Talk:Magnetic field

New comments at the bottom please

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It's been a while since I've done this stuff, and I don't have my textbook here to verify, but shouldn't we be using <math>\epsilon,\mu<math> instead of <math>\epsilon_0, \mu_0<math>? Or we can use H and D fields instead... am I correct?

Yes, you're correct. Using ε and μ would only be correct for linear media, using H and D would be more general. I fixed it by saying that the equations are only for free space -- generalities can stay at Maxwell's equations I think. -- Tim Starling 06:14 Apr 1, 2003 (UTC)

To Stephen: either magnetic field can be called "B" accurately, or we move some or all of this page to magnetic flux density. I don't like this "really it's H but we'll just call it B" business. It's too confusing. -- Tim Starling 08:49 11 Jun 2003 (UTC)

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Tim: Too bad, life (and the English language) is confusing. =) The point is, that people are not entirely consistent in their terminology, and an encyclopedia should describe this. (In most cases, μ=1 so B=H and the point is moot. It's only when you're talking about both at once that you need two different names. In this case, Jackson goes by the historical names of magnetic field for H and magnetic induction for B. Purcell writes:

Even some modern writers who treat B as the primary field feel obliged to call it the magnetic induction because the name magnetic field was historically preempted by H. This seems clumsy and pedantic. If you go into the laboratory and ask a physicist what causes the pion trajectories in his bubble chamber to curve, he'll probably answer "magnetic field," not "magnetic induction." You will seldom hear a geophysicist refer to the earth's magnetic induction, or an astrophysicist talk about the magnetic induction of the galaxy. We propose to keep on calling B the magnetic field. As for H, although other names have been invented for it, we shall call it "the field H" or even "the magnetic field H".

And this is just Purcell's take. As you say, Griffiths calls H the auxiliary field, and Jackson (the god of electromagnetism) uses the historical names only when he has to distinguish B and H.

- Steven G. Johnson

At no point does Purcell say that B is formally, technically, or more accurately called magnetic induction. We are not bound by historical nomenclature, and the historical terms are not a priori "correct". We no longer refer to refractive index as "refrangibility" or to Uranus as "George's star". Common usage is what goes in dictionaries, historical usage is for the history books. This is my point: if B is magnetic field in common usage, then that definition is as "correct" as any other. But if, as you claim, B is "more accurately" called magnetic induction, it would be inappropriate to write an entire article referring to B as the magnetic field. The current situation is confusing in that we claim that the entire article is inaccurate. A student learning the material wishes to hold accurate information in their head, therefore every time they see "magnetic field" on this page, they will be distracted by a little mental note telling them that this usage is not to be trusted. -- Tim Starling 00:14 12 Jun 2003 (UTC)

Tim, people aren't consistent in their usage, and that sucks, but both usages need to be reported; describing usage is not the same thing as describing "correctness." On the one hand, the term magnetic induction is a historical one for B (a fact that would arguably be worth mentioning by itself), and it remains in present day usage when people want to disambiguate B and H (e.g. in Jackson, one of the most respected advanced electromagnetism texts, but also in 2003 physics journal articles, as a quick literature search will tell you). On the other hand, many many people (including physicists and Jackson himself) call B the magnetic field, especially when μ=1. - Steven G. Johnson

The article looks good now. Thanks. -- Tim Starling 01:04 12 Jun 2003 (UTC)

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"...the magnetic field is the field produced by a magnet." Straightforward enough, and there is a nice handy link to magnet. However, in the magnet article, I'm told that a "magnet is an object that has a magnetic field". That's not useful! That's just frustrating in a very tired and cliche manner. Could someone make one of these articles more primitive than the other? Suggestion: Make it clear to a non-physicist like me why an electron does not count as a magnet (or why the force field associated with an electron is not a magnetic field, if you prefer that point of view). (Okay, so an electron only has one pole. But having two poles can't be part of the definition of "magnet", else the article on magnetic monopole makes no sense at all and someone should fix that.)

It's certainly not clear to me why an electron does not count as a magnet. Steven may well disagree -- he has some funny ideas about classical limits and pseudovectors. But an electron has two poles. It can be modelled as a very small current loop. It even has angular momentum.

Pfftbt. I agree that an electron is a magnet; it has a magnetic moment, after all. But it's not a classical magnet, since its moment is not a vector (and various other quantum funniness). (I don't know what you mean by the electron having "only one pole", though; it's a quantum dipole, after all.) —Steven G. Johnson

Speak of the Devil...  :) -- Tim Starling 00:20, Dec 18, 2003 (UTC)

Note also that, as soon we have things like μ and ε (as in the Wikipedia Maxwell's equations), one is talking about a macroscopically averaged field, as opposed to the rapidly-varying microscopic field generated by individual particles. Sophisticated textbooks like Jackson are careful to distinguish the two (Jackson even goes so far as to use different symbols—lower-case letters—for the microscopic fields). —Steven G. Johnson 05:07, 18 Dec 2003 (UTC)

A magnetic force is that force caused by moving charges. Whenever I say that, of course, someone has to add "or spin", which may be true but I happen to think it's pretty irrelevant at your level. -- Tim Starling 22:24, Dec 17, 2003 (UTC)

(Or by a changing electric field—this was Maxwell's big contribution, after all, and leads to wave propagation in vacuum.) Anyway, the historical understanding of the magnetic field came first from the Lorentz force law, and only later was an independent "physical existence" attributed to the field in its own right, and this is a reasonable pedagogical practice as well. (Freshman physics courses typically talk about the effects of magnetic fields before describing how they are generated, which is more complicated.) —Steven G. Johnson 00:13, 18 Dec 2003 (UTC)

Good point about the changing electric fields. -- Tim Starling 00:20, Dec 18, 2003 (UTC)

velocity with respect to what ?

In the equation

<math>

\mathbf{F} = q \mathbf{v} \times \mathbf{B} <math>

what frame of reference is v measured with respect to? If any inertial frame will do, then v can take on arbitrary values, which would change F and therefore the acceleration applied to the charged particle, which seems absurd. Is v measured with respect to the magnetic field flux lines? Is so, what does that really mean?

If an answer to this question is added to the article, perhaps the article on the Lorentz force should be updated to also include the answer or to point to this article. MichaelMcGuffin 21:36, 30 Aug 2004 (UTC)

Any inertial frame will do. When you change inertial frames of reference, of course, not only v but also B and E will transform ... and yes, the force will transform, too, according to relativity. (This is in contrast to the original conception of electromagnetism in Maxwell's equations, which did indeed postulate a "preferred" inertial frame, that of the ether.) Indeed, you can transform to a frame of reference where v is zero, and thus the magnetic force is zero...but in this frame of reference, there will generally be a non-zero electric-field force. (A famous thought-experiment along these lines shows how, in relativity, electric and magnetic fields are two aspects of the same thing.) —Steven G. Johnson 02:14, Aug 31, 2004 (UTC)

Thanks, I see now that a sketch of a thought experiment has been added to the article. Something still confuses me though. In the thought experiment, call the first observer A (i.e. the observer that is "stationary") and the second observer B (i.e. the observer moving with the lines of charge). The current description points out that, from A's point of view, B's clock ticks more slowly, thus A perceives the net force F_A between the lines of charge as weaker, i.e. F_A < F_B. This weakening corresponds to the magnetic field that A perceives, attracting the moving lines of charge and opposing the repulsive electric force. However, couldn't we also change our perspective to that of B, and say that from B's perspective, A's clock ticks more slowly, thus we should expect F_B < F_A ? Surely there's something basic about special relativity that I don't understand. We can't have both F_A < F_B and F_B < F_A. MichaelMcGuffin 18:01, 8 Dec 2004 (UTC)

Isn't that the classic "twin paradox" of special relativity? I do not know how to resolve that paradox (I hear that general relativity is necessary) but both inertial observers see the other's clock as ticking more slowly than their own. How can that be? However, the two lines of charge are moving along only with observer B. You can think of B and the two lines of charge are stationary and *all* that B observes is the electrostatic repulsion of the lines of charge. Since observer A is moving relative to B and the parallel lines of charge, that observer will not see it the same as B. r b-j 18:15, 8 Dec 2004 (UTC)

(General relativity is not required to understand the twin paradox. There are various ways to show conclusively what the observed result would be, but one of the simplest is to imagine keeping the twins in constant communication by having them send radio pulses of their respective clocks towards one another. At the end of the trip, they compare their cumulative "clock" counts, and you can see that they both agree that the stationary one is older. French (Special Relativity) has a simple discussion of this. —Steven G. Johnson 21:29, Dec 8, 2004 (UTC))

It's funny, because while the "moving" twin is at a constant velocity, there is no sense that he is moving and the other is stationary. How do they view each other's clock during that period? It's only that the twin that goes to the far away planet and returns younger relative to his brother, is experiencing acceleration over the stationary twin, that you can differentiate them. I didn't think that SR had anything to say about acceleration (other than the normal Lorentz transformation in SR) whereas GR has a lot to say about acceleration (and gravitation). Anyway, my 28 year old "Elementary Modern Physics" textbook says literally that the paradox is explained by use of GR (without explaining it). Steven, could you translate that French explanation and put it in the English version of Special relativity? r b-j 22:19, 8 Dec 2004 (UTC)

Let me clear up two confusions. First, you're right that the fact that one observer has to accelerate at some point is the key difference between them — one observer does not remain in a single inertial frame of reference. However, you don't need general relativity to explain what happens, because you can make the acceleration itself a negligible fraction of the trip (and even with acceleration, you can still use special relativity as long as you describe the acceleration from the rest frame...you only need GR if you want to make the laws of physics have the same form in the accelerated frames). Second, A. P. French is the name of the author (a former MIT professor who wrote many physics textbooks in the 60's); the explanation itself is in English. Many other modern textbooks on special relativity contain a similar explanation (e.g. Basic Concepts in Relativity by Resnick and Halliday). Further, French (1968) writes:

One last remark. It has been argued by some writers that an explanation of the twin paradox must involve the use of general relativity. The basis of this view is that the phenomena in an accelerated reference frame (including the behavior of a clock attached to such a frame) are regarded in general relativity as being indistinguishable, over a limited region of space, from the phenomenon in a frame immersed in a gravitational field. This has been interpreted as meaning that it is impossible to talk about the behavior of accelerated clocks without using general relativity. Certainly the initial formulation of special relativity, although it leads to explicit statements about the rates of clocks moving at constant velocities, does not contain any obvious generalizations about accelerated clocks. And, as Bondi has remarked, not all accelerated clocks behave the same way. The clock consisting of a human pulse, for example, will certainly stop altogether if exposed to an acceleration of 1000g — in fact, a mere 100g would probably be lethal — whereas a nuclear clock can stand an acceleration of 10¹⁶g without exhibiting any change of rate. Nevertheless, for any clock that is not damaged by the acceleration, the effects of a trip can be calculated without bringing in the notions of equivalent gravitational fields. Special relativity is quite adequate to the job of predicting the time lost. It had better be, for (as Bondi has facetiously put it), "it is obvious that no theory denying the observability of acceleration could survive a car trip on a bumpy road." And special relativity has amply proved itself to be a more durable theory than this.

When I first took a course in special relativity, some years ago, I distinctly remember my professor saying that the notion that general relativity was required to describe the twin paradox had been disproved years ago. —Steven G. Johnson 22:54, Dec 8, 2004 (UTC)

that must have been some time ago. anyway i'm looking at http://www.sysmatrix.net/~kavs/kjs/addend4.html as well as twin paradox here on wiki. i understand this stuff a lot less than signal processing and FFTs. not sure if you would say the same :-) r b-j 04:51, 9 Dec 2004 (UTC)

This twin paradox stuff is interesting, but it doesn't seem to address my original question. In the thought experiment I was asking about, neither observer ever accelerates or changes direction, so there is never any change of "simultaneity planes" as illustrated in the twin paradox article. Each one observes a net force, F_A and F_B, between the lines of charge, according to their frame of reference. Each observer's clock ticks more slowly from the other's point of view. If we want to conclude that F_A < F_B, and not F_B < F_A, I think there's some missing reasoning or logic that should be added to the explanation of the thought experiment. Or, at least, could someone add a reference to a textbook or academic paper/article that explains the thought experiment in more detail? Thanks. MichaelMcGuffin 18:50, 9 Dec 2004 (UTC)

this following is a quantitative expression of that thought experiment. i think the twin paradox is applicable. in addition i do not see any of this "F_A < F_B, and F_B < F_A" conclusion that you have brought up. think about perceived acceleration in a direction that is perpendicular to the lines and on the same plane for both observers. i do not think you get a_A < a_B, and a_B < a_A. i think you only get a_A < a_B . the observers are not qualitatively in the same situation. both can observe the other as moving and themselves as stationary, but both do not observe the lines of charge in the same way because one is moving relative to the lines of charge and the other is not. r b-j 21:45, 9 Dec 2004 (UTC)

The classical electromagnetic effect is perfectly consistent with the lone electrostatic effect but with special relativity taken into consideration. The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of lambda) and some non-zero mass per unit length of rho separated by some distance R. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static non-relativistic outward acceleration for each infinite parallel line of charge would be:

a = F / m = (1/(4*pi*epsilon_0)*2*lambda^2/R)/rho

If the lines of charge are moving together past the observer at some velocity, v, the non-relativistic electrostatic force would appear to be unchanged and that *would* be the acceleration an observer traveling along with the lines of charge would observe.

Now, if special relativity is considered, the in-motion observer's clock would be ticking at a relative *rate* (ticks per unit time or 1/time) of sqrt(1 - v^2/c^2) from the point of view of the stationary observer.   Since acceleration is proportional to 1/time^2, the at-rest observer would observe an acceleration scaled by the square of that rate, or by (1 - v^2/c^2), compared to what the moving observer sees. Then the observed outward acceleration for the stationary observer would be:

a = (1 - v^2 / c^2) * (1/(4*pi*epsilon_0)*2*lambda^2/R)/rho

or

a = (1/rho)*( (1/(4*pi*epsilon_0)*2*lambda^2/R)

- (v^2 / c^2)*(1/(4*pi*epsilon_0)*2*lambda^2/R) )

The first term is the electrostatic force (per unit length) outward and is reduced by the second term, which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors). The electric current, i_0, in each conductor is

i_0 = v*lambda

and 1/(epsilon_0 *c^2) is the magnetic permeability

mu_0 = 1/(epsilon_0*c^2)

because c^2 = 1/(mu_0*epsilon_0)

so you get for the 2nd force term:

(v^2/c^2)*(1/(4*pi*epsilon_0)*2*lambda^2/R)

=  mu_0/(4*pi) * 2*i_0^2/R

which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by R, with identical current i_0.

ugly math symbols

I really don't like the ugly <math>\mathbf{B}<math> in the text. Why not B for vector, or B for scalar. Much cleaner. dave 04:04, 6 Jan 2005 (UTC)

personally, i think it's cleaner if precisely the same symbol (except possibly smaller) that is used in the math equations are used in the text. whether it's <math>\mathbf{E}<math> or <math> E \ <math>. i think it's particularly cleaner if the Tex math is used for greek and exponents and subscripts and other special symbols than the kludge that many use. <math> x^y \ <math> vs. x^y . r b-j 17:36, 6 Jan 2005 (UTC)

Your point still has some validity, but don't cheat. Do it right—in text it should be x^y with italic symbols. A serif font would make it look better; I don't know if there's an acceptable way to do that in Wikipedia articles. The disparity in size often makes the Tex math look weird when written inline (the bottom of your E letters descending below the line is real ugly). Gene Nygaard 18:48, 6 Jan 2005 (UTC)

Depends on your screen resolution. I turn on "always render PNG" and it looks fine for me. Remember that you aren't the only person who views it. - Omegatron 02:41, Apr 26, 2005 (UTC)

moving with respect to what?

"moving electric charges (electric currents) that e"

One thing I guess I never learned. Because of relativity, don't we need to specify what the particles are moving relative to? Do they not see each others magnetic field if they are not moving relative to each other? I guess they would behave as if they were only electrostatically repelling or attracting each other? Oh man, now I have confused myself... - Omegatron 23:41, Apr 25, 2005 (UTC)

This is why the presence of a magnetic field depends upon your frame of reference. Consider a charged particle moving at a constant velocity (i.e. in an inertial frame). If I am in an inertial frame of reference where the particle is moving with respect to me, I see a magnetic field. If I am in the same inertial frame as the particle, so that it is at rest with respect to me, I see no magnetic field. In general, when you change frames of reference, magnetic and electric fields get mixed up (together, they form a rank-2 tensor). —Steven G. Johnson 02:04, Apr 26, 2005 (UTC)