Talk:Magnetic field

New comments at the bottom please

The only purpose of this header is to move the CONTENTS box to the top

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 $\epsilon ,\mu$ instead of $\epsilon _{0},\mu _{0}$ ? 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)

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)

"...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

\mathbf {F} =q\mathbf {v} \times \mathbf {B}

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)

But the current in a wire does not depend on the reference frame. It (and therefore the magnetic field) is uniquely determined by the relative velocity of the electrons and ions in the wire (which is unrelated to the velocity v in the Lorentz force). Also, the resultant electric field of the wire is obviously zero. So the definition of the Lorentz force is indeed ambiguous unless one specifies what v is referred to. I would think that this should be the center of mass of the current system producing the magnetic field, i.e. approximately the frame where the ions in the wire are at rest (see also my website http://www.physicsmyths.org.uk/#lorforce in this respect).--Thomas

Of course the current in a wire depends on the reference frame—it's electrons moving! These issues were resolved in physics in the early 20th century, and the resolution is part of the foundation of all of modern physics. I can explain in great detail if anyone asks, but I warn you the answer is not short. -- SCZenz 16:58, 5 November 2005 (UTC)[reply]

No, the total current is the sum of the currents due to the negative and positive charges and thus independent of the reference frame (if you are comoving with the electrons you are then moving relatively to the positive charges, which results in the same current). -- Thomas

Current is not a concept limited to wires. What of the current due to a single charged particle, or many charged particles, with no balancing charges? There the current must depend on the reference frame. In fact, in electrodynamics, under Lorentz transformations, current and charge transform into each other as part of the same four-vector. See, for example, J. David Jackson's Classical Electrodynamics. -- SCZenz 17:27, 6 November 2005 (UTC)[reply]

Obviously, the concept of a current is not limited to wires (in general not even to charges). The question is whether these 'bare' currents (consisting only of unbalanced charges) produce any magnetic field. But I think this question may go too far for this article. Fact is that the article uses the example of a current in a wire as an illustration, and in this case the current (and hence the magnetic field strength) does not depend on the reference frame. The velocity v in the Lorentz force requires therefore a physical definition. --Thomas

You're simply wrong. The mathematics of proving it in the specific case of a current loop would be very hard, but relativity is fundamental to electrodynamics; velocities, magnetic fields, and currents all transform in a way that keeps the equation correct in any frame. In absolute generality. To argue this further, you would need to know electrodynamics in some detail; I've cited a book above, which is the graduate text for almost every university in the U.S. (and likely abroad, although I don't know for sure). If after reading it you still disagree, I can help get you in touch with Professor Jackson. -- SCZenz 18:01, 6 November 2005 (UTC)[reply]

To clarify, here's an expanded version of the equation in the presence of an electric field:

\mathbf {F} =q\mathbf {v} \times \mathbf {B} +q\mathbf {E}

There is only one reference frame in which there is no electric field, the one where the wire was stationary, and it is in this frame that F=qvxB holds. In other frames, the velocity transforms, and the B field transforms into an E field, but if you use the equation above you still get the same force. Does that help? -- SCZenz 18:10, 6 November 2005 (UTC)[reply]

So you are suggesting then also that v in F=qvxB (which is the formula given in the article for the Lorentz force) refers to the stationary wire? -- Thomas

In this example, if you leave off the qE, yes it refers to the stationary wire. If you leave in the qE, it could refer to anything; the real point is that in the frame with the stationary wire, E=0, whereas in the others it doesn't. -- SCZenz 06:25, 8 November 2005 (UTC)[reply]

Well, this should answer then the original question of this topic. I have allowed myself therefore to add a corresponding statement to this effect in the 'Definitions' section of the article.--Thomas

First mention of the thought experiment

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)

Stop thinking about the thought experiment, talk of special relativity and the twins

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)

A lot of the trouble here is that the accelerated twin is assumed to have set up a bunch of clocks synchronized within his own reference frame. They are synchronized by exchanging light signals with his clock (and/or each other). When he suddenly changes velocity he finds that these clocks are way out of synch, the difference being proportional to their (signed) distance from him. The "paradox" always results from comparing one clock with a set that are all synchronized within an inertial frame. The "traveling" twin cannot compare a "stationary" clock with his; he compares it with a sequence of clocks he has synchronized beforehand (but after he was in motion); comparisons are done between clocks instantaneously in justaposition. General relativity is not needed for the analysis. Pdn 13:47, 28 July 2005 (UTC)[reply]

You've got it all right, I think. But it is actually quite a common mistake to make among non-GR-specialist physicists. -- SCZenz 15:44, 28 July 2005 (UTC)[reply]

now, back to the thought experiment, which is messed up in its current form

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)

For reference, here is the thought experiment in its current form, which I believe is wrong: "A thought experiment one can do to show this is with two identical infinite and parallel lines of charge having no motion relative to each other but moving together relative to an observer. Another observer is moving alongside the two lines of charge (at the same velocity) and observes only electrostatic repulsive force and acceleration. The first or "stationary" observer seeing the two lines (and second observer) moving past with some known velocity also observes that the "moving" observer's clock is ticking more slowly (due to time dilation) and thus observes the repulsive acceleration of the lines more slowly than that which the "moving" observer sees. The reduction of repulsive acceleration can be thought of as an attractive force, in a classical physics context, that reduces the electrostatic repulsive force and also that is increasing with increasing velocity. This pseudo-force is precisely the same as the electromagnetic force in a classical context."GangofOne 07:59, 28 July 2005 (UTC)[reply]

The problems: First it says "lines of charge having no motion relative to each other", but then you say they experience "repulsive force and acceleration", so they ARE moving relative to each other? Second, an infinite line of CHARGE against another infinite line of charge will feel and infinite force, but it also has an infinite mass, so who knows what the acceleration will be. The force will be infinite for either observer. You say below a = F / m = (1/(4*pi*epsilon_0)*2*lambda^2/R)/rho , if this is a force; but it is force /(mass/lenght), so it's acceleration / length, whatever that is. Jumping down a few paragraphs....GangofOne 07:59, 28 July 2005 (UTC)[reply]

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 repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance $R\$ ) for each infinite parallel line of charge would be:

a={\frac {F}{m}}={\frac {{\frac {1}{4\pi \epsilon _{0}}}{\frac {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 because of time dilation. 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 of the two infinite lines as viewed by the stationary observer would be:

a=\left(1-v^{2}/c^{2}\right){\frac {{\frac {1}{4\pi \epsilon _{0}}}{\frac {2\lambda ^{2}}{R}}}{\rho }}

or

a={\frac {F}{m}}={\frac {{\frac {1}{4\pi \epsilon _{0}}}{\frac {2\lambda ^{2}}{R}}-{\frac {v^{2}}{c^{2}}}{\frac {1}{4\pi \epsilon _{0}}}{\frac {2\lambda ^{2}}{R}}}{\rho }}={\frac {F_{e}-F_{m}}{\rho }}

The first term in the numerator, $F_{e}\$ , is the electrostatic force (per unit length) outward and is reduced by the second term, $F_{m}\$ , which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors).

jumping to here. The above shows you have an understanding of what's wrong, but missed it. The '2 lines of charge' is different than '2 current carrying conductors' (overall charge is 0). What your analysis ends up with is the result for 2 current carrying conductors.GangofOne 07:59, 28 July 2005 (UTC)[reply]

The electric current, $i_{0}\$ , in each conductor is

i_{0}=v\lambda \

and ${\frac {1}{\epsilon _{0}c^{2}}}$ is the magnetic permeability

\mu _{0}={\frac {1}{\epsilon _{0}c^{2}}}

because $c^{2}={\frac {1}{\mu _{0}\epsilon _{0}}}$

so you get for the 2^nd force term:

F_{m}={\frac {v^{2}}{c^{2}}}{\frac {1}{4\pi \epsilon _{0}}}{\frac {2\lambda ^{2}}{R}}={\frac {\mu _{0}}{4\pi }}{\frac {2i_{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}\$ .

Thought experiment needs to be reworded and analysis to fit. I leave it you, since I see you are competent. GangofOne 07:59, 28 July 2005 (UTC)[reply]

i'm still not clear as to what to fix without bringing in the quantative analysis. (is that what you want me to do?) even if there is outward acceleration between the two lines of charge, there is an instant of time where the relative velocity is zero (therefore not moving relative to each other). it is only this instant of time that i was referring to in the thought experiment.

two infinite lines of charge do exert an infinite force on each other, but the force per unit length is finite and if the mass per unit length is also finite, then the outward acceleration is determinable. no?

let's be specific about what needs to be fixed. i'm happy if it is fixed. r b-j 03:26, 1 August 2005 (UTC)[reply]

I have a number of comments and unclarities, maybe some are my own inadequacy of understanding. I will continue anyway. What you say directly above about the instanteaous acceleration, if it is infintesially short moment, what relevance does the speed of clock ticks? That would only apply in a finite interval of time. Anyway, my original idea was that these infinite lines of charge as an example would be better as an example as two parallel conductors, which I think your analysis is really referring to. But then the next question is , does this examplify that which it is supposed to be an example? I'm still thinking about this. But consider this: let's say we have a long but finite pair of lineal charges, just to work around the red herring of the the infinites involved. Assume the charged wires are contrained to only move away from each other, not arbitrarily freely. A force meter is put betweem them and reads some value. (The force is related to the acceleration it WOULD experience if free.) So another observer flies by in a differnt frame of reference and reads the same number on the meter. So, where's the relativity come in? GangofOne 05:26, 1 August 2005 (UTC)[reply]

i dunno how Lorentz transformation of special relativity would deal with the numerical force meter but i do know that, assuming the motion is in the x direction, Lorentz transformation does not change the perspective of distance in the y or z directions but does change how one observer views the other's flow of time and that the lengths in the x direction is changed. when the two lines are infinitely long, then they remain infinitely long after the application of length contraction.

it's just a thought experiement, not a real experiment. in a real experiment, to measure acceleration just by looking at position, we would need snapshots at at least 3 instances of time. in the thought experiment, we know there is this concept of instantaneous acceleration and we are asking what it would be at a single instance of time.

the issue about charge neutral conductors vs. these lines of charge can be dealt with by just thinking about it a little. it does not negate the physics of the thought experiment at all. the charge neutral conductors have a net magnetic field, but no net electrostatic field. the lines of charge have both electrostatic field and electromagnetic field for the "stationary" observer that observes the lines of charge moving past him/her. the "moving" observer that is moving alongside the two lines of charge observes no electromagnetic effect. it is simply electrostatic for him/her. i don't want to do this thought experiment with charge neutral conductors, but i can still use the results of current in conductors as a basis for determining the classical magnetic field. r b-j 05:07, 2 August 2005 (UTC)[reply]

ugly math symbols

I really don't like the ugly $\mathbf {B}$ 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

\mathbf {E}

or

E\

. 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.

x^{y}\

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)

According to my corrections instead of all this mumbo jumbo with all these letters and stuff, why dont you just say R=B+C?

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)

'O', the bottom line is that magnetic effects are a result of special relativity. That is, if you start with Coulomb's Law and then impose Lorentz invariance (or is it covariance?), you get magnetic effects. It turns out that there is an analog to magnetism in gravity - so called gravito-magnetism. I think it is also called 'frame-dragging'.Alfred Centauri 12:39, 2 September 2005 (UTC)[reply]

field line flow? field shells in open space?

I've added a little item about the orientation markers vs actual field flow. I've been unable to find anything that says if the magnetic field lines actually move or if they are simply static lines through space. (The arrow is just a reference to mark field orientation, as far as I've been able to determine.)

They don't move. In fact, they don't even exist, see below. -- Tim Starling 12:32, Jun 24, 2005 (UTC)

I'm not a professional physicist by any means, but I've also never determined if the field actually exists as separate concentrated lines/shells as iron filings demonstrate, or if the lines form only as a RESULT of the iron filings being there, building up into thicker lines as more iron filings enter the field. Is a magnetic field in empty space "ridged" as the filings suggest, or is a field in open space simply an even undifferentiated gradient from strong to weak? DMahalko 09:31, 24 Jun 2005 (UTC)

Field lines are just diagrammatic. The number drawn in any given diagram is arbitrary, the field is smooth, not ridged. Lines are just one way to show the direction and magnitude of a vector field at every point on a plane. Iron filings tend to mimic the shape of diagrammatic field lines, because adjacent lines of iron filings tend to repel each other, and because grains that are in contact tend to line up end-to-end with unlike poles touching. -- Tim Starling 12:32, Jun 24, 2005 (UTC)

Rotating magnetic field

I do not think that long quotes on Tesla's philosophy are appropriate for this article. -- SCZenz 18:13, 27 July 2005 (UTC)[reply]

I agree. I'm not sure if anything beyond the first paragraph in that section is appropriate for this article. Salsb 18:19, 27 July 2005 (UTC)[reply]

Ok, I'm paring that stuff down now... -- SCZenz 18:27, 27 July 2005 (UTC)[reply]

The people should be in the section. JDR 18:30, 27 July 2005 (UTC)[reply]

Why? The discussion strikes me as a historical interlude about an engineering application. Shouldn't it be in say an article about the history of motors as opposed to a basic article on magnetic fields? Salsb 18:37, 27 July 2005 (UTC)[reply]

It should be with the topic. The RMF concept is redirected here, so the history should be with that (as it is the history of the RMF). JDR 18:49, 27 July 2005 (UTC)[reply]

Yeah, people really aren't appropriate here. This article is about the concept of the Magnetic field, in great generality. My objection to your additions, Reddi, is that you are going into great detail on an issue you care about in particular, when no other sub-topic of magnetic field has such detail. A separate article really is more appropriate. -- SCZenz 18:41, 27 July 2005 (UTC)[reply]

SCZenz, people were in the see also section before. A separate article (which would be ideal) has been redirected here over and over again (lately by Salsb, WMC; earlier by Starling). JDR 18:49, 27 July 2005 (UTC)[reply]

People are appropriate for the see also section, not for the main article. As for the redirects, I think perhaps the issue is that the history of rotating magnetic fields, and the engineering applications, should be in separate places. In particular, I rather suspect that electrical motors and power plants have their own extensive articles already. But a separate article that brings it all together, if it's important to you to organize it like that, wouldn't be the end of the world to me. -- SCZenz 18:56, 27 July 2005 (UTC)[reply]

I removed the history except the two main, Ferriari and Tesla.

As for the redirects, the concept of "rotating magnetic field" is redirected here, and the engineering applications and it's history should have (in the least) a summary here.

A separate article on the RMF should exist. JDR

There is a place where this would fit in nicely: Electric_motors#AC_motors. As you point out, this is an important engineering principle, and your text seems to be part of the history of motors. Salsb 18:54, 27 July 2005 (UTC)[reply]

The concept redirects here. It wasn't even mentioned before, which it should have been, and it should have a brief summary, in the least. JDR 19:09, 27 July 2005 (UTC) (eg., The concept redirects here. The history of the concept should be at the article where the concept is.)[reply]

Since there is minimal history in this article, I disagree, I would include it in a nice history of engineering article but its not too important Salsb 19:28, 27 July 2005 (UTC)[reply]

The one other objection I have to this section, is the inclusion of the Earth's magnetic field from dynamo theory. This is about convection and rotation in fluids creating a magnetic field not about a rotating magnetic field. So if it should be placed in this article, which I am not sure it should be, it shouldn't be in this section Salsb 19:28, 27 July 2005 (UTC)[reply]

The Earth is one big dynamo (core rotor; atmosphere stator), directly related to rotating magnetic fields. JDR 19:41, 27 July 2005 (UTC)[reply]

As written it does not follow well, since the earth does not have a rotating magnetic field. There are articles on the Earth's magnetic field, on dynamos and on the dynamo theory, so I don't think there needs to be an additional discussion here, as opposed to a see also. Salsb 19:57, 27 July 2005 (UTC)[reply]

Salsb, you have to understand that you are talking to someone who thinks that the Earth's magnetic field is rotating because "it goes from south to north. THIS movement is the rotation of the field. It's rotating back and forth ... from the south pole to the north and back again. That's why compasses work." (see Talk:Rotating magnetic field.) —Steven G. Johnson 02:38, August 5, 2005 (UTC)

Stevie ... contrary to your negative implication via the link ... the inner core and outer core are rotating (akin to a rotor and stator) conducting (north-south) the magnetic field of the earth ... "THIS movement is the rotation of the field" and this action does make a compass work. JDR

Rotating magnetic field proposal

To find a way out of the impasse thus far, I propose that we have Reddi put his material for rotating magnetic fields into rotating magnetic fields, in place of the redirect. The concept isn't important enough for the space taken up in magnetic field, but it is true that a rotating magnetic field is a concept, independent of its applications, and that it could be useful. It might end up being a longer article than I would have written, but I don't think of that as a realy problem. But we certainly need a consensus somehow... thoughts? -- SCZenz 20:00, 27 July 2005 (UTC)[reply]

A rotating magnetic field in of itself is not special, its useful though its applications. I suggest that this text be added to the Motors article under AC motors, since there is a small history section there already which could be expanded. Salsb 20:05, 27 July 2005 (UTC)[reply]

What I'd consider even better, if Reddi would take the text on rotating magnetic fields and expand it into an article on the history of electric motors, that would be very cool indeed. Although admittedly a fair amount of work. Salsb 20:09, 27 July 2005 (UTC)[reply]

That sounds interesting ... I have been aquiring much of Edison's patents and Tesla's patents (as well as others). Alexanderson (sp?) built huge machines ...and this would include the early work in electrostatic machines. I'll kick this history of electric motors around in my head abit. Sincerely, JDR 18:19, 8 November 2005 (UTC)[reply]

I think the Magnetic Field article as it stands now is OK We'll keep a short section on rotating fields, and leave rotating magnetic field as a redirect. Sound ok? -- SCZenz 23:18, 27 July 2005 (UTC)[reply]

More velocity relative to what

I just reverted the edit made assigning v a specific reference frame in F = qvxB, because it was not correct. I then made an edit that I hope clarifies the situation. Here's a fuller explanation, for reference.

F = qv x B is always the correct force due to the magnetic field in any reference frame. However, the magnetic field can turn into an electric field as you change reference frames, so this equation will not remain constant for a system that produces a B field with moving charges (i.e. a current loop). However F = q(E + v x B) does remain constant in all reference frames; i.e. the total electromagnetic force can be defined unambiguously. F = qv x B is only the correct electromagnetic force when E = 0. In the case of a current loop, this occurs in the frame where the loop itself isn't moving.

However, the first part of the definition section doesn't refer to any example; it's a general example for force on a particle due to a B field. It doesn't say where the B field is from, so it is the correct equation for the force due to the B field in whatever frame you choose—you just have to pick the particle velocity and B field for the same reference frame. Hope that helps. -- SCZenz 21:12, 8 November 2005 (UTC)[reply]

I don't agree that the addition I made by referring v to the rest frame of the wire was incorrect. Given that F=qvxB in the article it was correct (I thought you agreed with this above). The total electromagnetic force is not written out here, so it should not be referred to either. As it is now, it is only counter-intuitive and in fact inconsistent (note also in this context my further comment under Talk:Lorentz force#Should the Lorentz Force Include the Electric Field?).--Thomas

The article states that the force due to the magnetic field is qvxB, which is correct in any reference frame. It's just not the same force in every reference frame, because you don't have the same magnetic field in every reference frame. -- SCZenz 19:58, 10 November 2005 (UTC)[reply]

How should the magnetic field depend on the reference frame if it is produced by the current in a wire? Assuming the wire is overall electrically neutral, the electric field is zero (in all reference frames) and the total current (and hence the magnetic field) is independent of the reference frame as well, as it depends only on the relative drift velocity of the ions and electrons in the wire (which has nothing to do with the test charge velocity v).--Thomas

Charge and current transform into each other under Lorentz transformations. Thus the magnetic field and electric field produced from them change under Lorentz transformations. A mathematical discussion of this is at an advanced undergraduate or graduate level, and non-trivial, so if you want a detailed proof I would do far better to refer you to a textbook than to try to reproduce the explanation myself. I recommend reading J. D. Jackson's Classical Electrodynamics, chapter 11, particularly section 11.9. I believe Griffiths' electromagnetism textbook also has a discussion; both texts are in the reference section of the Magnetic field article itself. -- SCZenz 19:13, 11 November 2005 (UTC)[reply]

Consider this heuristic, handwaving argument. You say: "Assuming the wire is overall electrically neutral, the electric field is zero (in all reference frames)...", but is it? If the e- are moving w.r.t. the wire (the positive charges), then there is a relativistic length contraction, and the charge per length of the e- is no longer the same as the charge per length of the positive charges, so there is a "net charge" per length, (or , better stated, something that acts like a charge, that can cause forces, etc). It "is" the magnetic field. The magnetic field is a relativistic effect. (And that's why the E and B are mixed up together, and mutually change, depending on the motions of the charges and point of space. (This could be better expressed, but maybe it's helpful. Continue your questions, maybe we can generate some explanations suitable for the article.) GangofOne 20:17, 11 November 2005 (UTC)[reply]

To continue, you say: "and the total current (and hence the magnetic field) is independent of the reference frame as well," No the current depends on the ref. frame. If I am static with the wire and "see" the e- move, I "see" one value for the current. If I move with the e- so it is the wire that moves, I "see" the opposite value for the current. If I split the difference, and move half the speed, so that the wire is moving one way and the e- the other, I see no current. So the current is NOT reference frame independent.

The idea is that from any inertial frame of reference, current, E, B, v, ALL change, but is such a way that F=qE +qvxB is still true. GangofOne 20:39, 11 November 2005 (UTC)[reply]

I am afraid what you are saying above is incorrect in several respects:

First of all, you seem to forget that electrons and ions are oppositely charged, so if you change the rest frame from one to the other, the velocity changes direction but also the charge, i.e. the current does not change sign (contrary to what you stated above).

Secondly, it would be quite a funny wire if the charge density of the ions would become different from that of the electrons depending on the reference frame: this would mean that charge neutrality would not be fulfilled anymore and hence corresponding electric force fields would be set up along the wire in addition to the electric field that you propose to be created perpendicular to it. So overall you would have a net force depending on the reference frame, which is physically not acceptable.

Thirdly, one should bear in mind that a magnetic fields doesn't do any work on a charged particle as the associated force is always perpendicular to the velocity. This property should obviously not depend on the reference frame. However, the proposed relativistic 'charging' of the wire however should always create an electric field directed towards to or away from it, i.e. it is only perpendicular to the velocity if the charge velocity is parallel to the wire. In all other cases, the resultant field would do work on the charge and change its kinetic energy. Again this is physically not acceptable.

For clarification, just consider the case of a charged particle in the magnetic field of a wire from a kinematical point of view: in the rest frame of the wire the orbit of the particle is a circle around the corresponding field line (assuming the velocity v to be perpendicular to the magnetic field and the field as homogeneous i.e. the Larmor radius as sufficiently small). Now if you view this scenario from a different inertial reference frame moving with velocity U, the latter velocity will simply be superposed to the Larmor circle which the charge traces out in the wire's frame i.e. we will have a Cycloid motion, which is however just the result of the superposed linear velocity but not of an electric field. --Thomas

You're right that GangofOne made an error in the bit about moving along with the electrons above. Ignoring relativity, you'd be right that the current wouldn't change with reference frame. However, because the electrons are moving at a different rate than the protons, they are affected differently by Lorentz transformations. Thus current does transform into charge (a fact which, even if we explain it poorly, is cited in every sufficiently advanced physics textbook on the subject).

Your second argument that, when the wire becomes charged, there is an electric field along the wire is rubbish; a charged wire has an electric field that points radially outward, as is solved in elementary E&M textbooks (see, e.g., Halliday, Resnick, and Krane's Physics, Volume 2, section 29-5). This removes the new, unexplained force and the phantom work you were concerned about.

A suggestion: if you want to keep trying to understand what's going on, that's great—the charged wire obscures the physics principles a bit, which makes it interesting. But you're not going to succeed in overturning the learned consensus of physicists on an issue that's been settled for over half a century, whether you win an argument with a lowly grad student like me on Wikipedia or not. -- SCZenz 18:49, 12 November 2005 (UTC)[reply]

I would agree that Wikipedia is probably not the right place to overturn learned consensus, but on the other hand, the contributors should also have some responsibility regarding the factual correctness of the articles. In this sense I think the points I mentioned are at least worth considering here.

I could actually go much further and discuss the Lorentz transformation in the first place, but I think this would go too far off topic here (for anyone interested, see my web page http://www.physicsmyths.org.uk/lorentz.htm ).

In any case, I don't agree that my argument connected with the charge invariance is rubbish: if the charge densities of the negative and positive charges are different in the wire, charge invariance requires that there is a corresponding surplus charge external to the wire, or in other other words, the negative and positive charges are merely distributed over different lengths, with the total charge being identical and unchanged. Now this may not make a significant difference if the wire is much longer than the distance to the wire, but consider the opposite case where the distance d is much larger than the length of the wire: in this case, the electric field is basically that of a point charge and does not depend on the charge density of the wire i.e. a different Lorentz contraction for the electron and ion distribution would not result in a net charge (there would only be a second order effect decreasing like 1/d^4 and which could therefore be made arbitrarily small compared to the magnetic field (which goes like 1/d^2 for the short wire)). So in this case the Lorentz contraction would not yield the required electrostatic force when changing reference frames.

Another point that I find very questionable in corresponding derivations is the fact that reference frame of the moving charge (i.e. the one where the v in F=qvxB is zero) is apparently being treated as an inertial frame when in fact the particle is accelerated due to the Lorentz force in the wire's frame (if you treat for instance the earth as an inertial reference frame and assume that the sun rotates around it, you would obviously arrive at vastly incorrect results regarding the force between the two). So there is actually no additional electrostatic force required in the frame of the moving charge as this frame is not inertial but itself accelerated by the force on the charge in the wire's frame (an exact definition of 'wire's frame' is still missing as well by the way, not only here but also in some textbooks that I checked out). --Thomas

Basically, what you're point out is that treating a current loop in the context of relativistic electrodynamics is complex, and you have to be very careful if you care about all the details. But nobody does care--there are far more interesting tests of relativity and electrodynamics. That's why textbooks don't discuss it. -- SCZenz 19:22, 15 November 2005 (UTC)[reply]

Suggestion to merge articles

I strongly disagree with the suggestion to merge this article with magnetic flux density. These are two different, though related, concepts. For starters, the two have different units of measures (dimensional decompositions). Magnetic field is amps per meter, magnetic flux density is webers per square meter. -- Metacomet 15:06, 31 January 2006 (UTC)[reply]

The article magnetic field discusses in detail both fields B and H. Therefore, there is no reason to have a separate article discussing B. (The units of B and H are indeed different. But the units of B from "magnetic field density" and of B from the article "magnetic field" are the same, and this is exactly the same quantity). Yevgeny Kats 17:13, 31 January 2006 (UTC)[reply]

In my opinion, there should be an overview article that discusses both B and H, including high-level definitions, differences between the two, and the relationship between them. Then there should be two separate, more detailed articles, one article on each of the two fields. For historical reasons, and because of the way electromagnetics is often taught in introductory courses, there is a lot of confusion between B and H. It would be a shame if WP were to reinforce the confusion, instead of trying to clarify the distinction and explain where it comes from. Again, these ideas are simply my opinion. -- Metacomet 17:28, 31 January 2006 (UTC)[reply]

As a suggestion, the three articles might be called:

Magnetic field (overview) to give high-level overview
Magnetic field intensity to give detailed discussion of H
Magnetic flux density to give detailed discussion of B

-- Metacomet 17:32, 31 January 2006 (UTC)[reply]

I think it's not possible to discuss H separately from B. It is possible, however, to discuss B without discussing H. Therefore, I'd suggest to divide it to

"Magnetic field", which will discuss only B.
"Magnetic fields in matter", which will discuss B and H in matter.

Each article will include a major link to the other.

Currenly we have some random division of the content between the articles "magnetic field density" and "magnetic field"

Yevgeny Kats 18:47, 31 January 2006 (UTC)[reply]

I disagree. I prefer the approach that I suggested above. You can easily discuss either field by itself, as long as you have a definition that does not involve the other, which is quite easy to do. I don't really want to get into a huge argument about it though. You are entitled to your opinion, and I am entitled to mine. -- Metacomet 19:56, 31 January 2006 (UTC)[reply]

Question of validity - biot savart law

\mathbf {B} ={\frac {\mu _{0}}{4\pi }}{\frac {q}{r^{2}}}\mathbf {\hat {r}} \times \mathbf {v}

is this equation right, or should r hat and v be switched (in the cross product). Cause i just had a physics HW problem that contradicted. Still doing the HW, so don't have time to check up on anything. User:Fresheneesz

The Biot-Savart Law is usually applied in differential forms, like in the following two expressions:

\mathbf {dB} ={\frac {\mu _{0}}{4\pi }}{\frac {dq}{r^{2}}}\mathbf {v} \times \mathbf {\hat {r}}

\mathbf {dB} ={\frac {\mu _{0}}{4\pi }}{\frac {i}{r^{3}}}\mathbf {ds} \times \mathbf {\hat {r}}

This law is important in studying magnetic fields. May somebody please add more detail about that in the article? Thanks.
- Alanmak 23:35, 12 February 2006 (UTC)[reply]

About the picture at the beginning of the article

This picture looks nice. But there are a few things that could be improved:

Magnetic field lines should form closed loops.
It is a widely accepted convention to use "B" instead of "M" as the symbol for magnetic field.
As the positive and negative ends of the electric wire has already been indicated, it is not necessary to specify that the current is "DC". It would be better to use "I", which is the common symbol for electric current.

Would anybody like to modify that picture for a little bit? - Alanmak 23:35, 12 February 2006 (UTC)[reply]

New magnet-theory?

I have made new theory of magnet field?

Basic idea is that;

magnet-field is not actually non-matter field, but it is, small matter-particles flowing process.

You can read my theory at this link;

http://www.moheitulkuri.fi/w/flow.ppt

br. Heikki. 18.2.2006. Heikki 10:26, 20 February 2006 (UTC)[reply]