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 $B=\mu H$).

  • $\begingroup$ the H field is important in electrical engineering. For example, if you take a microwave propagation in a cable, or just a plane wave propagating somewhere, the H field is an analog of a current. Say, V = RI for Ohms law. When E= Z*H, where Z is an impedance and it is is a complex version of a resistance. $\endgroup$ Nov 16, 2021 at 22:18
  • $\begingroup$ Also when considering boundary conditions, the H field is directly related to a surface current. So, it is useful. $\endgroup$ Nov 16, 2021 at 22:18

9 Answers 9


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.

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    $\begingroup$ 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... $\endgroup$
    – user541686
    Dec 26, 2016 at 8:40
  • 1
    $\begingroup$ 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? $\endgroup$
    – user541686
    Dec 26, 2016 at 9:16
  • $\begingroup$ See physicsforums.com/threads/… ? $\endgroup$
    – ProfRob
    Dec 26, 2016 at 12:14
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    $\begingroup$ It's a good answer; but lacking clarity. You need to define exactly what you mean by free and bound. $\endgroup$
    – BLAZE
    Sep 16, 2017 at 4:53
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    $\begingroup$ 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. $\endgroup$ May 3, 2018 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 $\vec{D}$.)

  • $\begingroup$ 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. $\endgroup$
    – user541686
    Dec 26, 2016 at 12:26
  • $\begingroup$ +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 $\endgroup$ Dec 26, 2016 at 16:07
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    $\begingroup$ @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. :) $\endgroup$
    – user541686
    Dec 26, 2016 at 19:15
  • $\begingroup$ @SergeiGorbikov: I just asked this related question, it might be helpful to follow it. $\endgroup$
    – user541686
    Dec 26, 2016 at 19:33
  • $\begingroup$ 10x. nice question, indeed. $\endgroup$ Dec 26, 2016 at 19:54

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.

  • $\begingroup$ 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. $\endgroup$ Dec 29, 2016 at 9:34
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    $\begingroup$ @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/… $\endgroup$
    – ProfRob
    Dec 29, 2016 at 9:40
  • $\begingroup$ ok, tnx. I got it. Bds/ds=B, that's why you call it magnetic flux density. +1 $\endgroup$ Dec 29, 2016 at 9:44
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    $\begingroup$ 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. $\endgroup$
    – AlQuemist
    Dec 30, 2016 at 10:30
  • $\begingroup$ @PhilosophiaeNaturalis Yes, that is a reasonable way to think about it, I am going to add that to the answer. $\endgroup$
    – ProfRob
    Dec 30, 2016 at 11:35

The fields $\bf E$ and $\bf B$ are the fundamental components of the electromagnetic field. They define the field via its effects, producing a force on a charge $q$: $$ {\bf f} = q ( {\bf E} + {\bf v} \times {\bf B}). $$ One can in principle do the whole of electromagnetism using just these fields.

In a material medium (e.g. glass, or a metal, or a semiconductor, or a gas, etc.) these fields are very complicated however. They vary at lot on the distance scale of the atomic spacing, getting huge near atomic nuclei and smaller elsewhere. So most of the time we do not treat the fields at a point, but rather we average over a region of space of the order of a few atomic spacings (e.g. a nanometre in a solid, or a larger region in a gas). This is where other things such as polarization $\bf P$ and magnetisation $\bf M$ come into play. If we divide the charge in any region into the part associated with little electric dipoles (called bound charge) and the rest (called free charge) then we have $$ q_{\rm tot} = q_{\rm bound} + q_{\rm free} $$ so the first Maxwell equation reads $$ {\bf \nabla} \cdot {\bf E} = \frac{q_{\rm b} + q_{\rm f}}{\epsilon_0} $$ where I used the subscripts $b$ and $f$ for 'bound' and 'free'. Now a basic result (which can be proved with a little standard manipulation, it is first year undergraduate level) is that $$ {\bf \nabla} \cdot {\bf P} = -q_{\rm b} $$ where $\bf P$ is the dipole moment per unit volume, called polarization. So it follows that $$ {\bf \nabla} \cdot (\epsilon_0 {\bf E} + {\bf P}) = q_{\rm f}. $$ Well look at that! The right hand side is nice and simple, because it only involves the free charge, and it often happens that we know from the start how much free charge there is. There might even be none at all! (e.g. a light wave travelling in glass in ordinary circumstances). So we choose to give the combination $(\epsilon_0 {\bf E} + {\bf P})$ a symbol of its own: we call it $\bf D$. This is how the field $\bf D$ gets introduced, with its associated equation $$ {\bf \nabla} \cdot {\bf D} = q_{\rm f}. $$

The story for $\bf H$ is similar. First we derive by analysis that the magnetization is connected to the part of the total current that is caused by little current loops, with other contributions coming from free current and changes in the dipoles. The central result of this derivation is that the total current can be divided up as $$ {\displaystyle \mathbf {j}_{\rm tot} =\mathbf {j} _{\mathrm {f} }+\nabla \times \mathbf {M} +{\frac {\partial \mathbf {P} }{\partial t}}}. $$ Next we use this in the fourth Maxwell equation, and we notice that a convenient way to write the equation is to give the combination ${\bf B}/\mu_0 - {\bf M}$ its own letter ($\bf H$) and we arrive at $$ {\bf \nabla} \times {\bf H} = {\bf j}_{\rm f} + \frac{\partial \bf D}{\partial t}. $$ Again, this is convenient because often we want to treat problems without worrying about what the magnetisation current and the bound charge is doing.

So, to conclude, fields $\bf E$ and $\bf B$ are the basic fields. (Together they make up a tensor called the field tensor, but you don't need to know that.) Fields $\bf D$ and $\bf H$ are introduced for reasons of mathematical convenience and the associated physical insight. They are particularly useful when thinking about capacitors and inductors and electromagnetic waves propagating in media when the free charge and free current are known.

The other two Maxwell equations do not involve the sources so they are not affected. They are $$ {\bf \nabla} \cdot {\bf B} = 0 $$ and $$ {\bf \nabla} \times {\bf E} = - \frac{\partial \bf B}{\partial t} $$ Note, for example, it is $\bf B$ not $\bf H$ which has zero divergence. So $\bf B$ lines always run in closed loops, but $\bf H$ lines do not need to if there is some magnetization around. In a similar way, if some medium carries no free charge then the $\bf D$ field runs in closed loops (or it may be zero), but the $\bf E$ field might not.

On usefulness

The fields $\bf D$ and $\bf H$ are useful when considering things like capacitors and inductors, but they really come into their own when considering electromagnetic waves in dielectric media. It would be hard work calculating things like reflection coefficients without them. And they also come into their own in the consideration of energy. The energy flow for example is given by the Poynting vector $$ {\bf S} = \frac{1}{\mu_0} {\bf E} \times {\bf B} - {\bf E} \times {\bf M} $$ ---a rather tricky formula. But how much easier it is in terms of $\bf E$ and $\bf H$: $$ {\bf S} = {\bf E} \times {\bf H}. $$

On relative permittivity and permeability

We always have ${\bf \nabla} \cdot {\bf H} = - {\nabla} \cdot {\bf M}$ and ${\bf \nabla} \cdot {\bf B} = 0$. But this means it will often not be possible to write ${\bf B} = \mu_0 \mu_r {\bf H}$, so answers which only make reference to that formula are missing a major part of the physics. In particular you usually can't use ${\bf B} = \mu_0 \mu_r {\bf H}$ when thinking about permanent magnets.

In the case of a permanent magnet you have a static case with no free current, so ${\bf \nabla} \times {\bf H} = 0$, which means the integral of ${\bf H}$ around a loop is zero, but this will not be true for $\bf B$, so there is no simple proportionality between them. However, in many simple amorphous media it happens that, at low fields, ${\bf M} \propto {\bf H}$. So in this case ${\bf B}$ is also proportional to ${\bf H}$ so we can introduce the relative permeability $\mu_r$ defined through the equation $$ {\bf B} = \mu_0 \mu_r {\bf H}. $$ This is a useful equation, but much more restricted in its validity than the other ones I have written above. (Actually we can also use this result slightly more generally, in non-linear materials where $\bf M$ is not proportional to $\bf H$ but is in the same direction; in this case $\mu_r$ will depend on ${\bf H}$.) Similarly, simple dielectric media will have polarization proportional to $\bf E$, and consequently $\bf D$ proportional to $\bf E$, so we define a relative permittivity $\epsilon_r$: $$ {\bf D} = \epsilon_0 \epsilon_r {\bf E}. $$

But what is the difference between $\bf B$ and $\bf H$ in physical terms?

${\bf B}$ is the field which gives the force on a moving charge, and it is the field which is induced by a changing electric field. It is the one involved in electromagnetic induction. Its integral over a surface is the flux.

$\bf H$ is the field which is easily calculated from a given amount of free current, and the component of $\bf H$ along a boundary does not change when you move from one medium to another (if there is no surface free current). This makes $\bf H$ useful in calculating what electromagnetic waves do, and it is also useful for tracking energy movements via the Poynting vector ${\bf S} = {\bf E} \times {\bf H}$.

A permanent magnet has $\bf B$ running in loops and $\bf H$ following $\bf B$ outside, but not inside, the magnet, in such a way that its integral around a loop is zero (unless there is a current flowing nearby), c.f. Direction of H and B inside and outside a bar magnet

A lump of glass with light waves propagating in it has both $\bf B$ and $\bf H$. If you have an inductor made of a solenoid with a fixed current, then when you slide a piece of glass into the cylinder (keeping the current constant) the value of $\bf H$ does not change but the value of $\bf B$ does. And if you slide in a piece of soft iron the value of $\bf B$ changes enormously. In this case the current supply providing the constant current will do some work, which provides the field energy.

A comment on units and physical dimensions

One of the puzzles of this area of physics is why $\bf B$ and $\bf H$ have different physical dimensions in the SI system of units, and so do $\bf D$ and $\bf E$. One should not hang too much on that. It is just a human choice about definitions. The people inventing the SI system could, with good logic and physical sense, have chosen to introduce the field ${\bf B} - \mu_0 {\bf M}$, giving it the symbol say $\tilde{\bf H}$. Then we would all be learning the formulae $$ \tilde{\bf H} = {\bf B} - \mu_0 {\bf M} $$ and $$ {\bf \nabla} \times \tilde{\bf H} = \mu_0 {\bf j}_{\rm f} + \mu_0 \frac{\partial {\bf D}}{\partial t}. $$ This way of looking at things has the handy result that $\tilde{\bf H}$ and $\bf B$ have the same physical dimensions, which makes a lot of sense, but it results in a $\mu_0$ in the formula relating $\tilde{\bf H}$ to current and the inventors of the system of units wanted to avoid that. So there we have it.

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    $\begingroup$ What a clear summary! Thank you for making the point about permanent magnets. $\endgroup$ Feb 1, 2021 at 10:58
  • $\begingroup$ $\mathbf{B}=\mu_0\mu_r\mathbf{H}$ still works fine for "reasonable" permanent magnets (ignoring frequency-dependence, saturation, hysteresis, etc.). The only reason you have a discrepancy between loop integrals of $\mathbf{H}$ and $\mathbf{B}$ is because of the change in $\mu_r$ between the inside and outside of the magnet. If your paths of integration are confined to fully inside or outside the magnet, then there is no discrepancy and $\mathbf{B}$ and $\mathbf{H}$ are proportional within each separate region. $\endgroup$ Oct 19, 2021 at 7:37
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    $\begingroup$ @hddh The trouble is if you ignore hysteresis then you are not dealing with a permanent magnet. But in the presence of hysteresis there is no one-to-one relationship between B and H, and there are many regions where a tangent to the curve of B vs H does not pass through the origin. In other words, they are not simply proportional to one another. $\endgroup$ Oct 19, 2021 at 8:02
  • $\begingroup$ Good point, I was thinking very loosely of "permanent magnet" as anything that isn't an electromagnet (e.g., the magnetic core of an inductor). $\endgroup$ Oct 19, 2021 at 8:15
  • $\begingroup$ I guess what I was trying to get at is that $\mathbf{B}=\mu_0\mu_r\mathbf{H}$ is still very useful. When dealing with magnets that are actually intended to be permanent (such as in a synchronous motor) we don't usually care about these aspects - it just produces some $\mathbf{B}$ field that is probably empirically known. In other cases, such as a inductor's iron core, we can still think in terms of $\mathbf{B}=\mu_0\mu_r\mathbf{H}$ and treat hysteresis as a non-ideality. $\endgroup$ Oct 19, 2021 at 8:19

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.

  • $\begingroup$ This is smth came to my mind after I've read the Wikipedia's article on the Maxwell's equations for a medium. $\endgroup$ Dec 24, 2016 at 19:53
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    $\begingroup$ 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. $\endgroup$
    – tparker
    Dec 24, 2016 at 23:00
  • $\begingroup$ @tparker OK, 10x for letting know ) $\endgroup$ Dec 25, 2016 at 13:36
  • $\begingroup$ Can't you just substitute $\mu_0\mu_r$ as needed, where $\mu_r$ is your relative permissivity? $\endgroup$ Dec 26, 2016 at 4:34
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    $\begingroup$ @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. $\endgroup$ Dec 26, 2016 at 8:13

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.

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    $\begingroup$ +1. wow, never knew H lines can be not closed. $\endgroup$ Dec 25, 2016 at 13:38
  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Dec 27, 2016 at 0:00

$\mathbf{B}$ is the magnetic field in a vacuum. It is what would be most properly referred to as "magnetic field", by analogy with the "electric field". It's what, along with the electric field, directly governs the motion of charges.

$\mathbf{H}$ is an effective magnetic field that comes about when we consider a magnetic field (i.e. $\mathbf{B}$) penetrating a material object, but at a macroscopic scale where we can ignore the fact that matter is composed of tiny particles moving about with vacuum between them. Because the particles are electromagnetically active. In effect, it's what the interactions between $\mathbf{B}$, all the atoms and molecules in the material, and subject charge, "looks like" to said charge when moving through it at a scale which is far larger than those constituents. It lets you calculate the magnetic effect on such without having to worry about the details thereof.

Likewise, the same holds for the electric fields $\mathbf{E}$ and $\mathbf{D}$.

$\mathbf{B}$ and $\mathbf{E}$ are the more fundamental entities, while $\mathbf{H}$ and $\mathbf{D}$ are derived, or emergent, entities.

  • $\begingroup$ +1 for the best theoretical explanation $\endgroup$
    – Jdeep
    Mar 24, 2021 at 17:14

It turns out that ampere law is still true in case of magnetics placed in external fields but we would have to include current due to magnetization.

$$ \oint B \cdot dl = \mu_o \left[ I + I' \right] \tag{1}$$

But now, we run into a problem, the magnetization current $I'$ is experimentally difficult to determine, how we can write equation of another field who's line integral over a loop is only determine by the current through the material. Applying the result of $ \oint J dl = I'$ on equation(1) and rearranging:

$$ \oint ( \frac{B}{\mu_o} -J) \cdot dl=I$$

Now, the quantity in integrand can be taken as:

$$ H = \frac{B}{\mu_o} - J$$

Now, this new vector field is independent of magnetization effects which makes it 'nice'.

Example of applying the result(page-181):

enter image description here

Consider a system of a long shape wire (the curve $\Gamma$ ) and an arbitary piece of paramagnetic through which the wire passes. Now, consider the line integral of magnetic field of over the loop $\Gamma$ notice that it is depending on if we place the paramagnetic or not. However, for the field $H$, the loop integral is same in both cases because it only depends on the conduction currents.

*:Based on discussion from page-178 to page-179 in I.E Irodov's Basic laws of electromagnetism.


$B=\mu H$

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.)

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    $\begingroup$ +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. $\endgroup$ Dec 26, 2016 at 8:17
  • $\begingroup$ Not true in general. $\endgroup$
    – ProfRob
    Dec 26, 2016 at 9:57
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    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Dec 26, 2016 at 23:58
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    $\begingroup$ Moreover any electrical motor employs ferro-/ferri-magnetic materials, especially magnets (e.g. any brushless motor, like those found in computer fans). Mankind has heavily relied on magnetic non-linearities for common activities for at least a a century now! $\endgroup$ May 3, 2018 at 3:01
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    $\begingroup$ Sorry, no. "Vast majority of cases" doesn't mean anything if you don't provide context. In nature in general linearity is a very rare thing. Only vacuum is truly linear, at least in the Maxwell's equation model, AFAIK. Whenever you deal with any physical medium, linearity is just an useful approximation which is valid in very specific conditions. Not even the resistance of an incandescent light bulb is linear (when it heats up its resistance increases, so the resistance depends indirectly on the current). $\endgroup$ May 3, 2018 at 11:09

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