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- Thread starter SamRoss
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In summary: It's just a vector with two components. See section 3.4.2, page 59, in this document: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.See section 3.4.2, page 59, in this document:

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etotheipi said:

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.

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.

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$$\vec{j}_m=\vec{\nabla} \times \vec{M}.$$

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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Vanadium 50 said:

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?

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Charles Link said:

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?

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SamRoss said: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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SamRoss said: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.

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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}##.SamRoss said:Does the derivation of F→1=∇(m→1⋅B→2) involve the Lorentz force? If not, what does it involve?

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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This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?

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Vanadium 50 said:This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?

B...but introductory electromagnetism

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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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I'm fine with integrals and vector operators.Vanadium 50 said:This thread is B-level. Are we sure we want to be explaining with integrals and vector operators?

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I've asked that the thread be made I-level.

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Done!Vanadium 50 said:I've asked that the thread be made I-level.

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See https://www.physicsforums.com/threa...use-a-translation-motion.981100/#post-6267508

See also the "link" in post 8 of the above thread.

See also the "link" in post 8 of the above thread.

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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.

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vanhees71 said: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.

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SamRoss said: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.

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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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}##SamRoss said:1. The ultimate source of any electric field is a charged particle.

$$(\nabla^2-\frac{1}{c^2}\frac{\partial}{\partial t})\vec{E}=\frac{1}{\epsilon_0}\nabla\rho+\mu_0\frac{\partial \vec{J}}{\partial t}$$ .

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).

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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}##.SamRoss said: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.

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alan123hk said:View attachment 272125

...and then infer the force of this magnetic field acting on the equivalent magnetization current of the other magnet.

Charles Link said: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.

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?

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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/SamRoss said: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?

and also see https://www.physicsforums.com/insights/permanent-magnets-ferromagnetism-magnetic-surface-currents/

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Charles Link said: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

I am happy if you find this usefulSamRoss said:Thanks, @alan123hk ! This seems to be EXACTLY what I was looking for!

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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.

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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?

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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.SamRoss said:are we then justified in using the right-hand rule to predict the force on imagined moving charges that aren't really there?

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Ibix said: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.

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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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