Talk:Hall effect

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It appears that the "w" symbol used in the derivation of the Hall voltage is not defined. Also, the q=-q and v=-v equations are nonsensical. It's understood that the charge and velocities are the inverse of a classical, positive charge. But, we need to introduce alternate symbols vs. simply negating the ones already in use. B is not defined in this article. — Preceding unsigned comment added by Andyberks (talk • contribs) 01:56, 22 September 2022 (UTC)[reply]

Hi fellow wikipedians! I've just uploaded a large, detailed illustration of the Hall effect to Commons - see Image:Hall_effect.png -- Peo from danish Wikipedia

Nice!!--Krackpipe 18:07, 10 May 2005 (UTC)[reply]

• Great diagrams! 17:33, 27 February 2006 (UTC)

The diagram is great! excellent addition to the article Kupojsin 11:59, 2 February 2007 (UTC)[reply]

What is the symbol for a Hall effect sensor (3 wire type) for use in a schematic diagram ? - Joshua Hrouda, 26/7/05

A full range of Hall Effect related articles at: http://www.bbautomacao.com/Sensors.htm

Some of the diagrams on the site belong to a company (Raztec). Something should probably be done about this.

most likely the now unavailable (on their site) diagrams were stolen from wikipedia

How does the Hall voltage build up if only one type of charge accumulates at one side of the conductor? For example, electric current is the flow of electrons so they deviate from their 'normal' path and 'rush' to one side of the conductor. But where does the positive charge to accumulate at the opposite side of the conductor come from? RokasT 11:32, 30 December 2006 (UTC)[reply]

All solid materials are made from neutral atoms locked together in a crystal lattice. A conductor is conductive because some of the electrons are free to move away from their 'parent' atoms and wander throughout the lattice. The electrons have a negative charge, so they leave behind positive ions that cannot move. Whenever electrons are forced to gather together in one region (against their mutually replusive instincts) concentrations of 'naked' positive charges must be left exposed in others. Hence the negative voltage on one side of a Hall probe is exactly matched by a positive voltage on the opposite face. The only difference is that the electrons are drifting along the face as a current, while the positve ions are stationary. StuFifeScotland 14:42, 1 April 2007 (UTC)[reply]

Thanks for your explanation. ;-) RokasT 20:54, 1 April 2007 (UTC)[reply]

One more question: If holes moving in say the +y direction are viewed as actually electrons moving in the -y direction, the polarity of the hall voltage generated is the opposite! Especially since electrons exist physically whereas holes only represent absence of electrons. This means n and p semiconductors should show identical polarity in the hall effect experiment, which is not so as demonstrated by experiment. What explains this? Thanks. -AK 220.225.214.2 (talk) 10:20, 24 January 2008 (UTC)[reply]

This confused me for awhile too. Holes moving in +y are really electrons moving in -y direction. However, those electrons actually have negative Effective mass and so are deflected in the opposite direction by the magnetic field. I found this paper http://www.journal.lapen.org.mx/jan09/LAJPE%20225%20Lianxi%20Ma%20Preprint%20f.pdf to be useful. I was considering making an edit to the page to explain this. Kiracofe8 (talk) 16:39, 23 March 2009 (UTC)[reply]

I think most people see the picture and think "But what about holes?" and incorrectly assume holes are coming from the same +x direction as the electrons. However, the current is the same, and because holes have a positive charge they wouldn't be coming from the +x direction, they would be coming from the -x direction. These holes would be feel a force in the -y direction initially, and so you have a buildup of positive charge at the -y end of the material, giving a Hall voltage with an opposite side. — Preceding unsigned comment added by 140.180.246.108 (talk) 15:40, 22 May 2013 (UTC)[reply]

Please read this paper here: http://gcdcc.hebut.edu.cn/ydzl/19-Correction%20to%20the%20classical%20two-species%20Hall%20Coefficient%20using%20twoport%20network%20theory.pdf

where Prof de Paor says "It came as a great shock...to discover that the classical expression for the two-species Hall Coefficient is incorrect due to unjustified neglect of terms in the derivation. However, it is of great interest to observe that the expression derived here approximates the old at sufficiently low magnetic field strengths, for that indicates that under certain conditions, clearly stated here, the terms neglected without comment by earlier authors are in fact negligible. From the educational point of view it seems surely desirable that students should be aware of the correct expression", and he of course derives the correct expression. Could someone please review that, and incorporate in the article? — Preceding unsigned comment added by Donn300 (talk • contribs) 20:26, 10 January 2014 (UTC)[reply]

Why isn't the Hall coefficient mentioned? R_H = 1/ne --82.43.144.131 18:01, 25 January 2007 (UTC)[reply]

Yeah, it's just wrong. The hall resistance has B but the coefficient doesnt. 2600:4040:24D4:3300:8EB2:4FE8:8EEE:7598 (talk) 00:32, 22 March 2024 (UTC)[reply]

This question is relevant to many articles in electromagnetism but here is as a good a place as any to bring it up: doesn't the Hall effect have to obey symmetry? E.g. if I reflect (in some horizontal axis) the system from that nice picture in the article, shouldn't the reflection also have the negative charge at the top of the Hall element? According to the theory, no (see parts B and C from that image). I know that the Right-Hand-Rule and vector cross-product come into this but these don't seem to be adequate explanations. I say this not to claim that all the physicists in the world are wrong and I'm right (I would far sooner take their word than mine!) but to ask whether it might be better explained (not necessarily in this article). There must at least some others who get confused about this point.--Ejrh 10:39, 15 April 2007 (UTC)[reply]

Good question. The reason is that the Magnetic field (being a pseudo vector) breaks the reflection symmetry you mentioned. Here's how it works:

Electric field and current are vectors, so they transform as you would expect under inversion/reflection. Magnetic field are pseudo vectors, so they don't change direction under inversion (which is taking every point ${\displaystyle {\vec {r}}}$ mapping to ${\displaystyle -{\vec {r}}}$. Magnetic fields are defined by ${\displaystyle {\vec {F}}=q{\vec {v}}\times {\vec {B}}}$, and you can check that it doesn't change to preserve that law. The flip that you mentioned is really an inversion followed by a 180 rotation, so basically, the magnetic field is now pointed downwards (from the rotation) from the reflection. Hence, the Hall effect doesn't break symmetry.74.102.181.37 06:37, 7 June 2007 (UTC)[reply]

Thanks very much! Your response lead me to Pseudovector which contains a nice example similar to this one. Ejrh 11:53, 1 July 2007 (UTC)[reply]

This comes up in conservation of parity and the Wu experiment. Gah4 (talk) 19:41, 16 September 2024 (UTC)[reply]

The Hall Resistance / Conductance is not defined in this article. I assume it's ratio of Hall (tranverse) voltage to current. It's important because the Quantum Hall effect does not reference to what is actually quantized.

I think the article "Hall Resistance" should be created.74.102.181.37 06:37, 7 June 2007 (UTC)[reply]

I came here to say the same. Hall Conductivity has a link to Hall effect in the Quantum Hall effect article, but it is not defined in this article. —Preceding unsigned comment added by 190.188.0.22 (talk) 17:59, 1 April 2010 (UTC)[reply]

I came here for the same reason. The phrase "Hall resistivity" does not seem to be common on the web, and I didn't understand it. 68.175.114.223 (talk) 19:19, 10 May 2014 (UTC)Josh Rubin[reply]

When I rotate the yellow image below, it apprears to match image "D" above, except the hall voltage polarity appears reversed. ````Harold````

It is mentioned that the magnitude of the effect is dependent on the type of material (as well as the thickness, current, etc.) I tried building one using a sheet of "1 ounce" copper on fiberglass. I was not able to observe any such effect. Is copper insensitive for this application? The commercial sensors seem to be made of silicon (for obvious reasons), but my understanding is that the material does not have to be a semiconductor. Plus, with a metal I can use much higher currents, which should result in a greater signal/noise ratio (sensitivity).

My point is: there should be information about the types of materials and their Hall output.

169.237.215.179 19:50, 20 August 2007 (UTC)[reply]

Try using bismuth. It has, by far, the highest Hall coefficient of any metal. Copper has a nonzero (but very low) Hall coefficient, so it might not be a good one to use: [1] Stonemason89 (talk) 00:42, 12 August 2010 (UTC)[reply]

Notice that the Hall coefficient has n (charge carrier density) in the denominator. For a given current, a metal has many charge carriers moving slowly, compared to a few moving fast in a semiconductor. But Hall did measure it on metals. Gah4 (talk) 05:18, 4 April 2018 (UTC)[reply]

Another advantage of hall devices is that they are devoid of the contact 'bounce' of mechanical switches. Switch bounce can cause confusion in digital circuits. I don't have a reference source on this, however. LorenzoB 17:45, 28 August 2007 (UTC)[reply]

But I think this is somehow well-known within the field of electronics. Does this exempt the citation?--113.252.82.168 (talk) 11:59, 17 April 2010 (UTC)[reply]

A spark accelerator is an alternative. It is a simple button for uphill climbs, relevant in electric vehicles. — Preceding unsigned comment added by 69.255.42.105 (talk) 19:57, 4 October 2011 (UTC)[reply]

There should be some mention of the use of hall effect sensors for the detection of lid closure in consumer electronics devices. The Nintendo DS & DS Lite for example use the fixed magnet in the right speaker to trigger a hall effect sensor near the Y button. MacBooks use the magnets that hold the lid closed to trigger a sensor near the bottom-left corner of the keyboard. —Preceding unsigned comment added by 128.114.60.119 (talk) 21:28, 17 December 2007 (UTC)[reply]

Why do we not list typical values of the Hall coefficient for various materials at room temperature? I've also had difficulty finding them online ... —Preceding unsigned comment added by Newagelink (talk • contribs) 04:46, 17 March 2008 (UTC)[reply]

For a specific metal, a web search for (metal name) hall coefficient should come up with it pretty fast. But a table with some common metals would be nice. Gah4 (talk) 20:56, 11 April 2018 (UTC)[reply]

It is written that hall effect shows why it makes sense to think of charge being carried by holes in semiconductors. How so? —Preceding unsigned comment added by Rainmanthe (talk • contribs) 15:19, 29 April 2008 (UTC)[reply]

The sign of the Hall voltage depends on the sign of the charge carrier. Holes are positive carriers and electrons are negative carriers. So when they get a Hall voltage for a p-type semiconductor (holes) that is opposite in sign from an n-type semiconductor (electrons), that indicates that in the p-type semiconductor the current is carried by holes which are positively charged. Bob K31416 (talk) 08:13, 6 July 2008 (UTC)[reply]

I found this very nice picture about Hall-effect ion engine thruster being test-fired for ESA's SMART-1 mission at ESA: http://www.esa.int/esaHS/ESAO4S0VMOC_exploration_1.html (there is a Back to article button too) Would someone write about it? --Ernobius (talk) 06:54, 10 July 2008 (UTC)[reply]

Its already been written about in the Wikipedia in the article Hall effect thruster. It looks like all that is needed is a brief mention of it, along with the link Hall effect thruster in the applications section. --Bob K31416 (talk) 01:17, 12 July 2008 (UTC)[reply]

The Hall effect diagram in drawing "A", for example, at the beginning of the article may not be a good representation of the flow of electrons in the Hall element. It shows electrons flowing in a curved narrrow beam rather than flowing throughout the Hall element with many microscopic collisions that have electrons moving in all directions, which is the case of a usual Hall element that is a solid. See for example the diagram of electron motion in the discussion of the Microscopic View of Ohm's Law. What the applied electric field does is cause a perturbation of the random electron motion between collisions so that there is a net electron flow that is parallel to the electric field and in the opposite direction. This net electron flow corresponds to the electron drift velocity. What the applied magnetic field does is add an even smaller perturbation that would result initially in a tiny net vertical component of electron flow that results in a Hall field. All this is happening while the electrons undergo nearly random motion from the many microscopic collisions that they undergo. --Bob K31416 (talk) 04:29, 28 July 2008 (UTC)[reply]

Would Hall Effect be seen in a perfect conductor? How Would Be The Electric Filed be Directed inside the Perfect conductor? —Preceding unsigned comment added by 59.95.68.188 (talk) 04:42, 11 March 2009 (UTC)[reply]

If carrier density goes to infinity, then Hall coefficient goes to zero. It might be that superconductors don't have a Hall coefficient, though. Gah4 (talk) 18:08, 12 April 2018 (UTC)[reply]

I just had a discussion with my professor, who stated that the hall effect is NOT due to the curving of the current path due to the magnetic field, but in fact the current path DOES NOT curve at all because the lattice of the conductor prevents it from doing so. Instead, the hall effect is due to a reorganization of the electrons in the conductor so that the total electromagnetic force is zero in the transverse (perpendicular) direction. in effect, this just means that there would be an asymetric current density as you look up and down the conductor. If this is the case, then this article needs to be severely modified (including the images), so i would appreciate it greatly if some of you would discuss this problem before i make any attempts at editing the article. Pjbeierle (talk) 19:48, 8 March 2010 (UTC)[reply]

The diagram is misleading and may have been motivated by one of the external links which shows the bending of an electron beam in a low-density plasma (an electrically conductive ionized gas).

• Interactive Java tutorial on the Hall Effect National High Magnetic Field Laboratory

Also, see my previous message in the above section Hall effect diagram.

I agree that the article needs work in the points that you mentioned. However, you might have misunderstood your professor when you wrote " current path DOES NOT curve at all because the lattice of the conductor prevents it from doing so." - It's the microscopic collisions that the electrons undergo, the induced transverse electric field (Hall field), and the smallness of the perturbation of the current path by the magnetic field, that causes the current path in a solid not to curve as shown in the diagram, although the microscopic paths of electrons between collisions are curved.

However, we need reliable sources for all of this and for what's in the article now. So maybe a trip to the library for books on the classical Hall effect is in order. Unfortunately, it may be difficult, if not impossible, to find these details in a reliable source. But I think we could still give a reasonable explanation that is based on reliable sources. --Bob K31416 (talk) 14:27, 16 May 2010 (UTC)[reply]

I added "between collisions" which should be an improvement, e.g. it now reads, "However, when a perpendicular magnetic field is applied, their paths between collisions are curved so that moving charges accumulate on one face of the material." --Bob K31416 (talk) 14:23, 17 May 2010 (UTC)[reply]

Thanks Bob, that is definitely a step in the right direction to the very least. also, since you agree that the image is misleading, i went on to remove the image. I had forewarned the image creator about a month ago, but he did not reply. Pjbeierle (talk) 22:01, 24 June 2010 (UTC)[reply]

I think the image should remain until you provide a better one. You could add a statement explaining what the current path really is. Q Science (talk) 08:40, 25 October 2011 (UTC)[reply]

i have a problem, we are talking that the hall voltage is produced in p type semi conductors due to the force acting on the holes, but how can consider that a force acting on an empty space (actually hole is an empty space, hole is an imaginary concept,(in p type semi conductors we consider the moving of electrons as moving of holes in opposite direction relatively.)). my question is how can we consider the holes as particles? and how a force acting on a hole?

please give me an explanation on this matter. jj shan. — Preceding unsigned comment added by Jj shan (talk • contribs) 05:11, 19 April 2011 (UTC)[reply]

For a hole to move, an electron must move to fill it, leaving a hole were the electron was before. Thus, all current in a semiconductor is created by moving electrons. The PN junction is created by the difference in energy necessary to fill a hole (P-type) or loose an electron (N-type), yet, in both P- and N-type material, it is really the electrons that move. Q Science (talk) 08:19, 25 October 2011 (UTC)[reply]

The "theory" section defines current as: "Current consists of the movement of many small charge carriers, typically electrons, holes, ions (see Electromigration) or all three." But as the previous two posters have noted, holes are a convienient fiction and current generated by 'moving holes' is really generated by moving electrons. I propose we remove 'holes' from this definition because it unnecessarily complicates the definition. -alex — Preceding unsigned comment added by 200.26.178.108 (talk) 16:08, 22 May 2012 (UTC)[reply]

You might read the electron hole article, but one explanation uses effective mass. But the whole reason it is important, is that p-type semiconductors have a positive Hall coefficient. You can also consider them as electrons with a negative effective mass, that that seems harder than as holes with a positive effective mass. A band that is more than half full is best described by a less than half full band holes, with the holes at the top. Also, some metals have both electron and hole bands, and sometimes positive Hall coefficient. Gah4 (talk) 05:07, 4 April 2018 (UTC)[reply]

According to Mirmin book, it should be H instead of B. Since the book is a legend on Crystalography and solid state physics, I believe that the error is on this page. Spiralciric (talk) 19:12, 1 December 2012 (UTC)[reply]

You assume that if H is correct than B is incorrect and vice-versa. I disagree. There is a formula for Hall effect in terms of H, and there is a formula for Hall effect in terms of B. Both formulas are correct. (The formulas are not quite identical but pretty close; there is a factor of μ difference.) Textbooks written by physicists are more likely to use B and textbooks by electrical engineers are more likely to use H, in my experience. --Steve (talk) 23:34, 1 December 2012 (UTC)[reply]

Hi! I created this diagram:

Which is based on this one: http://commons.wikimedia.org/wiki/File:Stromwandler_Zeichnung.svg

Not used to editing wiki pages so I'll leave that to someone else :D — Preceding unsigned comment added by Dracheschreck (talk • contribs) 19:34, 11 June 2013 (UTC)[reply]

In discussing equations, it is customary to indicate what all the variables represent just before or immediately after their introduction to the text. In this article, there is no mention that "B" represents the magnetic field until the section on "Hall effect in ionized gases", even though it is used in the first equation(s). Wleroydavis (talk) 19:52, 14 March 2016 (UTC)[reply]

Tangentially, when the variables are stated, having them be equations makes awkward line breaks (at least in mobile). I think they should be plain text if it's impossible to make them both equations and integrated into other text without newlines. Gvprtskvnis (talk) 18:34, 17 April 2016 (UTC)[reply]

Some variables are commonly used, and so sometimes some forget to indicate them. v for velocity, E for electric field, and B for magnetic field, as examples. Gah4 (talk) 21:20, 11 April 2018 (UTC)[reply]

The comment(s) below were originally left at Talk:Hall effect/Comments, and are posted here for posterity. Following several discussions in past years, these subpages are now deprecated. The comments may be irrelevant or outdated; if so, please feel free to remove this section.

When I take the yellow image below, and rotate it clockwise 90 degrees, it appears to resemble image "D" above. Except that the Hall voltage polarity is oppposite that of image "D". Under the "Discovery" section it says "The Hall effect was discovered in 1879 by Edwin Herbert Hall while working on his doctoral degree at the Johns Hopkins University in Baltimore, Maryland, USA. His measurements of the tiny effect produced in the apparatus he used was an experimental tour de force, accomplished 18 years before." This statement likely requires some explanation to put it into proper context. Further more the hanging "accomplished 18 years before." needs to be addressed. —Preceding unsigned comment added by 128.119.52.165 (talk) 19:09, 13 July 2010 (UTC)[reply]

Last edited at 19:09, 13 July 2010 (UTC). Substituted at 17:01, 29 April 2016 (UTC)

I've read enough about the Hall effect to know that this wikipedia article is simply wrong in many spots and any sane, literate person who has actually read a handful of references would know this too.

For example, see Solid State Physics (Ashcroft/Mermin 1976) pages 11-12 on the Hall effect. Or just google some other reference:

https://books.google.com/books?id=tUEoDwAAQBAJ&pg=PA102&dq#v=onepage&q&f=false — Preceding unsigned comment added by 143.215.119.98 (talk) 14:56, 3 March 2018 (UTC)[reply]

Magnetoresistance is different from the Hall effect, even though Hall worked on both of them. Gah4 (talk) 23:59, 22 September 2018 (UTC)[reply]

It seems that Beryllium, Cadmium, Cerium, Iron, Molybdenum, Tungsten, and a few other metals, have positive Hall coefficients. With one valence electron, it is most likely to have a half full band. (Complicated by crystal symmetry and such, but usually ...) With two valence electrons, it could be a full band (insulator), or two bands more and less than half full. Note also that the metals with positive Hall coefficient also tend to have high resistivity. With one hole and one electron band, the sign will depend on which one is more, and which one is less, than half full, and the mobilities in each. Aluminum has three valance electrons, which can partially fill three bands. So, some hole and some electron bands, and they change with magnetic field. At high magnetic field, Hc of Al goes positive, but is always the result of both holes and electrons. Gah4 (talk) 18:21, 12 April 2018 (UTC)[reply]

Can someone check the recent changes regarding sign? It should work out that for negative charge carriers (electrons) the Hall coefficient and Hall voltage are negative. I believe that is the usual definition for Hall voltage, but also that it differs from the picture. Gah4 (talk) 23:57, 22 September 2018 (UTC)[reply]

Hello, We are students from the last course of the Physics degree and we are studying Hall effect using matrix formalism. We have been asked by our teacher to modify the article about Hall effect as a way to evaluate us. We would like to add a new section at the theory section that is about matrix formalism in 2D system. It would be a pleasure to do it.

Regards --Alba Cazorla (talk) 09:53, 18 May 2020 (UTC)[reply]

Wikipedia is not a textbook. This is probably too specialized a contribution for a general-coverage encyclopedia such as Wikipedia. --Wtshymanski (talk) 02:13, 19 May 2020 (UTC)[reply]

The Hall Effect animation WebM file is not playing in the Wikipedia mobile App version in IOS 15.4.1

It plays as desktop/mobile websites but not within the app itself. Without being a developer, I’d lean towards it being the App itself not talking to the IOS decoders.

* Note *

I read in the FAQ about videos in IOS that there’s still issues with Apple IOS. I know Wikipedia uses open source such as WebM, but as there’s hundreds of millions of us using iPhones/iPads, maybe an H.264 video coding format is better.

2601:44:4380:E810:A02C:6484:8228:C57B (talk) 07:47, 19 September 2022 (UTC)[reply]

The "Applications" of the Hall effect can be simply said to be the Hall effect sensor. Thus everything that is currently in Hall effect#Applications can simply be cut and pasted into Hall effect sensor#Applications. The scope of this Hall effect article can then just focus on the theory of the Hall effect (and related effects like Quantum effect and Corbino effect). Em3rgent0rdr (talk) 01:02, 29 December 2023 (UTC)[reply]

I made the cut since I didn't see any objections, though maybe I should have waited longer, but seems a straightforward separation. Em3rgent0rdr (talk) 16:57, 30 December 2023 (UTC)[reply]