Can a magnet's magnetic field perform work on another magnet?

[gabbagabbahey]

Other then Om's points I'd like to say that the magnetic field/foce does work but INDIRECLTY.

You keep saying this as though you are under the impression that "does work indirectly" is well defined. I think most physicists will scratch their heads at that statement.

A force $\mathbf{F}$ does work on an object as it moves along a path $C$ if $\int_{C} \mathbf{F}( \mathbf{r} ) \cdot d\mathbf{r}$ is non-zero. Equivalently, a force does work on an object if it changes the object's kinetic energy. These are pretty much universally accepted definitions of what it means for a force to do work on an object, and I think the vast majority of physicists would agree.

I've never seen an analogous definition for what it means for a force to do work indirectly.

 

[Miyz]

You keep saying this as though you are under the impression that "does work indirectly" is well defined. I think most physicists will scratch their heads at that statement.

A force $\mathbf{F}$ does work on an object as it moves along a path $C$ if $\int_{C} \mathbf{F}( \mathbf{r} ) \cdot d\mathbf{r}$ is non-zero. Equivalently, a force does work on an object if it changes the object's kinetic energy. These are pretty much universally accepted definitions of what it means for a force to do work on an object, and I think the vast majority of physicists would agree.

I've never seen an analogous definition for what it means for a force to do work indirectly.

See here's the problem. How can I prove to you exactly magnet field/forces does work would be very diffecult but simple words could at least set you're in motion to think about this matter.

Based on Maxwell-Faraday's law "A changing magnetic field creates an electric field". Now that show's to you that a magnetic field/force DOES something alteast if we can't say that it does work. Without B , E = 0. B creates E so in a sense B as a force creates another force that does work... Pretty interestings stuff. Now about work being done indirectly that seems to confuse you. Read this point from Wikipedia:

"Because the magnetic force is always perpendicular to the motion, the magnetic field can do no work on an isolated charge. It can only do work indirectly, via the electric field generated by a changing magnetic field. It is often claimed that the magnetic force can do work to a non-elementary magnetic dipole, or to charged particles whose motion is constrained by other forces, but this is incorrect[19] because the work in those cases is performed by the electric forces of the charges deflected by the magnetic field."

The link.[/PLAIN]

Miyz,

 

[gabbagabbahey]

Based on Maxwell-Faraday's law "A changing magnetic field creates an electric field". Now that show's to you that a magnetic field/force DOES something alteast if we can't say that it does work. Without B , E = 0. B creates E so in a sense B as a force creates another force that does work

If I press a button which causes a rocket to launch into space am I (or the force I apply to the button) indirectly doing work on the rocket? Without me pushing that button, the rocket doesn't get launched.

 

[Miyz]

If I press a button which causes a rocket to launch into space am I (or the force I apply to the button) indirectly doing work on the rocket? Without me pushing that button, the rocket doesn't get launched.

Yes. That does support my point.

 

[cabraham]

That's what I keep posting in this strange thread over and over. The current density associated with magnetization is given by
$$\vec{j}_{\text{mag}}=c \vec{\nabla} \times \vec{M}.$$
Together with this definition also for the interaction between permanent magnets the Poynting theorem holds, and this clearly says that the electric field is responsible for the work done on charges. Of course there is the energy density of the electromagnetic field, $$T_{\text{em}}^{00}=\epsilon=\frac{1}{2}(\vec{E}^2+\vec{B}^2),$$
which in this three-dimensional notation is split into an electric and a magnetic part, and of course there is energy exchange between the electric and magnetic parts. Still, the power done on the charges is given by
$$P=\int \mathrm{d}^3 \vec{x} \vec{E} \cdot \vec{j}.$$
Thus, only the electric field is responsible for the exchange between energy of the electromagnetic field and charges.

You're misinterpreting the equation. The power described in the equation is indeed what results in work done. But the E force is not directed upwards lifting the lower magnet, B is that force, more strictly Fm, where Fm = q*(uXB). The magnetic component of Lorentz force lifts the magnet. But said B field gets energy from E field under dynamic conditions, so E interacts.

Why does E do work on an e- but not B? It has to do with direction of force vectors. In a loop where a time changing B induces an E & J in the loop, charges move around the loop (closed for this example). E acts tangentially to the loop so E force moves the e- around the loop. B force acts normal. Although E does the work, w/o B, there is no E at all. B²/2μ is energy density. If this quantity varies w/ time, we have non-zero power density. Hence it can be argued that the energy in the B field accounts for the work, which is true. But E is what does the work since the E force is oriented in a direction to do work on the e- whereas B is not.

No we look at the magnet being lifted. In this case E certainly can be equated with work per the integral above. To say that the equation you posted accounts for the work done would be correct. But only magnetic component of Lorentz force is oriented in a direction to lift the magnet. E.J provides energy for B to do the work.

These questions are as much about logic as they are about fundamentals. Also, some core concepts need to be emphasized. One thing I don't agree with is the idea that E does work but B cannot in a general sense. E & B are not one & the samr entity of course, but neither are they totally divorced. They are 2 sides of the smae coin. In dynamic conditions, neither is prsent w/o the other. When 1 does work, the other is exchanging energy with the 1 doing work. They are a pair working in tandem.

For the magnet, Fm points the right way, so it does the work, not Fe. Thanks, & BR,

Claude

 

[OmCheeto]

All the way to zero (for an ideal or point dipole). There's no problem computing the cross product of two vector fields at a point, provided both the vector fields are defined there.

Ok.

The microscopic details of what goes on inside a material are horrendously complicated, and not really well described by classical theory. Usually all we care about are macroscopic effects, and so you use the dipole moment per unit volume averaged over many thousands of atoms, the so-called magnetization $\mathbf{M}$ to calculate the macroscopic fields and forces.

In the presence of an external magnetic field, each magnetic dipole associated with the spin of an unpaired electron in a material will experience a torque which tends to align them with the external field (the force per atom holding the unpaired electron to its atom is typically much larger than this torque, so the electron stays bound to its atom/molecule and the external field then tends to flip the entire atom/molecue) this is the mechanism behind paramagnetism. Other internal forces and heat make it so this alignment is never 100%, and what you end up with is some average dipole moment per unit volume. Other processes such as diamagnetism (flipping of the orbital magnetic moment due to the change in speed of the orbiting electron) and ferromagnetism also will result in some average dipole moment per unit volume.

The net force and torque on an object in an external magnetic field can then be calculated by summing up (integrating) the force on each tiny bit of Magnetization.

This seems to make sense.
But what causes the torque? Should I trust the answers I've read from people I respect the most?

There are lots of these "Does the magnetic field do work" threads here at the forum, and many of the mentors seem to have implied that it is in fact the magnetic fields that interact and do work.

2004, 2006-8, etc.

If you can find that source again, maybe you can reference some more of it. I've never seen a classical situation where treating dipoles as current loops does not give the correct force and torque on a magnet.

It may have been me adding together the comments of several people:

Another example is the electron by itself. It has never exhibited any behaviour suggesting it is a composite particle. We can investigate a model where its magnetic moment arises from an extended distribution of charge, spinning at the appropriate rate. But we find that we end up with no coherent picture.

[ref]

The electron magnetic moment, which is responsible for ferromagnetism, is a relativistic QM effect that is not related to any current or "current".

[ref]

As I understand, it was Ampere's "model" that developed the virtual current as the equivalent needed to generate the magnetic dipoles. (ie. Nobody quote me on this!)

 

[gabbagabbahey]

Yes. That does support my point.

You don't find that (indirect work) to be an absurd concept? The energy it takes to launch a rocket into orbit is typically many many orders of magnitude higher than the energy it takes to push a button. Not only that, but the energy expent pushing the button does not get transferred to the rocket, it just temporarily completes an electrical circuit allowing a signal to propagate through.

 

[gabbagabbahey]

Ok.

This seems to make sense.
But what causes the torque? Should I trust the answers I've read from people I respect the most?

There are lots of these "Does the magnetic field do work" threads here at the forum, and many of the mentors seem to have implied that it is in fact the magnetic fields that interact and do work.

2004, 2006-8, etc.

I disagree with those viewpoints (classically, anyways) as they require you to treat magnetic dipoles as fundamentally different from the other two types of sources in classical electrodynamics - charges and currents.

It may have been me adding together the comments of several people:

[ref]

[ref]

As I understand, it was Ampere's "model" that developed the virtual current as the equivalent needed to generate the magnetic dipoles. (ie. Nobody quote me on this!)

So basically, the problem is that classical electrodynamics doesn't correctly predict/explain the quantum behaviour of permanent magnetic dipoles?

That certainly is a problem, and its one that requires quantum electrodynamics to solve. Treating magnetic dipoles as current loops is not the source of the problem, and treating them as a different type of source/sink does not solve the problem classically. I see no reason why these problems should be used as an argument for treating magnetic dipoles, classically, as being fundamentally different from current loops.

 

[OmCheeto]

I disagree with those viewpoints (classically, anyways) as they require you to treat magnetic dipoles as fundamentally different from the other two types of sources in classical electrodynamics - charges and currents.

Given that dipoles are dipoles, and charges are charges, I don't understand why you wouldn't treat them differently. That's like saying a stick and a ball are the same thing.

So basically, the problem is that classical electrodynamics doesn't correctly predict/explain the quantum behaviour of permanent magnetic dipoles?

That certainly is a problem, and its one that requires quantum electrodynamics to solve. Treating magnetic dipoles as current loops is not the source of the problem, and treating them as a different type of source/sink does not solve the problem classically. I see no reason why these problems should be used as an argument for treating magnetic dipoles, classically, as being fundamentally different from current loops.

So we have nothing more to discuss, as this is all opinion now?
Ok.
Thank you everyone. This has been most interesting.
I'm off to study Quantum Electrodynamics.

But I'll be watching.

 

[gabbagabbahey]

Given that dipoles are dipoles, and charges are charges, I don't understand why you wouldn't treat them differently. That's like saying a stick and a ball are the same thing.

Would you also treat a uniform sphere of charge as being fundamentally different from a uniform cube of charge? If so, your version of classical electrodynamics would have a whole lot of fundamental sources/sinks and force laws.

A magnetic dipole can be modeled (classically) as a specific type of current distribution, so why not treat it on the same footing as every other type of current distribution?

 

[harrylin]

If I press a button which causes a rocket to launch into space am I (or the force I apply to the button) indirectly doing work on the rocket? Without me pushing that button, the rocket doesn't get launched.

This has been discussed before - but perhaps not sufficiently explicit and elaborate?

The kinetic energy of the rocket is not supplied by you pushing the button.
In contrast, earlier we found that the kinetic energy of the magnets must be supplied by their magnetic fields; there is nowhere else that the energy could come from. Any induced currents are induced by those magnetic fields.

 

[Q-reeus]

There have been too many entries since my last one to start answering them all. Apart from OmCheeto, the rest of you are imho just not getting the real issue here. Despite what some maintain, intrinsic magnetic moments are very different in character to what otherwise might be an equivalent classical loop current (quite apart from the g factor issue). IF by some magic a classical loop current could resist responding to the solenoidal component of an -dA/dt E field, then sure it would act no differently to an intrinsic moment in terms of torque and force experienced in an external B field. But the majority here are insisting actual electrical work - integrating E.j over time and volume of magnetized media - is done on the constituent magnetic dipoles.

Yet no-one has come forth and met my challenge, in e.g. #45, #61, #66, #94, and even attempted to offer an explanation of where this so-called electrical work on a dipole appears. It does not. There is no E.j type work done on a real intrinsic dipole. If there were, the notional circular current in that dipole must respond to the solenoidal component of E by altering magnitude - SAVVY!? You know - loop current as inductor. And if the current alters magnitude, so also the dipole moment strength - SAVVY!? IT DOES NOT HAPPEN FOLKS - IT REALLY DOESN'T. There's this thing called quantization of magnetic moment, and it really does get in the way of what would otherwise be weak diamagnetism, as I have tried to make plain umpteen times now. The energy exchange between say two approaching bar magnets are drastically different to weak diamagnetic response precisely because no *actual* appreciable E.j type work is done.

By pretending those Amperian 'surface currents' via c∇×M are real, flowing currents, a purely formal equivalence between changed magnetic field energy and 'electrical work done' on those Amperian currents is found. An absurd, book-keeping-only balance given the induced E field never alters the magnitude of those Amperian currents, in say the case of two fully magnetized co-axial bar magnets that draw together - as per my unanswered challenge in #94. And the further absurdity is this 2-part balance conveniently ignores the additional player in the game - mechanical energy change. Since when do 3 equal amplitude scalar quantities balance out to a total of zero? It's only by recognizing that an intrinsic magnetic dipole moment totally ignores the solenoidal E in E.j, and thus has zero electrical work done on it, that one recovers the sane result there are only two real players in the energy balance: mechanical energy change + magnetic field energy change = 0. Well, as I have hinted at several times, there is a subtle internal energy issue, but not one germane to what is the issue here. So, a further round of madness ahead is it?

 

[Dale]

You are adopting a logically inconsistent position. You claim to agree with Maxwell's equations but disagree that E.j does work. The former implies the latter.

 

[Q-reeus]

You are adopting a logically inconsistent position. You claim to agree with Maxwell's equations but disagree that E.j does work. The former implies the latter.

That rubbish ignores the crucial role of quantization and you know it - otherwise you would have responded with the point-by-point rebuttal I asked of you much earlier but as expected that got conveniently ignored. Anyway, don't come back sniping at me with such silly claims, at least until you provide that retraction as per my first para #94. Clear that matter!

 

[Dale]

That rubbish ignores the crucial role of quantization and you know it

Yes, of course it ignores the role of quantization. Classical EM is a non-quantum theory. But it isn't crucial here, permanent magnets are accurately modeled by classical EM simply by using an empirically determined constituitive relationship between B and H, as described above and in Jackson and any other EM textbook.

If you believe that permanent magnets are not accurately modeled by the appropriate constituitive relationship then please provide good evidence to that effect. Meaning a rigorous derivation or a mainstream scientific reference.

Anyway, don't come back sniping at me with such silly claims, at least until you provide that retraction as per my first para #94. Clear that matter!

OK, I am glad to learn that I was in error with my mistaken belief that you did not accept the equations of classical EM.

Since you do accept classical EM and since classical EM logically implies that the work done is given by E.j then we must logically be in agreement.

 

[Q-reeus]

Yes, of course it ignores the role of quantization. Classical EM is a non-quantum theory. But it isn't crucial here, permanent magnets are accurately modeled by classical EM simply by using an empirically determined constituitive relationship between B and H, as described above and in Jackson and any other EM textbook.

So you keep saying. But see below.

OK, I am glad to learn that I was in error with my mistaken belief that you did not accept the equations of classical EM.

Well I'm formally obliged to accept that as a retraction. Fine - even; thank you. But, being suspicious minded I can't help but see a trap of sorts...

Since you do accept classical EM and since classical EM logically implies that the work done is given by E.j then we must logically be in agreement.

Nice try. If this 'inextricable linkage' is valid, here's my request to you. Consider the following simple scenario. We have two magnetic dipoles of equal moment m and orientation along x axis. One, labelled m_c, is a classical, perfectly conducting loop current. The other, labelled m_q, is an electron intrinsic magnetic moment. Assume for both some effective enclosed area a and mean circulating current I such that |m_c|, |m_q| = Ia. Somewhat iffy in the intrinsic moment case, but we have to somehow allow the conception of a possible E.j style work being done on it. A uniform ramping B field B_x = xB₀*kt is applied (x the unit vector along x), such that both m_c and m_q are nominally subjected to the same emf ζ = a(-dB/dt) = -aB₀k.

You must know from my frequently stated position what will be the contrasting behavior. m_c will act as a shorted perfect inductor - a ramp current I' = I₀'k't will be induced such that the enclosed flux ∅ = aB_av is invariant wrt time t - diamagnetic behavior. This is consistent with work being done on m_c - there is a continual building up of a magnetic field energy owing to the ramp I'. Hence m_c' = m_c(1+I'a/m) is time varying. I maintain zero E.j style work is done on m_q - there is no equivalent ramp current to I' in the classical loop current. So m_q is time invariant.

Your position logically insists E.j style work must be done on m_q. Explain please - really explain how.

 

[Dale]

Nice try. If this 'inextricable linkage' is valid, here's my request to you. Consider the following simple scenario. ...

The inextricable linkage is valid. Please see:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html
or (for a treatment involving the "macroscopic" quantities)
http://www.phy.duke.edu/~rgb/Class/Electrodynamics/Electrodynamics/node33.html

If you believe that it is not valid despite the proofs then it is up to you to provide good evidence supporting your belief. It is not up to me to chase down every unsubstantiated claim you can manufacture. My opinions are based on the best evidence I have available, if you have some additional evidence I have overlooked, then provide it.

 

[K^2]

Q-reeus, you are picturing an electron as an actual ball of something. Perhaps, an infinitely small one. But it's not. Regardless of how you decided to localize an electron, by either considering an electron around a nucleus, or placing it in a cavity, it is a distributed particle. And that distribution will have angular momentum and corresponding magnetic moment and current separate from intrinsic moment of an electron. When you start increasing an applied magnetic field, as you describe above, the electron's state will change, resulting in a current. Simplest case here is hydrogen atom. If you apply an increasing B field, you can cause the electron to transition to higher energy levels. These will have an overall magnetic moment. So in the end, you have exactly the same time-varying magnetic moment. It simply happens to increase in discrete jumps, because, well, Quantum Mechanics. If you do 3D particle in a box, you'll see same behavior. Though, it's a bit more difficult to describe mathematically due to symmetries.

Ultimately, any elementary dipole should be viewed as a particle field. You don't have just the spinor. You also have a spatial distribution of some sort. And whenever any work is being done on your elementary dipole, it will always be done via that spatial distribution. All a bare spinor can do by itself is precess in a magnetic field. No energy is transferred in spinor precession. So any work being done on an elementary dipole is still going to be E.j work done on the particle field.

 

[Q-reeus]

Q-reeus, you are picturing an electron as an actual ball of something. Perhaps, an infinitely small one. But it's not. Regardless of how you decided to localize an electron, by either considering an electron around a nucleus, or placing it in a cavity, it is a distributed particle. And that distribution will have angular momentum and corresponding magnetic moment and current separate from intrinsic moment of an electron.

Am I right in thinking you are referring to orbital contribution here? I recognize such does contribute, but in ferromagnetic media only to a relatively quite small degree. If you read back through my postings, I make it clear orbital contribution can be quite significant in ferrites. But there orbital contribution response to an applied B is strongly ferromagnetic in character, far from the diamagnetism expected if classical E.j work is done on a lossless, classical circulating current. if you meant 'dressed' vs 'bare' electron contribution to intrinsic spin, I'm not sure how this translates into explaining the known fact of magnetic saturation - implying rather strongly intrinsic moments are the chief contribution to ferromagnetism. Please elaborate.

When you start increasing an applied magnetic field, as you describe above, the electron's state will change, resulting in a current. Simplest case here is hydrogen atom. If you apply an increasing B field, you can cause the electron to transition to higher energy levels. These will have an overall magnetic moment.

Don't question that transitioning will occur, but to what extent in response to how great an applied B?

So in the end, you have exactly the same time-varying magnetic moment.

Not sure what you mean by that - magnetic susceptibility of hydrogen - say as liquid, is tiny and iirc right, diamagnetic. [oops - that refers to diatomic hydrogen, and you meant atomic. This sounds like spin-orbit coupling which gets tricky.]

It simply happens to increase in discrete jumps, because, well, Quantum Mechanics.

Which if you follow my #45, is a key argument imo against the notion that literal E.j work can account for energy changes, or rather, be equated to the negative of net magnetic field energy change. Pick an interval 'between jumps' and where can E.j work be invoked?

Ultimately, any elementary dipole should be viewed as a particle field. You don't have just the spinor. You also have a spatial distribution of some sort. And whenever any work is being done on your elementary dipole, it will always be done via that spatial distribution. All a bare spinor can do by itself is precess in a magnetic field. No energy is transferred in spinor precession. So any work being done on an elementary dipole is still going to be E.j work done on the particle field.

Doubtless you are far more clued up on QM here than my rudimentary grasp. Just basically, what is the net result then on the intrinsic moment that can be equated to E.j work done? Doesn't it get back to overall susceptibilities? Huge in ferromagnetic media, tiny and of the opposite sign in most other materials. As mentioned above, what I have read suggests intrinsic moment is overwhelmingly the dominant source contribution in ferromagnetic media. Classical loop currents, for which E.j follows readily enough, always predict weak diamagnetism, and it seems absurd to suppose one could ever construct a permanent magnet from such in the first place. But then, I do have much to learn, so fire away please.

 

[gabbagabbahey]

This has been discussed before - but perhaps not sufficiently explicit and elaborate?

The kinetic energy of the rocket is not supplied by you pushing the button.
In contrast, earlier we found that the kinetic energy of the magnets must be supplied by their magnetic fields; there is nowhere else that the energy could come from. Any induced currents are induced by those magnetic fields.

My point was that the concept of indirect work was, if not completely absurd, certainly useless.

As another example, consider pushing on an accelerating train car, perpendicular to its motion. Unless you have superhuman strength, the train car is not going to move off the tracks, but will continue along as if you weren't there. You are still transferring energy to the train car by pushing on it, but you are certainly not doing any work on it since the force you apply is perpendicular to its motion.

To me, it is useless to say "the magnetic field does work indirectly".

Much better is to simply say that the magnetic field does no work since it is everywhere perpendicular to the force it applies on a charge or current distribution. The concept of work is well defined. The concept of "indirect work" is not.

 

[cabraham]

My point was that the concept of indirect work was, if not completely absurd, certainly useless.

As another example, consider pushing on an accelerating train car, perpendicular to its motion. Unless you have superhuman strength, the train car is not going to move off the tracks, but will continue along as if you weren't there. You are still transferring energy to the train car by pushing on it, but you are certainly not doing any work on it since the force you apply is perpendicular to its motion.

To me, it is useless to say "the magnetic field does work indirectly".

Much better is to simply say that the magnetic field does no work since it is everywhere perpendicular to the force it applies on a charge or current distribution. The concept of work is well defined. The concept of "indirect work" is not.

But in the OP case the magnetic field is acting directly in line with the motion, upwards. E fields are induced once motion is realized, but said E forces do not act in the direction of the motion, but normal to it. Your train car analogy is exactly what I've been trying to illustrate, but with limited success. Since a force acts along the motion, it can be said to have done work. When a force acts normal to motion, it does not do work.

So all we have to do is, hold on, are you ready:

draw a picture!

Sorry to be so repetitive, but until that happens, we will argue in vain. Just so you & I are clear on this, I totally agree with your train car explanation. It is indeed the force acting with the motion that does work. Any force(s) acting normal to motion do no work. We agree at least on that point.

Now if we can only establish the direction of the various force vectors. I drew pics in the motor thread, & my opponents dismissed them w/o offering their own pics. I see little point in submitting pics since nobody will change their mind. Tonight I may be tempted to when I get home. In the meantime, if anybody can make a sketch, I would greatly appreciate it.

Claude

 

[gabbagabbahey]

But in the OP case the magnetic field is acting directly in line with the motion, upwards.

The magnetic field is parallel to the motion. The magnetic force applied is perpendicular, according to the Lorentz Force Law.

 

[cabraham]

The magnetic field is parallel to the motion. The magnetic force applied is perpendicular, according to the Lorentz Force Law.

Can you please illustrate? The lines of force around the magnet bend as they enter/exit the poles. Also, to compute the Lorentz force, Fm, per uXB, can you describe the orientation of u & B? Thanks.

Claude

 

[K^2]

Doubtless you are far more clued up on QM here than my rudimentary grasp. Just basically, what is the net result then on the intrinsic moment that can be equated to E.j work done?

None. When we are talking about one magnet interacting with another, no work is being done on the intrinsic moment. Like I said earlier, all a spinor can do is precess, and there is no work involved in that. The B.m angle remains constant. All of the work is done via actual currents. Yes, these don't contribute much to the magnetic field of a ferromagnetic in a steady state, but that's where all of the work will be done. Induced currents.

Of course, when you look at macroscopic picture, you can't really tell one from the other. So you can still describe it classically as classical magnetic moments with current loops.

So while you are absolutely right that an elementary dipole of a fermion cannot be adequately described as a classical current loop, it doesn't really lead to any violations. B field still doesn't do any work.

 

[Miyz]

B field still doesn't do any work.

So you agree the point that the B field creates/generates an electrical field that would do the work?

 

[K^2]

So you agree the point that the B field creates/generates an electrical field that would do the work?

As one possibility. In principle, work could be done entirely by an external E field, with B field merely assisting. Something along the lines of Hall Effect, for example, where there is no induced E field, yet amount of work done depends on whether the B field is present.

 

[Miyz]

As one possibility. In principle, work could be done entirely by an external E field, with B field merely assisting. Something along the lines of Hall Effect, for example, where there is no induced E field, yet amount of work done depends on whether the B field is present.

Im not sure what you're saying... Is that a yes or a no :tongue:. I am guessing that its a maybe.

 

[K^2]

I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.

 

[cabraham]

I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.

That makes no sense at all. If B cannot do work, only E can do work, then how can you say that B field does not have to induce an E field for work to be done in a system? Your 2 statements are contradictory.

When you say that "work can never be done by the magnetic field", that is the OP question under scrutiny. Why does B field never do work? Please tell me the reason, because the OP wants to know if & why, & so do I. Again, would you draw a picture showing the force that lifts the magnet? I don't believe the force lifting the magnet is Fe (electric component of Lorentz force), but rather Fm (magnetic component of Lorentz force). It is Fm that lifts the magnet.

Claude

 

[K^2]

Power (work per unit time) is given by F.v. The only force in electrodynamics is the Lorentz Force, F=q(E + v x B). So dW/dt = q(E + v x B).v = q E.v + q(v x B).v = E.j + q B.(v x v) = E.j

The term containing B field cancels, because any vector crossed with itself is zero. (I use property a.(b x c) = b.(c x a) = c.(a x b) )

That makes no sense at all. If B cannot do work, only E can do work, then how can you say that B field does not have to induce an E field for work to be done in a system? Your 2 statements are contradictory.

E that does work can be external. Again, think of Hall Effect as an example. Work is done by E field, but the amount of work done can depend on the B field.

 

[cabraham]

Power (work per unit time) is given by F.v. The only force in electrodynamics is the Lorentz Force, F=q(E + v x B). So dW/dt = q(E + v x B).v = q E.v + q(v x B).v = E.j + q B.(v x v) = E.j

The term containing B field cancels, because any vector crossed with itself is zero. (I use property a.(b x c) = b.(c x a) = c.(a x b) )


E that does work can be external. Again, think of Hall Effect as an example. Work is done by E field, but the amount of work done can depend on the B field.

But I said that the force doing work is Fm, not B. B does indeed cancel, but vXB does not. It should be well understood that B acting in the direction of motion will indeed do no work since the Fm vector points normal to B. Once again, are we defining work as force integrated over path? If so, it is universally known that B itself is not acting along the path of motion, but rather Fm is. B will drop out of the above equation, but Fm will not.

When I say that "B does the work", what I infer is that the force acting in the direction of motion is Fm = qvXB. Then as we know, B itself is normal to the motion. The Fm term does not drop out AFAIK. Am I mistaken? Also, nobody answered my previous question.

Which force lifts the magnet? It has an upward direction. Is it Fe = qE? Is it Fm = qvXB. If it is the former, where does E originate? Thanks in advance.

Claude

 

[harrylin]

But I said that the force doing work is Fm, not B.[..]

1. I find the induced current explanation suspect: a pair of magnets doesn't warm up noticeably and doesn't act like a shock damper.

2. At first sight Wikipedia agrees with you:
"In the physically correct Ampère model, there is also a force on a magnetic dipole due to a non-uniform magnetic field, but this is due to Lorentz forces on the current loop that makes up the magnetic dipole."
- https://en.wikipedia.org/wiki/Force_between_magnets

Regards,
Harald

 

[harrylin]

Now that I thought some more about this, I conclude that the induced field explanation is suspect for another, more fundamental reason. In my book, work is done by a force along the line of force. Work is not done by another, induced force, that replaces the driving force! (How could I not realize that earlier? ).

Moreover, at the start, at v=0, there can only be a magnetic force that drives the magnetic attraction between the two permanent magnets, as there can be no induced electric force at that time. Moreover, the magnetic force can't be instantly replaced by an equal induced electric force at the start of motion either - that's just nonsense.

Thus, the explanation according to which induced electric field does the work instead of the magnetic field cannot possibly be right.

 

[cabraham]

Now that I thought some more about this, I conclude that the induced field explanation is suspect for another, more fundamental reason. In my book, work is done by a force along the line of force. Work is not done by another, induced force, that replaces the driving force! (How could I not realize that earlier? ).

Moreover, at the start, at v=0, there can only be a magnetic force that drives the magnetic attraction, as there can be no induced electric force at that time. Moreover, the magnetic force can't be instantly replaced by an equal induced electric force at the start of motion either - that's just nonsense.

Thus, the explanation according to which induced electric field does the work instead of the magnetic field cannot possibly be right.

I said the same earlier. At the time up to the lower magnet ascending, the upward force can be only Fm, not Fe, because prior to motion, no E field has been induced yet. We seem to agree.

One thing I would like to mention is that in addition to laws of physics, we should observe rules of logic as well. I can't understand how anybody can say that the moving magnet's induced E field is the source of its own lifting force. The cause is happening after the effect it seems to suggest. Thanks for your input. BR.

Claude

 

[Miyz]

I'm saying that B field doesn't have to actually induce an electric field for work to be done in the system. But work will always be done by electric field, and never by the magnetic field. So an E field must be present in order for work to be done. Whether B field is the source of that E field is a different question.

That's just wrong K...

The magnetic field does create/generate an electrical field. A basic contradiction of you're words is a motor, without B, the electrical force that does work on the loop is not generate and no work is done! And based on Faraday-Maxwell's equation that states: "A changing magnetic field creates an electric field".