# What's the difference between magnetic fields H and B?

As per Wikipedia:
"The term (Magnetic Field) is used for two distinct but closely related fields denoted by the symbols B and H, where H is measured in units of amperes per meter in the SI. B is measured in teslas in the SI."

So, the two are closely related. Why do we need two, then? Could just one be used?

As I remember from the university, for vacuum the Maxwell's equations are written usually in terms of B, while for media in terms of H (and B=uH).

In layman's terms,

E and B are the total electric and magnetic fields.

D and H are the free electric and magnetic fields.

P and M are the bound electric and magnetic fields.

M would be the magnetic field caused by current loops in the material. In vacuum, like you said, B and H are proportional by a constant since there is no material. However, when you are not in a vacuum, you would need to incorporate M, leading to the equation B = H + M in natural units.

• This isn't layman's terms. Laymen don't work in "natural units", they work in basic physics terms with $\epsilon_0$ and whatnot. As a physics layman I actually thought your answer was nonsense (the units didn't even match up) until I realized you have a "1" with a dimension there, which (again, as a non-physicist) I find awful and confusing. Basically, this confused me more than anything... – Mehrdad Dec 26 '16 at 8:40
• By the way, it's also quite confusing to me why $\vec{E}$ and $\vec{B}$ are grouped together here; they don't seem analogous at all. My understanding is that $\vec{E}$ and $\vec{H}$ are analogous. Is this answer correct? – Mehrdad Dec 26 '16 at 9:16
• – Rob Jeffries Dec 26 '16 at 12:14
• It's a good answer; but lacking clarity. You need to define exactly what you mean by free and bound. – BLAZE Sep 16 '17 at 4:53
• It makes no sense to call D 'free electric field', as if it was electric field of free charges. It is not; D is defined by $\mathbf D=\epsilon_0 \mathbf E + \mathbf P$, using total electric field and density of electric moment. D does obey $\nabla \cdot \mathbf D = \rho_{free}$, but this is not enough to infer $\mathbf D$ is function of $\rho_{free}$; and in general, it is not. – Ján Lalinský May 3 '18 at 10:12

I love this question! Because I've struggled with it before, coming out frustrated that no one gave me the easy explanation. :-)

Now, I'm not a physicist, but I think I've managed to learn the correct intuition here:

• $\vec{D}$ and $\vec{B}$ are electric & magnetic flux densities.

• $\vec{E}$ and $\vec{H}$ are electric & magnetic field strengths.

The difference? Flux doesn't depend on the material, but field strength does — recall Gauss's law: $$Q = \oint_S \vec{D}\cdot \,d\vec{A}$$

Flux only depends on the charge inside your closed surface. (The "flow" must leave the volume!)
But naturally if you change the material then something is affected — and that's the field strength.

If you ever forget, just remember the units:

• $\vec{D}$ is in $\text{C}/\text{m}^2$, hence there's no $\epsilon$.

• $\vec{B}$ is in $\text{Wb}/\text{m}^2$, hence there's no $\mu$. (Though honestly I remember this by analogy with $\vec{D}$.)

• Can a physicist (or someone else who knows this better than I do) please confirm my answer is actually correct? I'm not 100% sure about it. – Mehrdad Dec 26 '16 at 12:26
• +1, i've got you statement re D&B. But would be nice to have explanation about "field strengths" (E&H). A layman like me may not understand. If you mean Lorentz force it seems to be B there, not H – Sergei Gorbikov Dec 26 '16 at 16:07
• @SergeiGorbikov: That's a really good question! Notice that if you look at the version that includes (hypothetical) magnetic charges, the full formula seems to be $\vec{F} = q_{\mathrm{e}}(\vec{E} + \vec{v} \times \vec{B}) + q_{\mathrm{m}} (\vec{H} -\vec{v} \times{\vec{D}})$, which implies the static force depends on the field strength, but the dynamic force depends on the flux density. I have no idea why this makes sense though, and I might have made a mistake canceling the $\mu$s... but now it looks more symmetric. :) – Mehrdad Dec 26 '16 at 19:15
• 10x. nice question, indeed. – Sergei Gorbikov Dec 26 '16 at 19:54

Write down the Ampere's Law in vacuum:
$$\nabla \times \bf{B} = {\mu _0}\left(J + {\varepsilon _0}{{\partial E} \over {\partial {\rm{t}}}}\right)$$ Divide both parts by ${\mu _0}$ and substitute $D$ for ${\varepsilon _0}E$ to get:
$$\nabla \times \bf{B \over {{\mu _0}}} = J + {{\partial D} \over {\partial {\rm{t}}}}$$ So, I guess, it was very convenient to "get rid" of ${\mu _0}$ by defining $\bf{H}=\bf{B \over {\mu _0}}$ to get the Ampere's Law for a medium: $$\nabla \times \bf{H} = J + {{\partial D} \over {\partial {\rm{t}}}}$$

No way I claim that this is how H (or B) appeared historically, but it is a way for me to remember the difference at least.

UPDATE: I received a downvote likely for stating that $\bf{H}=\bf{B \over {\mu _0}}$. So, disclaimer: this is, in general, not true. It was stated for vacuum.

• This is smth came to my mind after I've read the Wikipedia's article on the Maxwell's equations for a medium. – Sergei Gorbikov Dec 24 '16 at 19:53
• In general ${\bf D} \neq \epsilon_0 {\bf E}$ and ${\bf H} \neq {\bf B} / \mu_0$ - they differ by the electric polarization and magnetization respectively. – tparker Dec 24 '16 at 23:00
• @tparker OK, 10x for letting know ) – Sergei Gorbikov Dec 25 '16 at 13:36
• Can't you just substitute $\mu_0\mu_r$ as needed, where $\mu_r$ is your relative permissivity? – bright-star Dec 26 '16 at 4:34
• @bright-star, I'd love to do, but, as I know H can be not equal to B/u, in general. This is only true for s0-called linear materials. See, the Wikipedia's URL in my first comment to the post above (section Constitutive relations). I am not a big expert in the topic, so I preferred to keep on the safe side by using vacuum only formulas. The answer had a goal to remember and maybe to "feel/understand" the difference. From theoretical standpoint my answer is rather shaky, I think. – Sergei Gorbikov Dec 26 '16 at 8:13

Here is a reason.

The fourth of Maxwell's macroscopic equations says that $$\nabla \times \vec{H} = \vec{J} +\frac{\partial \vec{D}}{\partial t},$$ where $\vec{J}$ is the free current at a point. In general, it is not possible to rewrite this in terms of B-field without a detailed knowledge of the microscopic behaviour of the medium (with the exception of vacuum) and what currents and polarisation charges are present, either inherently, or induced by applied fields. Sometimes the approximation is made that $\vec{B} = \mu \vec{H}$, but this runs into trouble in even quite ordinary magnetic materials that have a permanent magnetisation or suffer from hysteresis and the general relationship is that $$\vec{B} = \mu_0 (\vec{H} + \vec{M}) ,$$ where $\vec{M}$ is the magnetisation field (permanent or induced magnetic dipole moment per unit volume). For these reasons, the auxiliary magnetic field strength $\vec{H}$ is invaluable for performing accurate calculations of the fields induced by currents, or vice-versa, within magnetic materials.

On the other hand, the Lorentz force on charged particles is expressed in terms of the magnetic flux density $\vec{B}$. $$\vec{F} = q\vec{E} + q\vec{v}\times \vec{B}$$ Indeed this can form the basis of the definition of B-field and can be used, along with the lack of magnetic monopoles, to derive Maxwell's third equation (Faraday's law), which does not feature the H-field. So, both fields are a necessary part of the physicists toolbox.

As Philosophiae Naturalis points out in a comment, the B-field can be thought of as the sum of the contributions from the (applied) H-field and whatever magnetisation (induced or intrinsic) is present. Often, we can only control or easily measure the applied H-field. In limited circumstances we can get away with using only one of the B- or H-field if the magnetisation is related to the applied H-field in a straightforward way. For other cases (and hence most ferromagnetic materials or permanent magnets) both fields must be considered.

• 10x for the answer. As I see, the way you wrote the Lorentz force F=qvB, here B is not a magnetic flux density, it the magnetic field strength. If correct, pls amend the wording. Also would be nice if you specify which field you call "auxiliary": B or H. – Sergei Gorbikov Dec 29 '16 at 9:34
• @SergeiGorbikov $B$ is the magnetic flux density. This is its correct name and it is defined by the Lorentz force law as such. Its units are Teslas (or Webers per square meter). Magnetic field strength $H$ I have described as auxiliary, though opinions differ about which is "more fundamental". For definitions: see en.wikipedia.org/wiki/… – Rob Jeffries Dec 29 '16 at 9:40
• ok, tnx. I got it. Bds/ds=B, that's why you call it magnetic flux density. +1 – Sergei Gorbikov Dec 29 '16 at 9:44
• Essentially, B is the total magnetic field produced by an external source and the magnetization of the material. In experiments, one usually controls the externally-applied H field. – AlQuemist Dec 30 '16 at 10:30
• @PhilosophiaeNaturalis Yes, that is a reasonable way to think about it, I am going to add that to the answer. – Rob Jeffries Dec 30 '16 at 11:35

The field B is the one that is all that matters. In vacuum both B and H are same except of course for the constant permeability. One can say that H was invented to make things simple that is with free currents one can calculate H. B is important when one considers fields in matter. That is where one has magnetic moments from matter. It would be wrong to consider B and H as separate entities. Note whereas the field lines of B are closes those of H in some situations is not.

• +1. wow, never knew H lines can be not closed. – Sergei Gorbikov Dec 25 '16 at 13:38
• Except that Ampere's law is expressed in terms of the curl of the H-field. You can only replace this with B using assumptions that are not true in general. – Rob Jeffries Dec 27 '16 at 0:00

$B=\mu H$

Where $\mu$ is the magnetic permeability of the material.

That's it. There's a lot more handwavium and complicated terminology, but that generally doesn't add anything of value.

(For a list of exceptions to this, look at the people screaming in the comments below.)

• +1, would be nice also to know when B does not equal to uH and why. It should be related to magnetic polarization, if I remember well. Wikipedia says the statement B=uH is only true for so-called "linear" materials. – Sergei Gorbikov Dec 26 '16 at 8:17
• Not true in general. – Rob Jeffries Dec 26 '16 at 9:57
• @SergeiGorbikov, expanded on why you don't need (for most purposes) to care about nonlinearity. – Stack Tracer Dec 26 '16 at 22:31
• Your edit is not true either. Many simple and common electromagnetic media are non-linear (most ferromagnetic materials or permanent magnets) and that is one of the reasons that B- and H-fields are needed as separate entities. – Rob Jeffries Dec 26 '16 at 23:58
• @RobJeffries, if you can do so much better, I'd love to see an answer from you. – Stack Tracer Dec 27 '16 at 5:09

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