# In magnetism, what is the difference between the B and H fields?

• jeebs

#### jeebs

hi,
I'm having trouble understanding what the difference is between the B and H fields. What my understanding is, if you have a magnetic dipole on its own, its going to have some B field around it.
Then say you bring some material into the field, and the field within the boundaries of the material will be modified right? it'll increase or decrease or something? And this field is called the H field, am I right?

So why then do we have the expression $$B=\mu_0 H$$ or $$B=\mu_0 (H + M)$$ where M is the sample's magnetization?

Shouldn't the permeability, which has units, really just be some unitless number?
Why does H have units of magnetization, ie. the magnetic dipole moment per unit volume?

Thanks.

PS. if this is in the wrong section, please say so and I'll put it elsewhere.

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• Molar, haloboy777 and gracy

Based on Maxwell's equations, electric fields are generated by changing B fields, while H fields are generated by changing electric fields.
In dc fields, static electric E fields create currents (magnetization currents) I (when σ>0) which in turn produce static H fields. A static B field (actually d/dt ∫B·n dA = 0) cannot produce an electric field E.

Even if we used natural units where μ0 = ε0 = 1, this distinction between B and H remains. In magnetic materials, B is not linearly related to H due to the magnetization term M. If B = H in space, then BH in magnetic materials.

If voltage = d/dt ∫B·n dA, then how can curl H= σ·E = σ·voltage/length if B = H?

So even if μ0 = ε0 = 1, they have to have units. Going without units is about as rational as claiming that h-bar = c =1. (I may get FLAK on this).

Bob S

hi,
I'm having trouble understanding what the difference is between the B and H fields. <snip>

The H field (the magnetic field) is the field in vacuum. This field can induce a magnetization of ponderable matter, and the total field (vacuum plus induced field) is the B field (magnetic induction). It is sometimes confusing that the conventions of E and D are different than B and H.

Constitutive relations are used to relate the H and B (as well as E and D) fields.

I'm having trouble understanding what the difference is between the B and H fields.

Hi jeebs! It's all to do with the difference between free charge and bound charge (which together make total charge).

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.​

So
E = D + P (except that for historical reasons E is defined differently, so we need to multiply it by the permittivity, and for some reason P is multiplied by minus-one ).

And
B = H + M (except that for the same historical reasons B is defined like E, so we need to divide it … why divide?? … by the permeability).

The latter equation says that the total magnetic field (divided by the permeability) equals the free magnetic field plus the bound magnetic field (the bound magnetic field is all those little loopy currents that make things magnetic).

When an electric current produces a magnetic field, it couldn't care what the field is going to be used for (ie for bold sweeping field lines or pokey little loops inside matter), so it produces a total field, which it's sensible to measure as B.

But once we put matter in the way, we can only measure the free field, H. As you say …
Then say you bring some material into the field, and the field within the boundaries of the material will be modified right? it'll increase or decrease or something? And this field is called the H field, am I right?

Basically, yes … (apart from the permeability factor, of course) B and H are the same away from matter, but in or near matter the matter soaks up some of the B, and all we measure is what's left, the H.

Finally, there's no reason in principle why B and H shouldn't be measured in the same units … but they're not! • Devraj, sulhar, Molar and 3 others
Only the B field is a real physical field and all of physics can be done with the B and E field alone. The H (and D) field is a mathematical trick to discard the consideration of induced magnetizations in materials.

Basically if you introduce magnetizable materials you have two choices.
1. Use B field only, but you have to integrate over all induced magnetic dipoles in materials and find a self-consistent solution
2. Use B and H field and find a self-consistent solution at the boundaries. No need to integrate over induced volume dipoles - you can ignore them.

The procedure for view (2) isn't usually as easy as saying "this object induces this field". You have to find self-consistent solutions of B and H field and which object created what you cannot tell in most cases.

In SI units B and H have different units.

Why does H have units of magnetization, ie. the magnetic dipole moment per unit volume?
This stems from the derivation of the "mathematical trick" which takes into account the induced dipoles. I can't remember which books have the derivation in them. Maybe
https://www.amazon.com/dp/0716703319/?tag=pfamazon01-20
In doubt, I can post it here.

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• sulhar
Only the B field is a real physical field and all of physics can be done with the B and E field alone. The H (and D) field is a mathematical trick to discard the consideration of induced magnetizations in materials.

Basically if you introduce magnetizable materials you have two choices.
1. Use B field only, but you have to integrate over all induced magnetic dipoles in materials and find a self-consistent solution
2. Use B and H field and find a self-consistent solution at the boundaries. No need to integrate over induced volume dipoles - you can ignore them.

The procedure for view (2) isn't usually as easy as saying "this object induces this field". You have to find self-consistent solutions of B and H field and which object created what you cannot tell in most cases.

In SI units B and H have different units.

This stems from the derivation of the "mathematical trick" which takes into account the induced dipoles. I can't remember which books have the derivation in them. Maybe
https://www.amazon.com/dp/0716703319/?tag=pfamazon01-20
In doubt, I can post it here.

Hmmm. You lost me with "only B is real, not H"! I'm having trouble with that one. Likewise for D & E. Two inductors, one air cored, and one ferromagnetic cored, are energized to the same B value, identical in all respects except for the core. The energies are vastly different as W = 0.5*B*H. One could argue that W = 0.5*(B^2)/mu, as well. Energy can be expressed only in terms of B and mu. But is mu "real"? It has to be or we have pure nonsense.

So here it is. Mu is real, and B is real. No argument so far? Since H = B/mu, then since B and mu are both real, how can their ratio, H, not be real as well? I'm just wondering.

D & H have to be every bit as real as E & B, or nothing makes sense. Since epsilon & mu are real, so are B, D, E, & H, as well. Am I missing something? Please enlighten me. You've engaged my curiosity. BR.

Claude

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So here it is. Mu is real, and B is real. No argument so far? Since H = B/mu, then since B and mu are both real, how can their ratio, H, not be real as well? I'm just wondering.
If course all vector fields exist. I meant real in the sense of "natural".

You can do physics with one simple law for the B field, but difficulties in the numerical solution.

Or you can do physics with complicating relations between B and H field, but easier numerics.

Nature doesn't care if the "numerical" solution is difficult or not. So I consider having only one simple law more "natural" and more "real".

However, for most practical reasons it's easier to do calculation with B and H fields despite their complicating interrelations.

I can try to work out these rules again. Only in the first case (B only) you can say that an object creates a particular field. With B and H field you have to make adjustments.

Btw, the most general law is
$$\vec{B}=\mu_0(\vec{H}+\vec{M})$$
Your relation with $B=\mu H$ is only valid for special media (linear,...).

• sulhar
magnetic energy density

Two inductors, one air cored, and one ferromagnetic cored, are energized to the same B value, identical in all respects except for the core. The energies are vastly different as W = 0.5*B*H. One could argue that W = 0.5*(B^2)/mu, as well.

Hi Claude! 0.5*B*H is the magnetic energy density, of course … J m-3 Hi Claude! 0.5*B*H is the magnetic energy density, of course … J m-3 Yes, of course. My apologies. I got sloppy and forgot to multiply by the volume. The 2 inductors I used in my example were identical in all respects except for the cores, air vs. ferromagnetic. With equal volumes, the unequal energy densities result in unequal energies. I stand corrected. Thanks.

Claude

A changing magnetic field creates an electric and 'a changing electric field creates a magnetic field.

Wikipedia has some relevant material here:
http://en.wikipedia.org/wiki/Electromagnetic_fields

For example, see:
http://en.wikipedia.org/wiki/Electromagnetic_fields#Dynamics_of_the_electromagnetic_field

In the past, electrically charged objects were thought to produce two types of field associated with their charge property. An electric field is produced when the charge is stationary with respect to an observer measuring the properties of the charge, and a magnetic field (as well as an electric field) is produced when the charge moves (creating an electric current) with respect to this observer. Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole — the electromagnetic field.

Once this electromagnetic field has been produced from a given charge distribution, other charged objects in this field will experience a force (in a similar way that planets experience a force in the gravitational field of the Sun). If these other charges and currents are comparable in size to the sources producing the above electromagnetic field, then a new net electromagnetic field will be produced. Thus, the electromagnetic field may be viewed as a dynamic entity that causes other charges and currents to move, and which is also affected by them

You can gain some insight from how using D and H have different properties than E and B.

(using natural units)

$$H = B - M$$

$$D = E + P$$

Looking only at the static situation for a second, curl of H deals only with free current, because we took out the curl of the magnetic dipoles (bound currents). Looks prettier eh.

$$\nabla \times H = \nabla \times B - \nabla \times M = J_f$$

Similar thing for electric field, only we want free charge instead of free current.

$$\nabla \cdot E = \rho_f - \nabla \cdot P$$

$$\nabla \cdot D = \nabla \cdot E + \nabla \cdot P = \rho_f$$

It's simpler because you don't have to take into account the currents and charges inside the the dipoles. Incidently, it's interesting that H can have non-zero divergence, like a magnetic source. And curl of D is non-zero, like magnetic current.

$$\nabla \cdot H = \nabla \cdot B - \nabla \cdot M = - \nabla \cdot M = \rho_m$$

$$\nabla \times D = \nabla \times E + \nabla \times P = \nabla \times P = J_m$$

When it comes to outside of a magnetic material both these fields are same. Inside a magnetic material they are completely different especially with regards to relative magnitude and direction. B field is dependent to a considerable extent on currents, both microscopic and macroscopic while the H field depends on microscopic currents. B field lines always form loops around the total current. In the case of H field the lines always loops around free current. They begin and end near magnetic poles.

Based on Maxwell's equations, electric fields are generated by changing B fields, while H fields are generated by changing electric fields.
In dc fields, static electric E fields create currents (magnetization currents) I (when σ>0) which in turn produce static H fields. A static B field (actually d/dt ∫B·n dA = 0) cannot produce an electric field E.

Even if we used natural units where μ0 = ε0 = 1, this distinction between B and H remains. In magnetic materials, B is not linearly related to H due to the magnetization term M. If B = H in space, then BH in magnetic materials.

If voltage = d/dt ∫B·n dA, then how can curl H= σ·E = σ·voltage/length if B = H?

So even if μ0 = ε0 = 1, they have to have units. Going without units is about as rational as claiming that h-bar = c =1. (I may get FLAK on this).

Bob S

My understanding is static electric field can cause FREE charges on conducting surface because of free electron movement. It also cause polarization of dielectric material which the charges on the surface consist of only BOUNDED charges.

Static electric field will not cause magnetic field and any surface current density. Am I getting this right?

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My understanding is static electric field can cause FREE charges on conducting surface because of free electron movement. It also cause polarization of dielectric material which the charges on the surface consist of only BOUNDED charges.

Static electric field will not cause magnetic field and any surface current density. Am I getting this right?

A static electric field on a conductor produces a dc current. You should think of the static (electric field) Maxwell equation for conductors as Curl H = σE = J, because E/∂t = 0, where J is current density. The integral form of this in vacuum is Ampere's Law.

Wind a N= 100-turn coil around a solenoid coil form. It should be at least 12 ohms. Make it about L = 0.1 meter long. Find a dc (static) voltage source, like 12 volts. Make sure the coil resistance R is at least 12 ohms. Attach voltage source V to coil. The current I (=V/R) times N divided by L is H = NI/L ampere turns per meter, sometimes called the magnetic intensity.

I am using the word static to mean non-changing, like E/∂t = 0.

Bob S

A static electric field on a conductor produces a dc current. You should think of the static (electric field) Maxwell equation for conductors as Curl H = σE = J, because E/∂t = 0, where J is current density. The integral form of this in vacuum is Ampere's Law.

Wind a N= 100-turn coil around a solenoid coil form. It should be at least 12 ohms. Make it about L = 0.1 meter long. Find a dc (static) voltage source, like 12 volts. Make sure the coil resistance R is at least 12 ohms. Attach voltage source V to coil. The current I (=V/R) times N divided by L is H = NI/L ampere turns per meter, sometimes called the magnetic intensity.

I am using the word static to mean non-changing, like E/∂t = 0.

Bob S

Sounds like we are talking about different things. Correct me if I mis understand this.

You are talking about putting a voltage across a conducting material where and electric field form inside the material and current flow even in DC. That I agree with you. As long as there is some dc resistance in the material that cause a voltage drop, an electric field will formed. And yes, when a current passing through, magnetic field will form.

What I am referring is a conducting material INSULATED in vacuum with electric field hitting it. Electric field only cause free charge to move ( eg, if the arrow of the electric field hitting the metal surface, electrons are drawn to the surface and leaving possitive charges on the opposite side of the metal.

What I am referring is a conducting material INSULATED in vacuum with electric field hitting it. Electric field only cause free charge to move ( eg, if the arrow of the electric field hitting the metal surface, electrons are drawn to the surface and leaving possitive charges on the opposite side of the metal.
If "electric field cause free charge to move", and "electrons are drawn to the surface", then there are currents. If there are currents, then by Amperes Law, there are magnetic fields.

Bob S

If "electric field cause free charge to move", and "electrons are drawn to the surface", then there are currents. If there are currents, then by Amperes Law, there are magnetic fields.

Bob S

But that is at the initial condition when the electric field is first apply, the current flow as charge more to the surface. Yes I agree there will be magnetic field generated initially. But as everything settle down, as long as the electric field is constant, the free charge remain constant and nothing change, no more current moving. The whole thing is in a stand still and no current flows and the magnetic field disappeared. It just stay there frozen until the field intensity change. That is the reason I call it surface charge.

Yes in varying electric field, it become current and magnetic field is generated.

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The real question here is why the electric field (E) is weaker inside matter while the magnetic field (B) is stronger!

I'm surprised that nobody bothers to explain that. I mean if we think of magnetic properties of materials as a collection of microscopic magnetic dipoles then the application of an external magnetic field (H) should have had a similar effect with the polarization case in dielectrics, where the inner electric field (E) is weakened from the opposed electric field (P/ε) of the microscopic electric dipoles.

PS: In case of magnetism the dipoles align parallel with the external field while in electrostatics they align antiparallel...

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Moreover if the alignment of those dipoles was indeed analog (e.g. ferroelectrics vs ferromagnets), the analogon with repect to the formulas for D and H would still remain "unsettled", because their derivation does not depend on the above mentioned behaviour of the matter! Particularly as you already know, H field is the interior magnetic field MINUS the magnetization while D field is the interior electric field PLUS the polarization.

What's the point of those differences? Are they a direct consequence of the native difference between electricity and magnetism expressed through the maxwell equations? Please note that their derivation has NOTHING to do with the strength of the interior forces and still the two theories are not compatible...

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B is magnetic flux density, whereas H is magnetic field intensity. H has units of amp-turn/meter, whereas B has units of weber/turn-meter^2. In non-ferrous materials they have a simple inter-relation given by B = mu*H. In ferrous media it is a non-linear relation and numerical and/or graphical methods of computation are used.

One is an intensity, the other a density. A rough analogy exists for E & D as well, although I advise against trying to correlate D/E with B/H. D is the electric flux density, and E is the electric field intensity. All 4 quantities are equally important. I don't think we can define things using just 2 of them.

Claude

Sorry but this is not a relevant answer!

Everyone is at least roughly aware of the nature of this fields... Both of my assertions deal with differing subjects.

One has to go back to how these quantities are derived through the macroscopic averaging process and note a fundamental difference both to the behaviour of dipoles as well as the maxwell equations between electrostatics and magnetostatics and if those two observations are somehow connected. I elaborated that already in my first posts.

Here is an interesting question:
Imagine there was a particle that has exactly the same properties as an electron except for it's charge. Instead of a negative charge this particle is a south monopole. Further imagine a particle nearly identical to a proton except that it is a north monopole. What happens if you replace every single electron and every single proton in the universe with those monopoles? Would the world come to an end?
-- disclaimer: the following statements are my personal opinions --
I say we wouldn't even notice the difference. The laws of physics and chemistry would remain the same. That is because the laws of physics that govern the behavior of electric and magnetic fields are 100% equivalent in every respect.
The different treatment that those two kinds of fields receive in maxwells equations is due to 2 reasons.
1. maxwell incorporated the fact that there are electric charges but no magnetic charges(monopoles) into his equations
2. different definitions are being used for things that are really equivalent and could be defined in an equivalent manner.

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@DrZoidberg What you are practically suggesting is that the existence of magnetic monopoles would make electicity and magnetism perfectly interchangeable when relativity is taken into account. E.g. a pure electric field could be "Lorentzly" transformed under this prerequisite into a pure magnetic field and vice versa.

That still doesn't explain the disanalogy of the H/D-field formulas! I've come to believe that the very reason we have given these forms in those general macroscopic quantities (namely D=E+ε$_{0}$P : weaker E-Field in matter --- H=B/μ$_{0}$-M : stronger B-Field in matter) is the fact that most of the electric materials are dielectric and not ferroelectric (which implies the weakening of the E-Field) and the opposite holding for magnetic materials. I do not see any other mathematical explanation since the definition ρ$_{P}$=-∇P used in the derivation of the D formula is totally arbitrary. Thus a similar assumption for the diamagnetic materials would yield H=B/μ$_{0}$+M instead!

PS: To make matters worse, polarization depends on the outer applied field M=χ$_{m}$B$_{a}$/μ$_{0}$ while magnetization depends on the inner reduced electric field P=χ$_{e}$ε$_{0}$Ε$_{i}$ ! I hate it when there is no symmetry where it should be...

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Trifis - The magnetic equivalent of the E field is (if I understand the physics correctly) the H field and not the B field. H and E are real fields, B and D are combinations of field and magnetization/polarization - a mathematical trick so to speak to simplify calculations.
Before you complain, let me explain what I mean by "real field".
Imagine a magnetic monopole and a very thin magnet so that the monopole is able to just shoot through the magnet. It then goes around in circles, shooting through the magnet over and over again. If the B field would determine the force on the monopole, it would pick up more and more energy with each revolution. The H field however points from N to S inside a permanent magnet so there is no energy gain for the monopole particle. I know there are no monopoles but the laws of physics don't rule out their existence. Since the H field determines the force on the particle it is the real field.
Btw. in the same way you can use conservation of energy to show that the E field inside an electret has to point form + to -.
So I don't see an asymmetry there.

D=ε0E+P : stronger D-Field in matter
B=μ0*(H+M) : stronger B-Field in matter

The asymmetry between those two equations is just because P is not defined analogically to M. If you defined M as µ0*magnetization you would get B=μ0H+M. It's all just a matter of definition.

I wished someone with better math skills than me could post a complete proof for the equivalent bahavior of electric and magnetic fields. I'm sure it can all be derived from the known laws of electromagnetism.

Hmmm that would clear up things if the E field was the real equivalent of H field instead of D field... BUT this is way too indistinguishable in the mainstream bibliography cause what is mostly stated is that D and H are the auxiliary fields introduced in order to simplify the maxwell equations in terms of FREE CHARGES/CURRENTS. And that too makes sense, because we tend to treat D and H field as the independent macroscopic fields which the respective E and B fields would be equal to in vacuum.

You are right to say that H field is the stable exterior exciter while B is the reduced interior. But think the case of electrostatics over. Consider a capacitor filled with dielectric material. The E field here is depended on the material and more often than not reduced due to the polarization. Thus it is the E field that is combined (by your definition) with P not the D field! D field on the other hand depends only on the free charges of the capacitor and we can thereby call it the exterior factor for the sake of argument... That again explains why we define ρ$_{P}$=-∇P and not ρ$_{P}$=+∇P as it should be! I think the founders have historically tangled up the concepts :P

Apropos the μ and ε assymetry is certainly not a problem for me!

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Yes, books on electromagnetism often state that D and H are the auxiliary fields. But that is not the same as saying that they are equivalent.

If you keep the charge on a capacitor the same than E depends on the material. If you keep the voltage constant then D depends on the material. But a capacitor is not a good example because there is no good magnetic equivalent to a capacitor due to the lack of monopoles.
Let's instead look at a ring. If you have an iron ring and there is a changing electric field going through its center perpendicular to the plane of the ring, then there is a magnetic field induced that magnetizes the iron. The H field in this case depends on the rate at which the electric field changes and is independent of the ring. The B field depends on the properties of the iron. Without the iron the B field would be much weaker.
Now take a ring made from a good dielectric, e.g. ceramic. If there is a changing magnetic field going through its center an E field is induced that polarizes the ceramic. Here the E field is independent of the material of the ring but the D field depends on it's material properties. It's much higher with the ring in place.

Yes, books on electromagnetism often state that D and H are the auxiliary fields. But that is not the same as saying that they are equivalent.
How do you explain the fact then that those two possesses the free charge/current dependency? Wouldn't be plausible that ∇E=ρ ? I was hoping that you/somebody would go into that, since I highlighted it...

If you keep the voltage constant then D depends on the material.
Can you prove that? I kept referring to our four fields with the qualifiers "interior" (E and B) and "exterior" (D and H), having implied an exterior field is produced by an outer "free" source and is thereby always stable.

But a capacitor is not a good example because there is no good magnetic equivalent to a capacitor due to the lack of monopoles.
I don't see why the monopoles play a role in this case... Why wouldn't a local homogenous H field be an equivalent?

Not to mention that I enjoyed the ring example.

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On an elementary level, there is only one electromagnetic field, given by the Faraday tensor $F_{\mu \nu}$ or, equivalently in the 3D formulation, $\vec{E}$ and $\vec{B}$. This relativistic point of view clearly shows that these two fields belong together since they build the one covariant antisymmetric Faraday tensor. They obey Maxwell's equations (in Heaviside-Lorentz units that are most natural), i.e., the homogeneous equations

$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$

and the inhomogeneous equations

$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E} = \rho.$$

If you have macroscopic matter, it's way too complicated (and also not necessary!) to solve these fully microscopic Maxwell equations, since here $\rho$ and $\vec{j}$ denote all the many charges of all atoms building the matter. It is impossible to solve these complicated equations, including all binding effects of the matter (which requires a fully quantum theoretical description by the way).

Instead one can solve approximate equations for macroscopic electromagnetic fields that are small compared to the microscopic electromagnetic fields between atomic nuclei and electrons forming, atoms, molecules, solids, etc. In this case one can describe the reaction of the medium in an averaged way by using the socalled linear-response theory. For a homogeneous isotropic medium, this yields to (induced and proper) electric polarization and magnetization. The microscopic charge and current distributions are "coarse grained" by taking the average over volume elements that are small compared to the macroscopic scales of change but large to the microscopic (atomic) scales (implying that for electromagnetic waves the wavelength of the macroscopic fields is large compared to atomic distance scales of the medium).

In this case one introduces the auxilliary fields $\vec{D}$ and $\vec{H}$ that form in relativistic notation also an antisymmetric 2nd-rank tensor. In non-relativistic approximation, the inhomogeneous Maxwell equations are then written in the form

$$\vec{\nabla} \times \vec{H}-\frac{1}{c} \partial_t \vec{D}=\sigma \vec{E}+\vec{j}_{\text{free}}, \quad \vec{\nabla} \cdot \vec{D}=\rho_{\text{free}}.$$

and the constitutive equations

$$\vec{D}=\epsilon \vec{E}, \quad \vec{H}=\frac{1}{\mu} \vec{B}.$$

Here, $\vec{j}_{\text{free}}$ and $\rho_{\text{free}}$ are the free electric current and charge densities, which are given by the coarse-graining procedure (averaging out the microscopic charge distributions) explained above. Further $\sigma$ is the conductivity and $\epsilon$ and $1/\mu$ the the electric and magnetic permeabilities. That the latter appear in the numerator and denominator of the constitutive equations, respectively, is an unfortunate historical heritage of the pre-relativity era of electromagnetic theory, we have to live with.

Trifis - The magnetic equivalent of the E field is (if I understand the physics correctly) the H field and not the B field. H and E are real fields, B and D are combinations of field and magnetization/polarization - a mathematical trick so to speak to simplify calculations.
I think I was a little confused here. If you just look at the math it can seem as if E is equivalent to H because that assumption can make part of the math more symmetric but looking at the physics of how magnetic dipoles differ from electric dipoles it's clear that E is equivalent to B.

How do you explain the fact then that those two possesses the free charge/current dependency? Wouldn't be plausible that ∇E=ρ ? I was hoping that you/somebody would go into that, since I highlighted it...
I'm not sure what you mean. ∇E=ρ is basically Gauss's law. Just without the ε0 but ε0 is just a unit conversion factor in this case.

http://en.wikipedia.org/wiki/Maxwell's_equations
I looked at the wikipedia article on maxwell's equations and found this
Wikipedia said:
For one reason, Maxwell's equations can be made fully symmetric under interchange of electric and magnetic fields by allowing for the possibility of magnetic charges with magnetic charge density ρm and currents with magnetic current density Jm.
So apparently the only reason maxwell's equations are asymmetric is because matter contains no monopoles.

the homogeneous equations

$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$

and the inhomogeneous equations

$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E} = \rho.$$

So that means when you have pure fields and no matter, j and p become 0 and you get

$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0$$
So it is only the properties of matter (e.g. the presence of electric charges) that create asymmetry? For pure fields the equations are completely symmetric?

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The H field (the magnetic field) is the field in vacuum. This field can induce a magnetization of ponderable matter, and the total field (vacuum plus induced field) is the B field (magnetic induction). It is sometimes confusing that the conventions of E and D are different than B and H.

Constitutive relations are used to relate the H and B (as well as E and D) fields.

In vacuum the H and B field are equivalent. You write as if the H field is somehow more fundamental, it certainly is not.

My understanding is that the H field is simply a mathematical tool, it depends only on the free currents and not on bound currents like the B field does, and so simplifies calculations. You can calculate the H fields from the known free currents, and then from that calculate the B field. At the end of the day though, it's the B field that we measure and is a "physical" thing, not the H field.

Plus, relativistic transforms show that in vacuum, an electric field transforms to the B field, not the H field.

Plus, relativistic transforms show that in vacuum, an electric field transforms to the B field, not the H field.

In vacuum B is not just equivalent to H it is identical. B=H (if you measure both in the same units e.g. gaussian units) so your statement makes no sense. Otherwise you are correct I think.

So that means when you have pure fields and no matter, j and p become 0 and you get

$$\vec{\nabla} \times \vec{E}+\frac{1}{c} \partial_t \vec{B}=0, \quad \vec{\nabla} \cdot \vec{B}=0$$
$$\vec{\nabla} \times \vec{B}-\frac{1}{c} \partial_t \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E} = 0$$
So it is only the properties of matter (e.g. the presence of electric charges) that create asymmetry? For pure fields the equations are completely symmetric?

Of course, if there is no matter, you have a free electromagnetic field, but Maxwell's equations for $\vec{E}$ and $\vec{B}$ are always valid, also at presence of matter. You only have to make sure that the charge and current densities denote the complete set of charges und currents of the elementary particles building up this matter (including the magnetic moments of all particles!).

As I wrote earlier, this complicated set of equations cannot be solved for macroscopic matter, and fortunately that's not necessary, but you can use approximations, of which the simplest and often applicable in everyday life is linear response theory. Then you get macroscopic classical electromagnetics as you find it in almost all textbooks on the subject, although these very often omit to stress the basic fact that on a fundamental level, there is only one electromagnetic field with components $\vec{E}$ and $\vec{B}$, while $\vec{D}$ and $\vec{H}$ are auxiliary quantities denoting the average fields over macroscopic small, microscopic large regions, containing the effective average charge and current distributions of matter. In these macroscopic Maxwell equations only the external charges and currents brought into the system from outside and not taken care of by the average charge and current distributions of the matter.

In vacuum B is not just equivalent to H it is identical. B=H (if you measure both in the same units e.g. gaussian units) so your statement makes no sense. Otherwise you are correct I think.

Yes well this is just a bit of semantics, but "equal" is more precise than "equivalent" I suppose.

It is good to know that there are some people willing to let us know what we already know...
BUT the primary question remains emphatically unanswered, so may I state it again:

D=ε$_{0}$E+P , ∇D=ρ$_{free}$

H=B/μ$_{0}$-Μ , ∇H=j$_{free}$

There is an obvious asymmetry here... To my eyes at least!

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