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2012-06-13 - Revised question in experimental format

(This is a thought experiment for which RF experts may have an immediate answer.)

I'll assume (I could be wrong) the possibility of creating a strongly insulating material with low permittivity and very high permeability.

From this material fabricate two identical 30 cm tubes with inside diameters of 1 cm, and wall thicknesses sufficient to provide robust blocking of magnetic fields.

A silver wire 30 cm in length and 1 cm in diameter is inserted into one of the tubes. Call this one the silver wire.

The other cylinder is placed inside within a vacuum chamber with an electron gun at one end and a conductive target at the other end. The electron gun is designed to fill the 1 cm interior with an even flow of electrons, so that it can simulate the flow of electrons in the wire as closely as possible. Call this one the vacuum wire.

Now for the experiment. Identical currents $\begin{align}\mathbf{I}=\frac{d\mathbf{Q}}{d\mathbf{t}}\end{align}$ are sent through both wires.

An experimenter then looks for induced magnetic fields around both wires. Will these magnetic fields be:

(a) Identical,

(b) Greater around the vacuum wire, or

(c) Greater around the silver wire?

My bet is (b). Anyone?

(See below for the history of this question. Again, it may turn out to be easy for RF experts who have to deal with weird combinations of permittivity, permeability, and conductivity on a daily basis.)

2012-06-10: Original version of question

Original title: Where are the electric field gradients in coil-generated magnetic fields?

Background

I tried to apply Feynman's SR-focused explanation of the relationship between electric and magnetic fields in wires to this, but in the end concluded he was addressing a rather different set of issues -- and even that only incompletely, since his purposes were for instructive purposes (his Lectures) than a complete analysis. My question is not about the mathematics of Maxwell's equations, but how such equations may be applied a bit to casually to situations that are actually quite different physically.

Case 1: Magnetic field induction in CRTs

In an old-style cathode ray tube (CRT) or television screen, the electrons that cross the vacuum of the CRT create a detectable electric field gradient between the electrons and the surrounding tube. This field can be roughly imagined by picturing the tube as a large capacitor (which it is; I know someone with the scar on his shoulder that proves just how large) in which the central vacuum area carries most of the negative charge and the sounding tube interior the positive charge. In the case of the CRT, the interior negative charges are also in rapid motion towards the screen.

Any accurate assessment of the above model requires explicit use of the modern version vector based version of Maxwell's equations. However, Maxwell was also fascinated by and made extensive conceptual use of hydrodynamic-inspired models of electric and magnetic fields. For example, Maxwell originated or at least popularized the term "flux lines," meaning flow lines, to describe both electric and magnetic field structure. The phrase "field lines" is more common these days, but means the same thing. The flux line model is still used in beginning courses on electromagnetics, where experiments using magnetic and non-conductive powders can make such lines starkly easy to visualize and comprehend. The flux line model can be defined with mathematical precision for classical velocities.

For the CRT example, electrons are surrounded by flux lines that extend out to the interior of the tube, and the orthogonal motion of those flux lines in turn generates a strong magnetic field with flux lines orthogonal both to the electric flux lines and to the direction of motion of the electric flux lines.

So, this is all quite straightforward: The component of electric flux perpendicular to the direction of travel generates a magnetic flux line that is perpendicular to both that electric flux and its direction of motion. The electric flux lines are in turn defined by a field gradient -- a voltage -- that extends from the electron to the interior of the tube. That electric gradient is quite real and easily measured. The resulting magnetic fields is equally real and measurable, and is in fact what is used to steer the electron beam and paint the screen with an image.

Case 2: Magnetic field induction in wires

Now as it turns out, you can also generate an very similar magnetic field using a rather different method. That method is to embed (in the ideal case) the same number of electrons as in the CRT case, moving at the same average velocity, within a conductive wire. The conductive wire would extend along the same path as the vacuum electrons, and electrons of similar number and velocity and moving along inside the wire can generate a field that, with careful physical adjustments, can be made identical in appearance and strength to that of the CRT case of the electrons moving through a region of vacuum.

Comparing the two cases

So, two cases give very similar magnetic field results: Electrons moving through a vacuum, and electrons moving through a wire. Both give strong circular magnetic fields that surround the path of the moving electrons, and both results can be estimated easily using Maxwell's equations.

One reflex reaction at this point may be (should be) "so what?" After all, moving electrons give magnetic fields, so why in the world shouldn't similar motions give the same magnetic fields?

The interesting point is that if you look at the two cases carefully, they are not the same experimentally, and here's why: One case (the CRT) has a large-scale set of electric field lines that are very explicitly associated with the corresponding magnetic field structure. For example, if you use a tight, wire-like beam of electrons reaching from the back of the CRT to the center of its screen, then at 20 cm out from the path of the electrons there will be a noticeable electric field gradient that in terms of the flux line model is "moving" and thereby generating the magnetic field that can be measured at the same location.

In the wire case, no such electric gradient exists. Because the charge of the electrons is cancelled out within the positively charged atomic background of the wire, the electric field makes no appreciable showing outside of the wire. Yet if you measure 20 cm out from the wire, you still see essentially the same magnetic field result, even though there are no longer any "moving electric flux lines" to generate the magnetic field locally.

The actual question

So, after all that preparation, my question is really quite simple:

Why do moving electrons seem to generate approximately the same long-range magnetic field regardless of whether their field gradients (electric flux lines) are cancelled nearby or very far away?

As often is the case in trying to ask questions like this, working through it has helped my look at my own question differently, so I now think I have some inkling of how to answer my own question. (And no, it's not SR based, since in this question is about why remote electric fields differ given the same "moving electron parts.")

So, is induction being taught accurately?

I'm asking anyway, in part because I'm not sure of my answer, but even more so because I think there needs to be some updating of how such situations are taught.

Specifically, the moving-flux-line model (which I believe is still used instructively and is certainly seductive) flatly does not give correct results. If it were really accurate, there would be no such things as electromagnets and electric motors. That's because the field gradients of the moving electrons in such devices all cancel out very quickly and very locally, on the scale of atoms, leaving no appreciable external electric fields at the ranges of the stable magnetic fields they generate.

References

Possibly related past Physics.SE questions have been asked by:

(1) The equivalent electric field of a magnetic field by Hans de Vries; asked 2012-04-19, answered 2012-04-19. A typically insightful question by Hans de Vries about the SR relationships of electric and magnetic.

(2) Mechanism by which electric and magnetic fields interrelate by Nitin Nizhawan; asked 2012-02-02, no final answer. Another interesting and mostly SR question.

(3) Moving conductors in magnetic fields: is there electric field or not? by Giuseppe Negro; asked 2011-05-14, answered 2011-05-14. A similar title, but not quite the same topic, I think.

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  • $\begingroup$ You're asking if $B$ is produced by $E$, then why does $\nabla E$ not affect it? Maxwell's equation shows that $B$ cares about the curl of $E$, not its gradient... $\endgroup$
    – Chris Gerig
    Commented Jun 11, 2012 at 1:55
  • $\begingroup$ And yes, obviously induction is being taught accurately, by simply stating the definition of induction. $\endgroup$
    – Chris Gerig
    Commented Jun 11, 2012 at 1:58
  • $\begingroup$ Yes, curl $E$, not $\nabla E$. But hopefully I'm not the only person who used to think it was OK to approximate curl was by visualizing "$E$ field lines" moving through space. That image fits well enough with the explicit $\nabla E$ field lines of the CRT example, so I'd never really though much about the complete absence of such gradients in the wire case. My curiosity is more along the lines of whether the equilibrium ideas from Maxwell's early mechanical models might provide a more concrete way to explain why the same magnetic fields form quite nicely with or without co-located $\nabla E$. $\endgroup$
    – Terry Bollinger
    Commented Jun 11, 2012 at 4:15
  • $\begingroup$ Hmm, since in retrospect Maxwell's mechanistic methods are likely not that well known... :), my point is that a stable magnetic field is an end state that does not come into existence instantly, but must instead grow outward as the electrons start moving. I'm pretty sure (didn't try) that growth works differently for the CRT and wire conductor cases, but ends in the same stable $B$ field. $\endgroup$
    – Terry Bollinger
    Commented Jun 12, 2012 at 2:57
  • $\begingroup$ All I hear is gibberish terms, but if you make everything precise, with formulas, then we'll all see the correct answer. $\endgroup$
    – Chris Gerig
    Commented Jun 12, 2012 at 2:59

2 Answers 2

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In a CRT, a power supply creates a high voltage DC electric field that accelerates an electron beam to a desired energy/velocity. The resulting DC beam current does not create that accelerating electric field. (Although space charge effects can modify that field to some extent, and means may be required to keep the beam focussed, since the electrons repel each other.) Also, the DC magnetic field produced by that current is small, and independent of the accelerating electric field.

The magnetic fields in your two cases are identical: within the current distribution, a circumferential field increasing proportional to the radius, which falls to zero when it hits your magnetic-field-blocking (superconducting) tubes. The magnetic fields are identical because the currents are identical.

...unless I'm completely missing your intent...

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  • $\begingroup$ Art, thanks. I think your answer (a), identical fields, is likely correct. My intent was this: moving electrons in the vacuum case have $\mathbf{E}$ fields that extend outside the magnetic shield, while moving electrons in the silver case do not. Does this excursion outside the shield of an $\mathbf{E}$ field that is associated with moving charge carriers translate into an increase in the $\mathbf{B}$ field in the same exterior region? I'm now thinking "probably not," because the $\mathbf{E}$ field outside the shield should be steady state for a constant current. Other comments, anyone? $\endgroup$
    – Terry Bollinger
    Commented Jun 15, 2012 at 2:37
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This is another of my hand waving answers, it is too long for comments.

Your high permeability material exists, it is called mu metal . We have used it around photomultipliers to shield them from stray magnetic fields.

You will have to use insulation inside the walls of your tubes.

Are you talking of the very weak field outside the mu metal tubes?

If you can make a thin enough and coherent enough electron gun to simulate the silver wire the result will be the same.

If you fill up the space with the electrons the result will be different due to the differing boundary conditions according to the distance of the electron to the walls, since there will be a metal mirror effect. It is b)

Even in the hypothetical complete insulator with high magnetic permeability the boundary conditions will be different for a silver wire and a full space, in my hand waving opinion.

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  • $\begingroup$ Nice analysis. Thanks especially for the specifics about mu metal. Your argument about it being (b) due to boundary conditions sound fascinating, very plausible, and tempting since it would make me sound right... but alas, I was trying to minimize such effects rather than rely on them, so I can't claim to have been intending my (b) answer that way. Since @ArtBrown answered (a), which I now think is correct, with the same basic points you also made, I'll wait a day for other comments and then mark his as the answer. $\endgroup$
    – Terry Bollinger
    Commented Jun 15, 2012 at 2:49
  • $\begingroup$ It is not clear to me whether the diameter of your free electron current is constrained to be the same as the wire. If it is not, I believe there will be a difference in the map of the field depending on the radius.At the impermeable walls it will be the same because the integral will have the same value I. $\endgroup$
    – anna v
    Commented Jun 15, 2012 at 3:48
  • $\begingroup$ My intent was "identical" electron paths, e.g. ballistic electrons traveling at the same velocity along identical paths; so yes, the diameter of the free electron current would be constrained to the same form factor and current densities as the wire. Both radii could be substantially smaller than the mu metal cylinder, since trying to send free electrons that close to matter would cause strong (and interesting) interactions that were not my intent. So: Your answer is very good, but since @ArtBrown was first to give a well-written and accurate answer, I'll stick to my guns and award to him. $\endgroup$
    – Terry Bollinger
    Commented Jun 15, 2012 at 21:33
  • $\begingroup$ Fair enough .I am not arguing for a check :) , just to understand your boundary conditions. $\endgroup$
    – anna v
    Commented Jun 16, 2012 at 4:12

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