In high-school level books (for example the german standard text: "Dorn-Bader") I have often seen an explanation of the Lorentz force as on the following picture:

enter image description here

The textbooks consider the superposition of the circular field of the wire and the homogenous field of the magnet (sure the homogenity doesn't matter here). Then the net field as you can see on the picture above on the right is larger on one side of the wire (here on the right) and smalle on the other one. So far so good.

However why does this explain the occurence and direction of the Lorentz force. Do do so, one would need another principle for example that the wire always wants to go to the weaker field regions or something like this. And this principle should be somehow more evident than the Lorentz-force itself (which you can "see" experimentally).

But how is this needed principle exacly forumlated? Why it is correct? Is there any good reason that it is more evident than just taking the Lorentz-force as an experimental fact?

Would be great if someone could clarify the logic of this, evaluate the soundness of the argument and embedd it conceptually and mathematically in the big picture of electromagnetic theory.

Additionally I want to know if the above cited "explanation" has any common name and if there are university level textbooks which proceed in a similar way. I feel that this argument goes back to Michael Faraday (just by the style of reasoning) - so if someone has a reference to the orgin of the argument I would be interested in it, too.

By the way: The magnetic field in the above cited book ("Dorn Bader") is introduced by the interaction of permament magnets...

  • $\begingroup$ I am really happy to see a book introducing it like this. What is the exact title? Dorn Bader Mittelstufe, or Dorn Bader SI/SII? $\endgroup$ – lalala Jan 15 '19 at 13:36
  • $\begingroup$ For example in the red "Physik Sek II" (ISBN 3-507-10724-4) on page 208 "Wie kommt die Lorentzkraft zustande?" Maybe in other versions this occurs as well. Why would you be happt to see an introduction like this? $\endgroup$ – Julia Feb 5 '19 at 6:45

This is an interesting question. I did a bit of research and I think the principle at work here is that there is a pressure associated with an energy density.

From the Wikipedia article "Energy density":

Energy per unit volume has the same physical units as pressure, and in many circumstances is a synonym: for example, the energy density of a magnetic field may be expressed as (and behaves as) a physical pressure

To the right of the wire, the magnetic fields add while to the left, they subtract. Thus, the magnetic energy density is greater to the right than to the left.

From the article "Magnetic pressure and tension":

The magnetic force (per unit volume) in the equation for fluid motion may be re-expressed as

$$\mathbf J \times \mathbf B = \frac{1}{\mu_0}(\nabla \times \mathbf B) \times \mathbf B = -\nabla \frac{B^2}{2\mu_0} + \frac{1}{\mu_0}(\mathbf B \cdot \nabla)\mathbf B$$

$\frac{B^2}{2\mu_0}$ is the magnetic pressure and the term $-\nabla \frac{B^2}{2\mu_0}$ is the magnetic pressure gradient or magnetic pressure force.

The term $\frac{1}{\mu_0}(\mathbf B \cdot \nabla)\mathbf B$ has a component that cancels the magnetic pressure force in the direction parallel to the magnetic field lines so the magnetic pressure force acts perpendicular to the field lines. The remaining component is the magnetic tension force.

  • $\begingroup$ Actually, in the article you quote, $\mathbf B$ is total magnetic field and $\mathbf J$ is total current density. $\mathbf B$ cannot be just external field acting on the current density $\mathbf J$, because this would imply $\nabla \times \mathbf B = \mathbf 0$, while in the above derivation it is assumed $\nabla \times \mathbf B = \mu_0 \mathbf J$. $\endgroup$ – Ján Lalinský May 1 '14 at 19:57
  • $\begingroup$ @JánLalinský, I added that section at the last minute and went out to cut some grass. During that time I came to the same conclusion. I corrected some additional inaccuracies and hopefully have not introduced others. $\endgroup$ – Alfred Centauri May 1 '14 at 20:17
  • $\begingroup$ Actually you are spot on. The general principle is called 'Maxwell stress tensor': en.wikipedia.org/wiki/Maxwell_stress_tensor one finds that the force on any charge and current can be described by a divergence of a tensor (+ radiation from poynting vector). $\endgroup$ – lalala Jan 15 '19 at 13:35

While the field is $\vec F = q\vec v \times \vec B$, it is not a terribly intuitive process. The closest one comes to a vector-product in the real world is the Coralis force, where the wind goes clockwise around a low, and anticlockwise around a high.

A moving charge sets up a circular magnetic field, which is one direction or the other, depending on the sign of the charge. When this moves into a magnetic field, the fields add on one side, and subtract on the other side, so that charge is pushed perpendicular both to the field, and to the direction of travel.

In the case of a conductor, like this example, the charge is bound to a rod, and the rod is moved outwards (or inwards) to the magnet, as the current is flowing one way or the other.

Where the charge is not restricted to a mechanical device (like a peice of wire), the charge travels in a circle, and such devices are known as cyclotrons.

Permanent magnets are used, because these are able to produce a constant $\vec B$. An electromagnet is necessarily an changing-flux and changing-current thing, and thus can not make a constant flux-field.

The vector-product is not symmetric, ie $\vec A \times \vec B = -\vec B \times \vec A$, and since it is a parity thing in both 2d and 3d, one can use either a left-hand rule or a right hand rule. But it must be used consistantly, like it's ok for everyone to drive on the left or on the right of the road, but everyone's got to do the same. So we ram the right-hand rule, for rotating A onto B gives A×B. It's just a way of keeping tabs on the symmetry-breaking process.

Jeffimenko's "Electricity and Magnetism" gives the formula alone, but vector relations occupy the full first chapter. Like Heaviside, J supposes a healthy dose of vector arithmetic is needed before electricity is mentioned.

The text i used at uni in the seventies (Grant and Phillips 'Electromagnetism'), shows essentially the diagram in two parts (the poles, and then some pages later the wire), but both indicate three orthogonal vectors. No name is attached to it, but it pretty much represents the notional way (legal metrology), that one would define this event: bereft of the complexities where the geometry adds additional factors to the product.

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    $\begingroup$ Thanks, but I don't see, how this solvs my problem. In the first part you are just citing the usual formula for the Lorentz force on a moving charge and using this to verify that the conclusion drawn in the book is correct in this case. But it doesn't really answer my question. $\endgroup$ – Julia Mar 31 '14 at 11:13
  • $\begingroup$ @Julia Actually, one supposes a vortex around a moving charge, and that vortex adds with the magnetic field, to push the charge away from the sum and towards the difference. The right-hand rule just sets the direction of $B$ around the cahrge and the field, so they give the correct parity. This is all in the second paragraph. You ask about permanent magnets, the third para does this. The fourth para simply means we have to write it as $qv\times B$ and not $B \times qv$, to keep the parity right. The last two are examples of college books and how they bring Lorentz-force in. $\endgroup$ – wendy.krieger Mar 31 '14 at 11:40
  • $\begingroup$ But why does it push the charge away from the sum and towards the difference (see my original question!)? $\endgroup$ – Julia Mar 31 '14 at 11:45
  • $\begingroup$ @Julia The theory used of magnetism has its roots in the vortex theory associated to eddies in liquids. But in essence, when a charge moves, it creates a magnetic presence, like a charge, of the order $qv$. This magnetic presence is then subject to fields, but like eddy currents, the field acts at right-angles to the field. One can better think of it as the charge wanting to get to a position where there are fewer lines of force from its own eddy field, and the field produced by $B$. This is orthogonal to the flow of flux (ie $B$), and the circulation around the $qv$. $\endgroup$ – wendy.krieger Mar 31 '14 at 12:03

Suppose that you knew about the Coulomb interaction between charges. Now imagine a wire with a current, and let the wire be at rest. This means that the ions in the wire are at rest, but the conduction electrons have a net velocity. The wire is electrically neutral. Then if there is a charge in motion outside the wire, there should be no force on it, you would say.

But suppose that you also know about special relativity. Then in the rest frame of the conduction electrons, because of length contraction, the distance between conduction electrons will be greater and that between ions smaller. (The ion-ion distance in the wire rest frame is the proper length, so it is shorter in any other frame.) So in a different frame the wire is charged. In this frame there must be an electric force between the wire and the charge outside! There was no electric force in the lab frame, so in the the lab frame this force would be something else: the magnetic force.

If you do this quantitatively you can find that the magnetic force is exactly what you expect it to be (I assume you have met with the field from a thin wire.)

This argument also shows that in relativistic theory you should not think about the electric force and the magnetic force, but rather the electromagnetic force, since different observers will disagree on what is electric and what is magnetic. In fact it was this situation that led Einstein to discover special relativity. Einstein's argument is sort of the reverse though: he showed that given that the equations of electromagnetism should be the same for all observers, we should expect special relativity. The argument above is that given that our universe is described by Einstein's relativity, it is not consistent to have only an electric force. This is historically backwards as mentioned, but I think it's conceptually the better way to think.

  • $\begingroup$ Sure, I know this relativistic argument, but I don't see, how it solves my question, which was to analyse the cited argument and not about another or better argument... $\endgroup$ – Julia Mar 31 '14 at 11:09
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    $\begingroup$ Downvote: does not answer the question. $\endgroup$ – garyp Mar 31 '14 at 13:19

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