How do moving charges produce magnetic fields?

Question

I'm tutoring high school students. I've always taught them that:

A charged particle moving without acceleration produces an electric as well as a magnetic field.

It produces an electric field because it's a charge particle. But when it is at rest, it doesn't produce a magnetic field. All of a sudden when it starts moving, it starts producing a magnetic field. Why? What happens to it when it starts moving? What makes it produce a magnetic field when it starts moving?

nothing happens to the particle to make it produce a magnetic field as it starts moving: electric and magnetic field are components of the electromagnetic field, which is a single entity, similar to how energy and momentum are components of 4-momentum; in a charged particle's rest frame, the magnetic components vanish, as does its 3-momentum, and only the time-like ones (the electric field and the energy, respectively) remain — Christoph, May 21 '13 at 10:18
@Christoph: You used lot of new words I don't understand. I don't major in physics. Could you suggest something (simple) to read? — claws, May 21 '13 at 10:23
you'll need to read up on special relativity; if you wait for a bit, I'll expand my comment into a proper answer... — Christoph, May 21 '13 at 10:47
Related Veritasium/Minutephysics video: http://www.youtube.com/watch?v=1TKSfAkWWN0 — Qmechanic, Jul 10 '14 at 17:31
I highly recommend the work by Professor Oleg D. Jefimenko titled: "Causality, Electromagnetic Induction, and Gravitation: A Different Approach to the Theory of Electromagnetic and Gravitational Fields, 2nd ed." In this work the good professor solves Maxwell's equations and shows electric current and magnetic fields are not causal sources of each other but are a dual entity caused simultaneously from a common source. — Yokai, Apr 22 '18 at 11:39
A moving magnet will also have a electric field. If you accelerate a charged particle or magnet you will produce a electromagnetic wave, since the two fields are a manifestation of the same force. — mesompi, Jun 13 '20 at 21:31
Why without acceleration? If it moves with acceleration , then what happen? — SHYAMANANDA NINGOMBAM, Aug 28 '20 at 15:13

score 257 · Accepted Answer · answered May 21 '13 at 19:31

257

If you are not well-acquainted with special relativity, there is no way to truly explain this phenomenon. The best one could do is give you rules steeped in esoteric ideas like "electromagnetic field" and "Lorentz invariance." Of course, this is not what you're after, and rightly so, since physics should never be about accepting rules handed down from on high without justification.

The fact is, magnetism is nothing more than electrostatics combined with special relativity. Unfortunately, you won't find many books explaining this - either the authors mistakenly believe Maxwell's equations have no justification and must be accepted on faith, or they are too mired in their own esoteric notation to pause to consider what it is they are saying. The only book I know of that treats the topic correctly is Purcell's Electricity and Magnetism, which was recently re-released in a third edition. (The second edition works just fine if you can find a copy.)

A brief, heuristic outline of the idea is as follows. Suppose there is a line of positive charges moving along the $z$-axis in the positive direction - a current. Consider a positive charge $q$ located at $(x,y,z) = (1,0,0)$, moving in the negative $z$-direction. We can see that there will be some electrostatic force on $q$ due to all those charges.

But let's try something crazy - let's slip into $q$'s frame of reference. After all, the laws of physics had better hold for all points of view. Clearly the charges constituting the current will be moving faster in this frame. But that doesn't do much, since after all the Coulomb force clearly doesn't care about the velocity of the charges, only on their separation. But special relativity tells us something else. It says the current charges will appear closer together. If they were spaced apart by intervals $\Delta z$ in the original frame, then in this new frame they will have a spacing $\Delta z \sqrt{1-v^2/c^2}$, where $v$ is $q$'s speed in the original frame. This is the famous length contraction predicted by special relativity.

If the current charges appear closer together, then clearly $q$ will feel a larger electrostatic force from the $z$-axis as a whole. It will experience an additional force in the positive $x$-direction, away from the axis, over and above what we would have predicted from just sitting in the lab frame. Basically, Coulomb's law is the only force law acting on a charge, but only the charge's rest frame is valid for using this law to determine what force the charge feels.

Rather than constantly transforming back and forth between frames, we invent the magnetic field as a mathematical device that accomplishes the same thing. If defined properly, it will entirely account for this anomalous force seemingly experienced by the charge when we are observing it not in its own rest frame. In the example I just went through, the right-hand rule tells you we should ascribe a magnetic field to the current circling around the $z$-axis such that it is pointing in the positive $y$-direction at the location of $q$. The velocity of the charge is in the negative $z$-direction, and so $q \vec{v} \times \vec{B}$ points in the positive $x$-direction, just as we learned from changing reference frames.

answered May 21 '13 at 19:31

31

Thank you very much for telling me what I don't know instead of directly answering my question. Thank you for the book recommendation. Thank you for using such simple language to explain things new to me. Loved your answer :) – claws May 21 '13 at 19:40
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Wow! (+1) One of the legendary damn good answers to an apparently simple question, we see here from time to time. It reminds me that of the kinetic energy and pieces of clay of R.M. – Eduardo Guerras Valera May 22 '13 at 06:28
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I don't think that's the whole story. The electromagnetic field can only be reduced to an electric one in case of $P<0$. In case of $P>0$, your description breaks down and you'd be forced to consider magnetostatics the fundamental interaction that gets boosted to different frames... – Christoph May 22 '13 at 08:02
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@Christoph I'm not trying to get rid of the magnetic field, as that would involve transforming into the rest frame of the charges producing it, and clearly such a global frame will not exist for most current distributions. I'm transforming into the frame of the test charge, which can always be done and in which there is no effect of magnetism on the charge. – May 22 '13 at 17:33
@ChrisWhite: thanks for the clarification, I failed to understand the premise of your setup; a non-vanishing magnetic field will generally have a effect on the test charge, though (alignment of intrinsic magnetic moment); can that be modelled sensibly as well, eg by making the rest frame of the test charge rotate? – Christoph May 22 '13 at 18:50
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@Christoph Intrinsic spin - now that's a complication I'll admit to not having thought all the way through. – May 22 '13 at 19:28
Nice try, but you really should have copied Purcell's argument where he considers an electron current moving in a wire where the electric fields cancel one another in the lab frame, but don't in the electron's frame. – Larry Harson May 25 '13 at 00:47
@ChrisWhite I had a question about your answer: Shouldn't the space between current charge remain constant? I ask that because, from Bell spaceship paradox, my impression is that for 'stationary' observer who sees two bodies separated by a space 'x' accelerate to a velocity 'v', then he/she would not observe a difference in the gap between the bodies, although they themselves would contract. – mehfoos Jul 19 '13 at 23:36
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@user1218748 I have actually never heard of that paradox before. It seems to hinge on the subtleties of how the acceleration is implemented (specifically when it is stopped). Accelerating frames are tricky, especially when trying to extend them over long distances. Certainly you have to be careful, because your interpretation would mean length contraction cannot happen at all - just view all objects as collections of their atoms. In any event, this sounds like the basis for a separate question. – Jul 20 '13 at 01:09
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@ChrisWhite: Suppose your charge were approaching the current instead of moving parallel to it. What do you predict then? There's more to E&M than Coulomb + SR. Please review Jackson 12.2. – Art Brown Aug 06 '13 at 16:56
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FYI, Griffiths gives this a nice treatment as well in his book. – Mar 07 '14 at 03:47
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Perhaps I'm taking too literally your words, but I think that the sentence “Magnetism is nothing more than electrostatics combined with SR” is false. The easiest example that you cannot treat with electrostatics, is the one of an accelerating charge. As Feynman teaches in lecture 26, you can obtain the whole fields from Couloumb's law if you add the ideas that (1) the potential $(\rho , A_x,A_y,A_z)$ is a tetra-vector and (2) that it only depends on positions and velocities at the delayed time. – pppqqq Jul 10 '14 at 18:10
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Is there an analog for the other forces, eg: gravity? I don't see any reason why there shouldn't be but at the same time I'm not aware of anything like it. – alexdavey Apr 02 '16 at 23:22
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I'm afraid this answer is wrong. It repeats a myth. You cannot derive rotational magnetic motion from length contraction in a linear wire. See this question and the associated discussion. – John Duffield Aug 04 '16 at 15:35
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@JohnDuffield On the question you linked, it seems to me that the top voted answer agrees with this answer, or did I misread something? – Shufflepants May 30 '17 at 21:32
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180 votes but it's wrong and I am asking you to give your answer a major edit. The question whether magnetism can be wholly derived from electrostatics plus relativity has long been considered and it certainly cannot. One reason is that it will not tell you the effects of acceleration. Also, $\bf B$ doesn't merely contribute a force on a moving particle, it also acts as a source to $\bf E$ by induction. How do you explain an electromagnetic wave from electrostatics? It can't be done. Finally, whenever ${\bf E} \cdot {\bf B} \ne 0$ there is no frame in which $\bf B$ vanishes. – Andrew Steane Feb 26 '19 at 23:09
2
I'm layman, but this answer bothers me because of two things:
- The drift speed of a electron through a copper wire of cross-sectional area 3.00 x 10-6 m², with a current of 10 A will be approximately 2.5 x 10-4 m/s, a very low relativistic speed. The electric field will be in the velocity of light tough.
- You can also produce a electric field moving a magnet. You can even produce a electromagnetic wave you accelerate the magnet, the same behavior that a electric particle have.
Don't these questions are better answered with quantum mechanics and it's theory of fields?
– mesompi Jun 13 '20 at 22:11
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I just want to point out that it is a bit unclear why test charge q is required to move. The explanation makes sense (whether wrong or right) when test charge is not moving, unless I am missing something. – user1700890 Jul 11 '21 at 12:04
Why won't that explanation predict "gravitational magnetic field"? Replace charges with point masses and Coulomb repulsion with gravitational attraction; wouldn't you see the same effect? In fact, the effect of the masses spaced closer together would be compounded by the relativistic increase of the individual masses. – Michael Aug 24 '21 at 19:13
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Unfortunately, this answer is wrong. The idea that the magnetic field is merely the relativistic effect of electric field has been considered decades ago; I believe Feynman mentioned it somewhere for the purpose of explaining why it's wrong. Lorentz transformations explain why the effects electromagnetic field are consistent regardless of the coordinate frame. Also it would seem is toy examples that electrostatics + Lorentz is all that's needed to explain things, but no, generally speaking that's not a complete picture. – Michael Nov 18 '21 at 17:35
@EduardoGuerrasValera I'm interested in the kinetic energy and clay pieces thing you mentioned. Do you have a link? – Some Guy Feb 23 '23 at 19:58

score 34 · Answer 2 · edited Mar 14 '16 at 15:24

Electric and magnetic fields are what the electromagnetic field 'looks like' from a particular (inertial) frame of reference.

Take a charged particle: In its rest frame, it appears to generate an electric field only and no magnetic field at all. From a different frame of reference (in particular one in relative motion), we'll see the charge moving, thus a current which generates a magnetic field as well.

This does not mean that setting the particle in motion somehow flipped a switch within the particle - rather, it's an artifact of our choice of frame of reference: Observers in relative motion will measure different strengths of electric and magnetic fields the same way they measure different velocities and momenta.

There are however invariants of the electromagnetic field, i.e. things all observers can agree upon, and in particular $$ \begin{align*} P &= \mathbf {B}^2 - \mathbf E^2 \\ Q &= \mathbf E\,\cdot\mathbf B \end{align*} $$

Let's take a nonzero em field with $P,Q=0$, i.e. $\mathbf E^2=\mathbf B^2$ and $\mathbf E\perp\mathbf B\;.$ An example would be a plane electromagnetic wave, which will look like a plane wave for everyone.

Now, let $P\not=0$ but $Q=0\;.$ Then, we can find frames of reference where either the electric (in case of $P>0$) or the magnetic field (in case of $P<0$) vanishes. The rest frame of our charged particle would be such a one.

For more details, you'll need to look into the literature on special relativity.

So does that mean a charged particle produces no electric field in motion ? — Jdeep, Aug 04 '20 at 12:30
How can we subtract $B^2$ from $E^2$ when they have different units? — Asher2211, Apr 11 '22 at 07:04
@Asher2211 the units being used here are cgs units, which are not uncommon for theoretical electromagnetics. in SI Units the invariant is $\mathbf{B}^2-\frac{\mathbf{E^2}}{c^2}$ — Cyrus Tirband, May 05 '22 at 06:51

score 23 · Answer 3 · edited Mar 07 '14 at 13:32

Although Chris White’s answer to the question “Why Moving Charges Produce a Magnetic Field?” posted by a High School teacher (Claws) last year, was selected as the best answer, I think it contains several pitfalls. Chris White imagines a stream of positive charges flowing in the $+z$ axis direction, while a test charge $+q$ initially located at $(1,0,0)$ is moving in the opposite $(-z)$ direction with speed $v$. Next he intends to prove that when the observer locates himself in the frame of the moving test charge, he will see, in addition to the regular electrostatic Coulomb (repulsion) force acting on the test charge, an additional repulsion in the $+x$ direction whose origin is entirely relativistic. This happens, he says, because the original separation $Δz_0$ between the charges (when seen from the Lab rest frame) is now contracted to $Δz = Δz_0\sqrt{(1-v^2/c^2)}$ (The “famous” Lorentz contraction).

Consequently all the distances of the flowing charges to the test charge become smaller (as if the charge density increased) and, hence, Coulomb repulsions also increase. This excess of repulsion is the “illusory” magnetic force that the Lab observer sees when the test charge moves in the $–z$ direction with speed $v$.

In short: there is no intrinsic magnetic force. All is Coulomb force, seen from the Lab frame (pure electrostatic force), or seen from the moving charge frame (electrostatic plus more Coulomb repulsion). We can bypass here all the quantitative details which White also omits, but we cannot overlook the pitfalls:

First there is a verbal contradiction: to notice the contracted $Δz$, smaller than $Δz_0$, the observer must locate himself at rest with the charge $q$ (i.e., moving with the charge). But then at the end, White says that the new “anomalous force seemingly experienced by the charge” (i.e., the defined magnetic field), occurs “when we are observing it not in its own rest frame” (emphasis mine). So, what’s the deal? To predict the extra Coulomb (magnetic) force we have to adopt the frame of the moving charge. But to observe it we have to remain in the Lab frame, which is NOT the moving charge frame.
In the same vein there is a numerical pitfall: the new (contracted) charge separation Δz observed from the frame of the moving charge is calculated as $Δz=Δz_0\sqrt{(1-v^2/c^2)}$ where $v$, says White, is “$q$’s speed in the original frame”. He should have put not $v$ but $2v$, since the relative velocity between the charge stream going up, $v$, and the test charge going down, $-v$, is $v-(-v) = 2v$. So the contraction factor should be $\sqrt{1-4v^2/c^2}$.
Furthermore, if we use the heuristic strategy used by White, we reach a contradiction: Start with all charges at rest: the $z$ axis full of charges and the test charge at $(1,0,0)$. Call $Δz_0$ the separation between all charges at rest. Now allow the $z$ axis charges to move as before, with a speed $+v$. Already the Lab observer AND THE TEST CHARGE $q$, will see a contraction of the separation according to $Δz = Δz_0\sqrt{(1-v^2/c^2)}$. Hence by the same maneuvers as before, special relative must predict an additional “Coulomb” repulsion due to the compacted charge density. So the “magnetic” force, thus predicted, must act on the RESTING charge at $(1,0,0)$. And this is not observed. To the best of my knowledge, no current along the $z$ axis can ever produce a magnetic force on a resting charge at the origin.

In conclusion: contrary to what White says, magnetism is NOT JUST electrostatics plus special relativity. Such reductionistic view converts magnetism into a superficial play-game between frames of reference.

Nick Stauner: thanks for the editing but the relativistic equations are all wrong. I used superscripts in the original for v^2 and c^2 and also for the square root I used ^1/2 but now the powers appear as subscripts and the square root became the fraction 1/2. Please see that the equations are restored. Thanks — Francisco Muller, Mar 07 '14 at 03:42
As far as I can tell, this is no valid critique on White's answer. The only valid point is 2., where he indeed did a numerical mistake, but this does not change the qualititve outcome of his explanation (and by the way, your version is still erroneous). All White is saying is, that all magnetic effects observed in any inertial frame, can be explained by only Coulomb forces in the rest frame of $q$, and I see no point of yours contradicting this. Especially the "additional force" you are describing in 3. acting on $q$ is no additional force in $q$'s rest frame. Only when we leave it. — M. Winter, Jan 23 '18 at 10:19

score 8 · Answer 4 · answered May 21 '13 at 18:29

Charge produces a field that acts on other charges. But action of this field looks different from different reference frames.

By definition,

electric field is something that accelerates other charges, and
magnetic field is something that rotates other charges.

Consider charge at rest. It produces only electric field in its rest frame. In this frame it acts on other charges by accelerating them in the direction of the electric field $\textbf E$. What we see in the rest frame of the charge is that the momentum vectors of other charges in this frame are "boosted".

However, if we will look at this from the moving frame, we will see that momentum vectors of other charges are not just "accelerated", but also "rotated".

This is simply because "pure" acceleration in one frame looks like combination of acceleration and rotation in other frame.

To account for this "new effect" - rotation of the momentum vector - physicists say that in the second frame (that is moving w.r.t. the charge) there is a magnetic field (in addition to the electric field that (by definition, see above) only accelerates other charges).

-1 A magnetic field also accelerates charges. – Larry Harson May 24 '13 at 14:29 — Larry Harson, May 24 '13 at 14:29
Rotation is acceleration, isn't it? – Murod Abdukhakimov May 25 '13 at 13:50 — Murod Abdukhakimov, May 25 '13 at 13:50

score 7 · Answer 5 · answered Mar 10 '16 at 00:26

7

A simple "first answer" would be to use the analogy of a boat in a lake. When the boat moves on the surface of the water, it perturbs the water and creates ripples. When it does not move, it does not.

Similarly, when a charged particle moves through the "pervasive" EM field (space), it perturbs the EM field and generates a magnetic field perpendicular to the direction of the particle's motion.

Then you can use any or all of the other answers that you have received to go into more details.

answered Mar 10 '16 at 00:26

Guill

2,493

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Excellent answer. The $\mathbf{E}$-field can be thought of as a type of "fluid" that fills space. A charged particle in motion through this fluid creates perpendicular ripples that can be interpreted as the $\mathbf{B}$-field. They are linked: You cannot have one without the other. – Joel DeWitt Feb 23 '19 at 00:35
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@Guill sorry for the dumb question, but is the magnetic field simply the movement/wave (caused by the motion of charged) in space (that's made of the electric field which is like fluid)? – JinSnow Nov 11 '19 at 20:23

R. Romero · Answer 6 · 2019-02-26T21:23:42.217

Suppose you have two charges. One is at the origin of our coordinate system. The other is at some arbitrary position $(x,y,z)$ and lets assume some magical force keeps it there, whatever EM fields might happen there.

Suppose the charge at the origin is moving in a straight line at constant speed. The target charge only receives updates of the moving charge's location at the speed of light. It will respond to the moving charge, not according to where it is now, but where it was some time in the past.

As the moving charge gets closer to the target charge, some of the effect will cancel out the effect due to the charge earlier in its trajectory. The reverse happens as the charge moves away. There will be some canceling of the field due to the overlapping effects with this cancelling happening to the component parallel to the direction of motion.

A moving charge impinges on a target from a different distance over time. A moving charge impinges on a target from a different direction over time. The changing effects have a delay before reaching the target.

I like your explanation, since it does not require two separate coordinate systems. Are there any detailed explanations how interaction between charges changes as one charge moves. It would be nice to have some discrete approximation. — user1700890, Jun 15 '20 at 20:05
@user1700890: https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential
It's an electric potential for a moving charge that takes into account relativistic effects. The magnetic potential is taken by multiplying it by the velocity then dividing by the square of the speed of light. There's various "correction factors" that appear in the expressions for the fields making it differ from the static case. — R. Romero, Jun 19 '20 at 05:11

score 1 · Answer 7 · answered May 21 '13 at 11:06

You may want to say "Electric field of a charge at rest appears as an electric field and a magnetic field when viewed from a moving frame of reference." The comments out it right, a charge is associated with an electromagnetic field. It appears as an electrostatic field when viewed from a frame in which the charge it at rest.

How do moving charges produce magnetic fields?

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