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I want to ask about the workings of the Hall effect. Why do the electrons come to rest on the edge of the wire? The magnetic field pushes them up, and the electric field pushes them forward. Shouldn't they reach the top of of the conductor and then stop moving up but continue moving forward?

Why does the magnetic field stop them from continuing to flow? The magnetic force is up so why would it effect their velocity forwards?

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  • $\begingroup$ Why do you think they stop moving forward? $\endgroup$ – Phil Frost Feb 12 '14 at 15:54
  • $\begingroup$ @PhilFrost I have no clue. I thought about this for a while and couldn't come up with an explanation. I don't see why a vertical force ever affect horizontal velocity. $\endgroup$ – dfg Feb 12 '14 at 15:55
  • $\begingroup$ Let me clarify: you ask "Why does the magnetic field stop them from continuing to flow?", but I don't think the magnetic field does stop them from continuing to flow. I'm wondering why you think it does. $\endgroup$ – Phil Frost Feb 12 '14 at 16:08
  • $\begingroup$ @PhilFrost Yup, that's what I'm asking. I think that because if the magnetic field didn't stop them, they would continue to move and there wouldn't be a buildup of electrons. $\endgroup$ – dfg Feb 12 '14 at 16:10
  • $\begingroup$ @qmechanic is it a question? He seemed pretty sure that he wanted to ask about the working of the Hall effect. I'd punctuate that with a "." not a "?". $\endgroup$ – Phil Frost Feb 12 '14 at 17:08
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I think you are confused about what the Hall effect does. You ask: "Why does the magnetic field stop them from continuing to flow?" The answer is: it doesn't. Take this setup:

Hall effect experimental setup

Here, the magnetic field is pointing up, in the +z direction. The conventional current in the purple conductor is flowing towards us (electrons going in the opposite direction). This current is unaffected by the magnetic field. That is, the current is in the +x direction, magnetic field or no.

However, the electrons experience a force orthogonal to this, in the -y direction. This is a consequence of the Lorentz force:

$$ {\mathbf {F}}=q\left({\mathbf {E}}+{\mathbf {v}}\times {\mathbf {B}}\right) $$

In particular, the ${\mathbf {v}}\times {\mathbf {B}}$ term. This doesn't stop the current or change it. Initially, electrons will be pushed to the left, but very soon the system will reach equilibrium, and the "electron gas" will be denser at the left so as to cancel the force they experience due to the magnetic field $\mathbf B$. Then, the net force experienced by the electrons is in line with the conductor, just as it would have been in the absence of a magnetic field. The result of this charge redistribution, the electrons being more dense at once side, is a measurable voltage, $V_H$, between the sides of the conductor.

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  • $\begingroup$ But if they just "slosh" and don't buildup, wouldn't the field they generate disappear once they flow out of the portion of the conductor, leaving no field for the next batch of electrons? $\endgroup$ – dfg Feb 12 '14 at 16:44
  • $\begingroup$ @dfg They slosh, and then they've built up, and then they stay that way. But, you seem to be stuck on thinking of individual electrons as moving around, which is the wrong way to do it. In a practical circuit, the electrons are moving at a nearly glacial pace anyhow. Think of it like a fluid, with the individual molecules/electrons moving around all over the place, all the time, but the features (waves, gradients, fields, etc) retaining the same shape because as one molecule/electron leaves, another one enters. $\endgroup$ – Phil Frost Feb 12 '14 at 16:54
  • $\begingroup$ @PhilFrost, Nice picture and explanation. I think it's wrong to say the electrons are moving at a glacial pace. They are flying around due to thermal motion. (because they are so light.) And having collisions with the lattice all the time. In between the collisions they are subject to several forces, (as you described). There's the electric field along the wire that drives the current, there's the vXB term, and also an electric field across the wire due to the excess charge that comes from the vXB term. $\endgroup$ – George Herold Jul 26 '14 at 3:54
  • $\begingroup$ @GeorgeHerold Sure, more accurately, the drift velocity is glacial. You aren't wrong, but the very fast motion of individual electrons is usually irrelevant, and in my experience an impediment to a good intuitive understanding of how most electronics work, and that's what I'm trying to convey here. We can say the same thing about air or water or any fluid (very fast particles), but this isn't relevant to understanding hydraulic or pneumatic systems. In fact, if you thought that hydraulic fluid shot around the circuit at the speed of sound, you will get a lot of things wrong. $\endgroup$ – Phil Frost Jul 26 '14 at 4:05
  • $\begingroup$ Hi @PhilFrost, I think a few weeks ago I would have agreed with you. But ignoring the thermal motion can lead to problems. One assumption of electron motion is that the scattering time/ mean free path does not depend on the applied electric field. If there was no thermal motion then the scattering time would change quadratically with the field! And maybe all the thermal flying around will help dfg get unstuck with his understanding. :^) $\endgroup$ – George Herold Jul 26 '14 at 4:42

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