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

1. Jan 18, 2010

### 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.

Last edited: Jan 18, 2010
2. Jan 18, 2010

### Bob S

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

3. Jan 18, 2010

### Andy Resnick

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.

4. Jan 18, 2010

### tiny-tim

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 …
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!

5. Jan 19, 2010

### Gerenuk

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/Electromagnetic-Fields-Waves-Paul-Lorrain/dp/0716703319
In doubt, I can post it here.

Last edited by a moderator: Apr 24, 2017
6. Jan 20, 2010

### cabraham

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

Last edited by a moderator: Apr 24, 2017
7. Jan 20, 2010

### Gerenuk

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,...).

8. Jan 20, 2010

### tiny-tim

magnetic energy density

Hi Claude!

0.5*B*H is the magnetic energy density, of course … J m-3

9. Jan 21, 2010

### cabraham

Re: magnetic energy density

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

10. Jan 21, 2010

### Naty1

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

11. Jan 22, 2010

### kcdodd

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.

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

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

12. Nov 6, 2010

### kinley

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.

13. Nov 6, 2010

### yungman

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?

Last edited: Nov 6, 2010
14. Nov 6, 2010

### 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

15. Nov 6, 2010

### yungman

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 refering 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.

16. Nov 6, 2010

### 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

17. Nov 6, 2010

### yungman

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.

Last edited: Nov 6, 2010
18. May 27, 2012

### Trifis

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...

Last edited: May 27, 2012
19. May 27, 2012

### Trifis

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...

Last edited: May 27, 2012
20. May 27, 2012

### cabraham

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