic materials. The reason is that $H$ tells you the magnetic field due to free currents in a material. This allows us to split the magnetic field due to the magnetization (loosely speaking the "intrinsic" magnetic field of a material due to effects like ferromagnetism) and the magnetic field we induce by passing current through a material (the free currents, which source $H$).

Answer 2

Consider the expression for Gauss law $\nabla\cdot\vec{D} = \rho_f$. Evidently, $\vec{D}$ depend on free charges. But in the laboratory, we can usually control only the total charge. As a result, $\vec{D}$ is not a very useful quantity but $\vec{E}$ is. One can also write $\nabla\cdot\vec{E} = \rho/\epsilon_0$, where $\rho$ is the density of all charge and $\epsilon_0$ is the permittivity of free space.

The analogous equation for magnetic fields is Ampere's law $$\nabla\times\vec{H} = \vec{J}_f + \frac{\partial\vec{D}}{\partial t}.$$ In an experiment, one can usually control the free current not the total current. Therefore, $\vec{H}$ is more useful in the laboratory than $\vec{B}$. The $\vec{B}$ field depends on free current density $\vec{J}_f$ as well as magnetisation current density $\vec{J}_m$.

It is probably not a good idea to consider $\vec{D}$ to be "caused" due to $\vec{E}$ and $\vec{H}$ due to $\vec{B}$ (or the other way round). Rather, it is more convenient to treat $\vec{H}$ and $\vec{D}$ as no more than "book-keeping" fields. They isolate the fields contributed by free currents and free charges.

Some part of the confusion, I believe, is also contributed by SI units. The book-keeping is clearer in gaussian units $$\begin{align}\vec{D} &= \vec{E} + 4\pi\vec{P} \\ \vec{H} &= \vec{B} - 4\pi\vec{M}.\end{align}$$ The lhs of both equations has "field due to free sources", while the rhs has "field due to all sources" and "field due to bound sources".

Note that a moving charge experiences effect of all sources, free and bound. As a result, the force on it is written in terms of $\vec{E}$ and $\vec{B}$. In SI units it is mentioned in Andrew's answer. In gaussian units, it is $$\vec{F} = q\left(\vec{E} + \frac{\vec{v}}{c} \times \vec{B}\right).$$

Answer 3

A comment to my previous reply asked for an explanation for the statement about experiments favouring $\vec{H}$ over $\vec{B}$. I attach a paragraph from David Griffith's book that provides an excellent explanation of the point.

As it turns out, $\mathbf H$ is a more useful quantity than $\mathbf D$. In the laboratory you will frequently hear people talking about $\mathbf H$ (more often even than $\mathbf B$), but you will never hear anyone speak of $\mathbf D$ (only $\mathbf E$). The reason is this: To build an electromagnet you run a certain (free) current though a coil. The current is the thing you read on the dial, and this determines $\mathbf H$ (or at any rate, the line integral of $\mathbf H$); $\mathbf B$ depends on the specific materials you used and even, if iron is present, on the history of your magnet. On the other hand, if you want to set up an electric filed, you do not plaster a known free charge on the plates of a parallel plate capacitor; rather, you connect them to a battery of known voltage. It's the potential difference you read on your dial, and that determines $\mathbf E$ (or at any rate, the line integral of $\mathbf E$); $\mathbf D$ depends on the details of the dielectric you're using. If it were easy to measure charge, and hard to measure potential, then you'd find experimentalists talking about $\mathbf D$ instead of $\mathbf E$. So the relative familiarity of $\mathbf H$, as contrasted with $\mathbf D$, derives from purely practical considerations; theoretically, they're all on equal footing.

You can find this paragraph on p. 271 in section 6.3 of the third edition of his book: Introduction to Electrodynamics.