# Magnetism Variables

1. Nov 2, 2011

### Harrisonized

This isn't really a homework problem or a problem that I can just figure out. I'm wondering why the letter B represents the magnetic field. Also, what does H represent?

(For example, F represents the "force" field, E represents the "electric" field, V represents the scalar voltage. I'm actually not sure what q represents, but I think it means "quanta" so that makes some sense.)

2. Nov 9, 2011

### Harrisonized

It's been a week. Am I allowed to bump threads?

3. Nov 20, 2011

### Harrisonized

4. Nov 20, 2011

### physicsjock

H is an easy vector to use to determine magnetic fields.

It's called afew different things, i was taught its name was just the H vector, its the magnetic equivalent of the D vector in electrostatics.

So its the field due to the free charge.

Using H makes it easy to determine fields as H = M/x (x being magnetic susceptibility) < there you go.

So H is used to easily determine the free current enclosed

The surface bound current and the volume bound current can be found by.

K= M x n, all vectors, n being normal vector to surface at the point where K is calculated

J = Curl (M) curl is nabla x M all vectors,

Again these are only due to the bound current which is the current produced by the electrons already inside the material,

under the influence of a field these electrons move and shift around forming a current, which is not really an actual current per say, its just the movement of the electrons within the atoms of the material in a manner that makes it a current,

maybe somone else can explain it better,

Which can then be used to determine the magnitsation due to the bound charges.

you can also use H to directly find the B field,

B = uH where u is the relative permeatability times the permeatability of free space. the relative is unique for each material

Last edited: Nov 20, 2011
5. Nov 20, 2011

### Harrisonized

I know what these variables are. I'm in an E/M course and we do all the calculus that comes with it. I guess I wasn't exactly very clear in my first post, so my apologies.

I just want to know why they were named as they are. For example, D stands for dielectric, and M stands for magnetization. That's why I didn't ask for those.

The only thing related to magnetism that I can think of that starts with a B is Biot, after Jean-Baptiste Biot, but I'm not exactly sure it's named after him, since the law for the B field is named after both Biot and Felix Savart, not just after Biot alone. Also, I have no clue what could start with H.

Last edited: Nov 20, 2011
6. Nov 20, 2011

### physicsjock

Oh right,

In D is displacement vector and dielectric is just dielectric.

We use Biot-Savart Law, so maybe thats why B but theres heaps similar forumlas.

I really don't know hey, the only H I can see is from Henry's.

I've never really thought about the naming of each thing, I always just thought they named it that way because it was easy and didn't make stuff confusing.

I can't find much on who made the H vector

7. Nov 20, 2011

### Harrisonized

Oh right. I forgot about Henry.

Also, since you bothered to do this, I couldn't just leave it alone:

M = χmH, where χm is the magnetic susceptibility (you still spelled it wrong, despite your edit).

This is χ, the Greek letter "chi", NOT x. x is already used in the standard xyz coordinate system. If we used x for anything else, it'd be horribly confusing whenever we want to calculate integrals.

Anyways, we have that B = μ0(H+M), so it follows that:

B = μ0(1+χm)H

In standard practice, we let μ = μ0(1+χm) and we get

B = μH = μ/χmM

Since they are related in a linear fashion, we don't really mathematically treat the H and M fields much differently than we do for the B field. I don't see how you can claim that the calculations are any easier when all that's happening here is the rescaling of B.

What? I'm not sure I'm understanding you here. K = ∫M×dS ?

Is there even a meaning in this integral? Do you mean this?

M.dS

The above integral better be zero, or we're in trouble, because that would mean that B is a divergent field since B = μ/χmM, and

B.dS = ∫∫∫(∇.B) dV = 0

You mean that j = curl(B)?

i = ∳B.dl = ∫∫(∇×B).dS (by Stoke's theorem)

i = ∫∫j.dS

Therefore,

∇×B = j

But B = μ/χmM.

∇×(μ/χmM) = μ/χm∇×M

Therefore,

μ∇×M = χmj

I'm not sure when you would ever use this.

-----

If you want to state facts, please get them straight.

As someone who is passionate about physics, I think your attempt at an intro is terrible. Physics is about deriving beautiful results from a few simple observational facts. Jumping straight to the end is an injustice, and someone who didn't already know about these fields would still not know anything about these fields after reading your intro. I appreciate you trying to teach, but keep in mind that flaunting basic results is not the way to do it.

Furthermore, I know about this subject, but let's just pretend that I don't. In my first post, I mentioned that I knew what B is. Then, if you're trying to explain the H field, it's by far better to make a comparison of H to B than giving some random crap about electrons moving about.

Anyways, just saying.

Last edited: Nov 20, 2011
8. Nov 20, 2011

### physicsjock

Everything I have stated is correct,

Yes there was a spelling error because I didn't bother spelling it right, I fixed it for you

Each of the equations I wrote were taken from Griffiths Intro do electro dynamics. Maybe you should have a read of it or something it's the book they use through first year EM at the uni i attend.

So what if I used x instead of chi? I defined it after..

The K and J are for bound currents, which is what I said..

I don't know why you thought they were integrals, or why you integrated them.

You would have to integrate K as well not just the RHS and i'm not sure what you were doing with the J.

The whole point of those equations is to make it easier to determine M and B..

If there were integrals I would have wrote int, but I avoided writing those equations.

If you said you had already dealt with it before, I wouldn't have bothered.

Yes you said you knew what B is, but really, anyone in any field of first year science knows what it is.

I was just trying to explain what the point of it was. What really is just an easy method to find B and M.

What would the point of giving a derivation be in a online physics forum if you could just look on wikipedia or in a book?

Why would I give a full derivation of H and all the calculus identities for a small question about "what H represents"?

The reason no one has answered your post is because it's the letter H is just convenient. If there was a reason for it, it would be all over the internet. The only reason i answered this post was because I miss interpreted your question as a valid question for what it actually is.

Which is just an easy way around finding the enclosed free current and the B field.

So in other words,

The only thing I didn't do was write it using tex?

Last edited: Nov 20, 2011
9. Nov 20, 2011

### Harrisonized

Don't type in tex. It lags my browser. Just plain typing is fine.

Okay. In that case, please explain to me me what K=M×n is. I've never heard of it. What is it used for? I integrated them because you said

"K= M×n, all vectors, n being normal vector to surface"

All vectors implies an integral, since n can vary across the surface. The only way to account for this is by

K = ∫M×dS,

kind of like how F = ∫i dl×B for a wire varying in space. For comparison purposes, the above integral is equivalent to K = -∫ dS×M, but I've never seen anyone integrate a surface this way. It doesn't even make sense. This is like a line integral with a cross product. Maybe you can integrate around the boundary of an open surface? I don't know. Usually, when we work with surfaces, we put them in regular surface integrals (or double integrals) to find the flux, but that results in a scalar. As you've defined it, it seems like K is a vector, which is why I got confused.

I thought by J, you meant j, the current density vector. (Or at least some variation of j, like jmag, which is different from the j that equals curl(B).)

Um. I'm still not sure how it's easier. All three fields, B, M, and H are linearly related to each other. If two things have such simple relationships, then they are obviously described fully by the same mathematics.

And I'm saying you did a bad job at explaining. :I

That's why it's a good starting point if you genuinely felt like teaching the subject.

I don't know. Why would you mention anything at all if you're not going to make sure the person you're explaining to would fully understand what you're talking about? Isn't the point of PF to provide a medium for people to teach others for free? Are you teaching here or just flaunting knowledge?

There's a reason for everything. If the answer were all over the internet or it were obvious, I wouldn't have asked. I know you misinterpreted, but I couldn't resist myself from posting further because you did such a bad job at your pseudo-teaching. I'm not saying that I'm the best teacher myself, but at least you could try.

I use Fundamentals of E/M by Arthur Kip. It's a very good book that is very well thought out, easy to learn from, and covers some very deep mathematics. I've looked at Griffiths before. I like Kip by far, but I don't think the comparison is fair, since I've been learning out of Kip and not out of Griffiths.

Last edited: Nov 20, 2011
10. Nov 20, 2011

### physicsjock

Ok soooooooooo

that K equation, its for the surface bound current at a given point. Right?

the n is the normal vector at that point. which i already wrote

You can use it to find the surface bound current at any point, relatively easily.

by saying all vectors i meant K, M and n were vectors. No where does it say it is an integral is present.

You could integral K around the surface. What you have been doing is only integrating one side of the equation, which doesnt make sense.

The J i wrote was BOUND current density, and i wrote that.. haha

H makes it alot easier, like we were given questions to use biot-savart law, to find somthing initially which was a page of working or somthing

THEN they taught us H, which could be used in two lines to find the B field, precisely because its a simple linear relationship. Do you get what i mean?

You only have to deal with an easy integral to find the H vector where as biot-savart can be very messy.

Once you know H you know B in a second.. aswell as M

haha and fair enough!!!!!!!!!!!! It was a poor explanation but when i wrote it i imagined explaining it to a first year or highschool person who didnt know what H was and just wanted to know why it is used and what it represents

oh and good luck with your exams if you have them soon, mine are next week
you seem like you understand the calculus very well

11. Nov 20, 2011

### Harrisonized

Okay. That was good clarification.

My book calls J the amperian surface current and represents by jmag.

M.dl = imag = ∫∫jmag .dS

There's nothing similar to K though. Hmm....

Thanks. I have them in two weeks. Sucks that you have them right before Thanksgiving, but good luck. The calculus is what makes the subject beautiful. Vector calculus was a pain to learn, but it was worth it.

By the way, if you respond anymore, can you please not put

lines

in

between

every

sentence?

Thanks.

Last edited: Nov 20, 2011
12. Nov 20, 2011

### physicsjock

Hmm

This is a bit weird for me, If fmag is the same as my K, then this integral should have H instead of M, since that integral is the free current enclosed. Copying your code haha :

H.dl = Ifree enc

I think the best thing to do its just to forget about it, in the end it will end up being the same thing just that we learnt with different symbols!

Yea I liked my vector calculus course, it flowed smoothly with electro and magneto statics but it was linked with complex analysis which was a bit of a drain haha

13. Nov 20, 2011

### Harrisonized

Well, the thing is... my book doesn't mention K = M×n at all. That's why I was trying to figure out what it was. Is there a relationship between K and jmag?

Judging from your words, you said that jmag is the bound current density and that K is the surface bound current. Does that imply that ∫∫jmag .dS = K ? But ∫∫jmag .dS = imag, so is K just imag?

Last edited: Nov 20, 2011
14. Nov 20, 2011

### physicsjock

Did you guys learn the magnetic vector potential?

In Griffiths that's the path they used to get to those two equations.

Last edited: Nov 20, 2011
15. Nov 20, 2011

### Harrisonized

Last edited: Nov 20, 2011
16. Nov 20, 2011

### physicsjock

Soo I just realised that offer was a little bit not allowed.

Could you PM an email or something I can send it to to avoid problems?

17. Nov 20, 2011

### Harrisonized

Yep. Just sent.

18. Nov 20, 2011

Ditto