68

From 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 university, for vacuum, Maxwell's equations are usually written in terms of $B$, while for media in terms of $H$ (and $B=\mu H$).

Qmechanic
  • 201,751
  • 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. – Pierre Polovodov Nov 16 '21 at 22:18
  • Also when considering boundary conditions, the H field is directly related to a surface current. So, it is useful. – Pierre Polovodov Nov 16 '21 at 22:18

9 Answers9

42

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.

SmallRed
  • 553
  • 14
    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... – user541686 Dec 26 '16 at 08:40
  • 4
    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? – user541686 Dec 26 '16 at 09:16
  • See https://www.physicsforums.com/threads/in-magnetism-what-is-the-difference-between-the-b-and-h-fields.370525/ ? – ProfRob Dec 26 '16 at 12:14
  • 6
    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 04:53
  • 4
    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 03 '18 at 10:12
  • @JánLalinský: Isn't it actually the other way around than that answer says? In other words, isn't it E that is analogous to H instead, and D analogous to B? – Outis Nemo Mar 21 '24 at 11:28
  • 1
    @OutisNemo it depends on context, there is no unique universal analogy linking E to H or E to B. – Ján Lalinský Mar 21 '24 at 16:43
24

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 $\vec{D}$.)

user541686
  • 4,131
  • 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. – user541686 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
  • 1
    @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. :) – user541686 Dec 26 '16 at 19:15
  • @SergeiGorbikov: I just asked this related question, it might be helpful to follow it. – user541686 Dec 26 '16 at 19:33
  • 10x. nice question, indeed. – Sergei Gorbikov Dec 26 '16 at 19:54
  • If you 'change the material', both $\mathbf E$ and $\mathbf D$ will change. – Ján Lalinský May 03 '18 at 10:18
  • D⃗ and B⃗ are electric & magnetic flux densities. E⃗ and H⃗ are electric & magnetic field strengths. The difference? Flux doesn't depend on the material, but field strength does. BUT E & H doesn't depend on material. "magnetic field strength" designated by H. It can be defined by the relationship H = B/μm = B/μ0 - M and has the value of unambiguously designating the driving magnetic influence from external currents in a material, independent of the material's magnetic response.. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfield.html – CrownedEagle May 24 '18 at 15:32
20

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.

Andrew Steane
  • 58,183
  • 2
    What a clear summary! Thank you for making the point about permanent magnets. – Philip Wood Feb 01 '21 at 10:58
  • $\mathbf{B}=\mu_0\mu_r\mathbf{H}$ 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. – PersonWithName Oct 19 '21 at 07:37
  • 1
    @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. – Andrew Steane Oct 19 '21 at 08:02
  • 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). – PersonWithName Oct 19 '21 at 08:15
  • 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. – PersonWithName Oct 19 '21 at 08:19
  • "anything that isn't an electromagnet" - no that doesn't make sense - if a coil of wire with an iron core isn't an electromagnet then I don't know what is. I was just thinking of any magnetizable material at all - not sure why my brain associates that with permanent magnet... – PersonWithName Oct 19 '21 at 08:25
  • 1
    Great article. I finally understand after a lot of time. – Darkmatter Oct 01 '23 at 03:47
  • I don't see how B is the fundamental like E is, if it's actually H and E that are the fundamental field intensities. Seems more like H and E are fundamental while B and D** are secondary, no? – Outis Nemo Mar 21 '24 at 11:31
  • @OutisNemo What can I say? My answer is what it is. The force is $q {\bf v}\times {\bf B}$. The other field ($\bf H$) is introduced as a convenience when dealing with materials and magnetization. – Andrew Steane Mar 21 '24 at 21:41
  • @AndrewSteane: Well, that makes sense, since B would still be the actual magnetic force experienced by the particle...but it still seems in some ways to be a derived quantity from the more fundamental field intensity H. I did however have a fairly long discussion recently where it was explained to me that both E and B represent the total fields while D and H only represent the fields due to free charges and currents, so I'm still trying to process exactly how they all relate. – Outis Nemo Mar 21 '24 at 23:13
14

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.

ProfRob
  • 130,455
  • 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 09:34
  • 1
    @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 https://en.wikipedia.org/wiki/Magnetic_field#Definitions.2C_units.2C_and_measurement – ProfRob Dec 29 '16 at 09:40
  • ok, tnx. I got it. Bds/ds=B, that's why you call it magnetic flux density. +1 – Sergei Gorbikov Dec 29 '16 at 09:44
  • 1
    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. – ProfRob Dec 30 '16 at 11:35
  • @AlQuemist, the $H$ field is also total in the sense in general it depends both on free currents and magnetization of the material. Indeed it can be controlled in some experiments, such as toroidal body wrapped in a toroidal coil, but this is not a general property of $H$. In general, the $H$ field depends both on free currents and magnetization currents. – Ján Lalinský May 03 '18 at 10:38
  • @RobJeffries, it is possible to rewrite it, in terms of $B$ field, as $\nabla\times\mathbf B = \mu_0 \mathbf j_{total} + \mu_0 \epsilon_0 \frac{\partial \mathbf E}{\partial t} $, no detailed knowledge of $\mathbf j_{total}$ is needed. This is equally valid formulation of the Ampere-Maxwell law. – Ján Lalinský May 03 '18 at 10:43
  • @JánLalinský I know you can write it like that, but you then do need detailed knowledge of the total current if you want to calculate the B-field. I see though that you are picking me up on what I said in the first paragraph - an edit is in order. – ProfRob May 03 '18 at 11:23
  • @JánLalinský Actually, how did you write this? Are defining $\vec{J}_{\rm total} =\vec{J} + \nabla \times \vec{M} + \partial \vec{P}/\partial t$? – ProfRob May 03 '18 at 11:33
  • @RobJeffries, of course one would need to know the $\mathbf j_{total}$ to calculate $\mathbf B$, but not to formulate the equation. I think the situation is similar with your formulation - there, one would need to know both $\mathbf J$ (free current distribution) and $\mathbf D$ (displacement field) in all space to calculate $\mathbf B$ or $\mathbf H$. – Ján Lalinský May 03 '18 at 12:59
  • @JánLalinský Yes and that is usually better known. Hence H is useful. Especially in non-LIH media. – ProfRob May 03 '18 at 13:05
  • I did not dispute that, only that one cannot write the equations using $\mathbf B, \mathbf j_{total}$. Which formulation is more useful depends on the situation. – Ján Lalinský May 03 '18 at 13:44
8

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.

Firebug
  • 105
  • 1
  • 4
  • 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
  • 3
    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 04:34
  • 2
    @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 08:13
  • Ahhh of course, it starts to be a problem of nonlinear relations and tensor functions. That makes sense. – bright-star Dec 26 '16 at 19:20
3

$\mathbf{B}$ 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.

2

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

SAKhan
  • 1,365
0

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.

-3

$B=\mu H$

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
    +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 08:17
  • Not true in general. – ProfRob Dec 26 '16 at 09:57
  • @SergeiGorbikov, expanded on why you don't need (for most purposes) to care about nonlinearity. – Stack Tracer Dec 26 '16 at 22:31
  • 3
    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. – ProfRob 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 05:09
  • " it's thoroughly nontrivial to get significant non-linear effects from a material," completely false in modern world. Any modern power-supply uses switching technology which relies on having inductors with ferro-/ferri-magnetic cores, which are heavily non-linear even in "normal" conditions (no need of ridiculous field levels to run into non-linearities). Even if we diregard this, trivial mains power transformers, which changed the world of electricity since decades, have Si-Fe cores, which are non-linear. ... – LorenzoDonati4Ukraine-OnStrike May 03 '18 at 03:01
  • 1
    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! – LorenzoDonati4Ukraine-OnStrike May 03 '18 at 03:01
  • @LorenzoDonati very interesting, but for the vast majority of cases (basically, if you have to ask whether the system is linear or not, it is), the system is linear. – Stack Tracer May 03 '18 at 03:17
  • 1
    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). – LorenzoDonati4Ukraine-OnStrike May 03 '18 at 11:09
  • We live in a mostly non-linear world. The linearity "illusion" is what allows us to do simpler calculations, but linearization techniques in science and engineering have great importance because most systems need to be linearized to be treated as linear (under very specific working conditions, as I said before). – LorenzoDonati4Ukraine-OnStrike May 03 '18 at 11:09