Magnetic attraction / repulsion from the Lorentz force

  • #1
SamRoss
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I'm trying to understand how the Lorentz force can explain why magnets attract and repel. The explanations that I have found have mostly involved the magnets moving in a way that decreases the forces between them ( ) but I have not been able to find any intuitive explanation involving the Lorentz force. Is there one?
 

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  • #2
Treat the bar magnets as magnetic dipoles ##\vec{m}_1## and ##\vec{m}_2##, find the magnetic fields ##\vec{B}_1## and ##\vec{B}_2## produced by these two dipoles [look up the form of this field, I can never remember the exact equation!], and then the force on one dipole due to the other dipole is ##\vec{F}_1 = \nabla (\vec{m}_1 \cdot \vec{B}_2)##, and vice versa for ##\vec{F}_2##.
 
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  • #3
SamRoss
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Treat the bar magnets as magnetic dipoles ##\vec{m}_1## and ##\vec{m}_2##, find the magnetic fields ##\vec{B}_1## and ##\vec{B}_2## produced by these two dipoles [look up the form of this field, I can never remember the exact equation!], and then the force on one dipole due to the other dipole is ##\vec{F}_1 = \nabla (\vec{m}_1 \cdot \vec{B}_2)##, and vice versa for ##\vec{F}_2##.

Thank you for your response. I am looking for a more intuitive answer based on the Lorentz force (if one exists). For example, we know that a positively charged particle will create an electric field pointing a way from it and a negatively charged particle will create an electric field pointing toward it (electric field lines are drawn in blue below). Furthermore, according to the Lorentz force, F = qE + qv x B (the second term is not really needed for this example), a negatively charged particle will feel a force (drawn in purple) which points in the opposite direction as the electric field. We can put these concepts together to understand why unlike charges attract and like charges repel.
1604184333506.png
1604184676526.png

Is it possible to draw a simple diagram like the one above to understand why magnets attract and repel? A good diagram would show 1. the magnetic field lines of the magnets 2. the velocity vectors of the relevant particles in the magnet (perhaps, rather, vectors representing spin?) 3. the force that is felt by the particles in each magnet due to the qv x B term of the Lorentz force.
 
  • #4
Charles Link
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I think post 2 is about as good as you can get for a simple answer. It is also interesting that the dipole energy gradient analysis can give a result where the force is parallel to ## B ##, where if you just look at the Lorentz force with the cross product, you might wonder if such a parallel force is possible.
 
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  • #5
vanhees71
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I don't know whether this is more intuitive, but if you take mathematical identities as intuitive (I definitely do) then it's very intuitive to note that effectively for macrocsopic electrodynamics the magnetization of a permanent magnet (which on a microscopic level cannot be understood in classical terms but is an effect of the spin of the electron and its associated fundamental magnetic moment as well as the exchange-force effect, which is a quantum effect due to the indistinguishability of the electrons being fermions) is equivalent to a current density, the "magnetization-current density"
$$\vec{j}_m=\vec{\nabla} \times \vec{M}.$$
Now you can describe the force on magnet 2 due to the presence of magnet 1 by the force on the magnetization-current density of magnet 1.
 
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  • #6
Vanadium 50
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I'm afraid that you will not be successful in your question. The Lorentz force is a force on a charge. A magnet feels a force not because it is charged but because its internal charges are in motion,
 
  • #7
SamRoss
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I'm afraid that you will not be successful in your question. The Lorentz force is a force on a charge. A magnet feels a force not because it is charged but because its internal charges are in motion,

Isn't the Lorentz force a force on a charge which is in motion? Isn't this what the v in qv x B is all about?
 
  • #8
SamRoss
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I think post 2 is about as good as you can get for a simple answer. It is also interesting that the dipole energy gradient analysis can give a result where the force is parallel to ## B ##, where if you just look at the Lorentz force with the cross product, you might wonder if such a parallel force is possible.

Does the derivation of ##\vec{F}_1 = \nabla (\vec{m}_1 \cdot \vec{B}_2)## involve the Lorentz force? If not, what does it involve?
 
  • #9
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Isn't the Lorentz force a force on a charge which is in motion?

Sure. And is a magnet charged? No. What you have is a bunch of charges in a complex motion, and you are demanding that that be simple. It won't work out.
 
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  • #10
Does the derivation of ##\vec{F}_1 = \nabla (\vec{m}_1 \cdot \vec{B}_2)## involve the Lorentz force? If not, what does it involve?

See section 3.4.2, page 59, in this document:
https://www.damtp.cam.ac.uk/user/tong/em/el2.pdf

You can imagine the dipole to be a loop of current, which in a magnetic field experiences a Lorentz force$$\vec{F} = \int_{\Omega} d^3 x \vec{J}(\vec{x}) \times \vec{B}(\vec{x})$$With the assumption that ##\vec{B}## varies slowly in the neighbourhood of the current loop, you can perform some quite complicated manipulations to end up with the desired result.
 
  • #11
vanhees71
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Does the derivation of F→1=∇(m→1⋅B→2) involve the Lorentz force? If not, what does it involve?
Sure. From a classical point of view you can derive it by calculating the force on a current-density distribution within a compact object in a given (external) magnetic field. The magnetization of a permanent magnet in this context is effectively equivalent to a current-density, ##\vec{j} = \vec{\nabla} \times \vec{M}##.

The only issue, which gets wrong in this context is when you try to take this picture too literally and make the connection between the magnetic moment and the angular momentum of the object. The classical model gives a gyro-factor ##1##, i.e., ##\vec{m}=q \vec{L}/(2\mu)##, where ##q## is the total charge and ##\mu## the mass of the particle and ##\vec{L}## the classical orbital angular momentum of the classical current density.

The issue is that the usual permanent magnets we have in everyday life get their magnetic moment not from classical molecular currents but mainly from the spins of the electrons, and their gyrofactor is close to 2 rather than 1. That was a famous blunder by Einstein, who convinced his experimental colleague de Haas that the gyrofactor he should get in the famous Einstein-de Haas effect should be 1. Indeed de Haas neglected experimental results clearly giving gyro factors larger than 1, and it has been found shortly thereafter by others that indeed it's 2. At this time (1915) of course, nobody had an explanation for this nor was the notion of spin clear at all. To the contrary there was a big puzzle related to the (anomalous) Zeeman effect for some years thereafter. It was finally solved with the discovery of electron spin in 1925 and the theoretical understanding of the gyro factor of close to 2 came with the discovery of the Dirac equation.
 
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  • #12
Vanadium 50
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This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?
 
  • #13
This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?

B...but introductory electromagnetism is the study of integrals and vector operators, with a bit of Physics sprinkled on top :wink:
 
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  • #14
vanhees71
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To really realize the qualitative explanation I gave in #11 you need vector calculus, including integrals over charge-current distributions. After all what we are discussing is the Biot-Savart Law for the force between current distributions and the dipole approximation.

I've no idea, how to simplify this calculation avoiding this level of complication, but isn't on this level the qualitative explanation good enough? It's maybe a bit dissatisfying that I cannot argue in a qualitative way, why you get the final result ##\vec{F}=\vec{\nabla} (\vec{m}_1 \cdot \vec{B}_2)##.
 
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  • #15
SamRoss
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This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?
I'm fine with integrals and vector operators.
 
  • #16
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I've asked that the thread be made I-level.
 
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  • #17
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  • #19
SamRoss
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A slightly off-topic question - are the following statements true?

1. The ultimate source of any electric field is a charged particle.
2. The ultimate source of any magnetic field is either the movement of a charged particle or the intrinsic spin of a charged particle.
 
  • #20
vanhees71
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I'm not sure, what you mean by "ultimate source". But of course the sources of the electromagnetic fields are charge-current distributions of any kind. This includes moving charged elementary particles which also often have a magnetic dipole moment like an electron or a proton. Of course "ultimately" such an object cannot be described within classical physics but you need quantum theory.
 
  • #21
SamRoss
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I'm not sure, what you mean by "ultimate source".

When I say "electric field line", I mean an imaginary line which can be drawn such that any charged particle will be accelerated in a direction parallel to the line.

When I say "magnetic field line", I mean an imaginary line which can be drawn such that any moving charged particle will be accelerated in a direction perpendicular to both the line and the original direction of the moving particle.

When I say "ultimate source", I mean that if I see particles behaving this way, I will be able to find electric charges as the source of the electric field lines and moving (in my reference frame) electric charges or electric charges with intrinsic spin as the source of the magnetic field lines.
 
  • #22
alan123hk
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I'm trying to understand how the Lorentz force can explain why magnets attract and repel

I used to have the same idea as you, that is, how to use Lorentz force to explain the attraction and repulsion between magnets. I also tried to find relevant information on the Internet, but could not find it. Therefore, I used an intuitive drawing method to solve this problem, as shown in the figure below.

Magnetic Force 1.jpg

Just as when we calculate the attractive force between the two plates of a parallel plate capacitor, we only consider the electric field generated by the charge on one plate, and then calculate the force that this electric field exerts on the charge on the other plate. Therefore, the point here is that in order to correctly apply the Lorentz force, we should consider the magnetic field generated by one of the two magnets, and then infer the force of this magnetic field acting on the equivalent magnetization current of the other magnet.

Using the same method, when we consider the magnetic field generated by magnet 1, we will also draw the conclusion of attraction, and when the polarity of one of the two magnets is reversed, we will also draw the conclusion of repulsion.
 
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  • #23
Delta2
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1. The ultimate source of any electric field is a charged particle.
I just want to comment on this from a classical physics point of view. A charged particle generates an electric field when it is stationary, but when it is moving, it generates additional electric field, as long as its movement is with acceleration. That is because from Maxwell's equations we can infer the following equation for the electric field ##\vec{E}##
$$(\nabla^2-\frac{1}{c^2}\frac{\partial}{\partial t})\vec{E}=\frac{1}{\epsilon_0}\nabla\rho+\mu_0\frac{\partial \vec{J}}{\partial t}$$ .
The RHS of the above equation is the source term and it tell us that the sources of electric field are charge density ##\rho## and time varying current density ##\vec{J}##.
So, a current density (moving charges) ##\vec{J}## generates a magnetic field, but when it is time varying (i.e ##\frac{\partial \vec{J}}{\partial t}\neq \vec{0}##) it generates additional electric field as well (together with the magnetic field).
 
  • #24
vanhees71
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When I say "electric field line", I mean an imaginary line which can be drawn such that any charged particle will be accelerated in a direction parallel to the line.

When I say "magnetic field line", I mean an imaginary line which can be drawn such that any moving charged particle will be accelerated in a direction perpendicular to both the line and the original direction of the moving particle.

When I say "ultimate source", I mean that if I see particles behaving this way, I will be able to find electric charges as the source of the electric field lines and moving (in my reference frame) electric charges or electric charges with intrinsic spin as the source of the magnetic field lines.
Field lines are always lines, whose tangents point in direction of the field. So your definition of field line is in accordance with the usual use for electric fields, because the force on a point charge due to the electric field is ##F_{\text{el}}=q \vec{E}##, but for a magnetic field it's ##\vec{F}_{\text{mag}}=q \vec{v} \times \vec{B}##, which is not parallel to ##\vec{B}##.
 
  • #25
Charles Link
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I want to comment on @alan123hk 's post 22. Very good, but one comment is for uniform magnetization, the equivalent currents are not on the endfaces, but rather along the outside cylindrical surface, as if the magnet were a solenoid. Otherwise, a very good explanation and diagram.
 
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  • #26
SamRoss
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View attachment 272125
...and then infer the force of this magnetic field acting on the equivalent magnetization current of the other magnet.

Very good, but one comment is for uniform magnetization, the equivalent currents are not on the endfaces, but rather along the outside cylindrical surface, as if the magnet were a solenoid.

Thanks, @alan123hk ! This seems to be EXACTLY what I was looking for!

A couple of quick questions to make sure I'm interpreting everything correctly...
1. The black arrows representing the equivalent magnetization current would be the v in the F = qv x B formula (and, of course, the red arrows would be the B and the blue arrows would be the F). Is that correct?
2. There is not really any current in the magnet (at least, none that contributes significantly to the magnetic field). The current depicted in the diagram is the result of the spins of the electrons in the magnet cancelling out on the inside, resulting, as @Charles Link said, in a current which is only on the outside of the magnet as in a solenoid.
1604518038407.png

Of course, the electrons are not literally spinning either but each of them produces a magnetic field as if they were. Is all of this correct?
 
  • #27
Charles Link
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Of course, the electrons are not literally spinning either but each of them produces a magnetic field as if they were. Is all of this correct?
Yes. It might interest you that the equivalent magnetic surface current per unit length ## K_m=\vec{M} \times \hat{n} ## in units where ## B=\mu_o(H+M)##. See also https://www.physicsforums.com/threads/a-magnetostatics-problem-of-interest-2.971045/
and also see https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/
 
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  • #28
alan123hk
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I want to comment on @alan123hk 's post 22. Very good, but one comment is for uniform magnetization, the equivalent currents are not on the endfaces, but rather along the outside cylindrical surface, as if the magnet were a solenoid

Thank you for pointing out the problem with the diagram.
I plan to fix this when I have the opportunity to upload it to the internet next time :rolleyes:

Thanks, @alan123hk ! This seems to be EXACTLY what I was looking for!
I am happy if you find this useful :biggrin:
 
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  • #29
Charles Link
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See also https://www.physicsforums.com/threa...perature-relationship-in-ferromagnets.923380/
I don't want to break PF rules about hijacking a thread, but I think the OP @SamRoss might find the above thread of interest. The standard textbook calculations are somewhat lacking for the forces between two permanent magnets, (it is more complex than simple electrostatic type forces), but he may be interested in seeing some calculations involving permanent magnets that are rather straightforward.
 
  • #30
SamRoss
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...but I think the OP @SamRoss might find the above thread of interest.

All new knowledge is of interest. Thank you :)
 
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  • #31
SamRoss
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@alan123hk @Charles Link

I was thinking about this thread again and one nagging thought occurred to me. I understand (hopefully) the equivalent current as the current that would produce the same magnetic field as is seen around the magnet and is in reality produced by the spin of the electrons. This current, then, is fictitious. I'm okay with imagining a fictitious current in order to "explain" the magnetic field, but are we then justified in using the right-hand rule to predict the force on imagined moving charges that aren't really there?
 
  • #32
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are we then justified in using the right-hand rule to predict the force on imagined moving charges that aren't really there?
Perhaps a better way to think of it is, for any magnet, you could replace it with a custom designed solenoid and get the same field. Then obviously you can use the usual rules for currents and fields for that equivalent solenoid.
 
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  • #33
SamRoss
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Perhaps a better way to think of it is, for any magnet, you could replace it with a custom designed solenoid and get the same field. Then obviously you can use the usual rules for currents and fields for that equivalent solenoid.

I guess the jump from two things producing the same field to those two things reacting in the same way to a nearby object is not so obvious to me. After all, the forces are supposed to act on charges, not fields. To make a weird analogy - if two people happen to draw the same waves in a Japanese-style sand garden, it doesn't mean they will each react the same way to some nearby music.
 
  • #34
Charles Link
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In the magnetization currents, there is no actual charge transport, but the magnetic field that is calculated from them using Biot-Savart is, in fact, precisely correct. In Griffith's E&M book, he does a derivation to compute the magnetic vector potential ## A ## from an arbitrary distribution of microscopic magnetic moments, and the system acts as if there were bulk currents given by current density ## J_m=\nabla \times M ## and surface current per unit length ##K_m=M \times \hat{n} ##.
To answer another question, for the system that creates the magnetic field, at least for magnetostatics, e.g. between two magnets, the force that is experienced will indeed be the same if one or both are replaced by the equivalent solenoid. This is because they will experience equal and opposite forces.
 
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  • #35
vanhees71
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The problem is that the magnetization arising from fundamental dipole moments of particles like in a permanent magnet, where the dipole moments of the electrons are aligned in macroscopic domains, cannot be fully understood within classical electrodynamics, because they are generic quantum phenomena related to the spin of the particles. However, quantum theory tells you that the magnetic moments of the particles and also the resulting macroscopic magnetization are equivalent to the magnetization current discussed above. This follows from the Hamilton operator describing the interaction of the particles, taking into account both their charge and their magnetic moment. The same Hamiltonian of course also describes the interaction of the particles with the electromagnetic field and the magnetization current leads precisely to the same Lorentz force law as usual currents of moving charges.
 
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