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

[chingel]

The car paint scenario is not analogous. It would be, if for example you could prove that you only pushed the paint and that pushing the paint couldn't possibly do the things claimed (for example cigarette burns on the seats), then you have a case that you didn't cause the cigarette burns. Same with the current, the magnetic field can only push electrons and because it pushes them in such a way that the force is always perpendicular to motion, it cannot possibly do work on them or on anything else, because it doesn't affect anything else.

 

[cabraham]

No, what I said was:



An you then claimed to have proved otherwise with some sketches you drew:



So, to clarify, which part of my quoted statement are you claiming to have disproved and why?

Here is your quote from before:

"The Lorentz force doesn't act on the loop. Rather, it acts on the charge carriers in the loop. Its direction is perpendicular to each of the charge carrier's motion, so it does no work.""

Actually, its direction is not perpendicular to the loop motion, but rather along the direction of loop motion. Therefore Fm is doing that work. Fe, OTOH, acts normal to loop motion, along the direction of charge motion. The only logical conclusion is that Fe does work on the charge carriers, but not on the loop. Fm does work on the loop, but not on the charge carriers.

What my sketches disproved was your claim that Fm being normal to the charge carriers, it does no work on them, which is correct, but you then concluded that Fm likewise does no work on the loop, which is not supported by physics. My sketches agree w/ classical physics that Fe does work on the charge carriers in the loop, but Fm does work on turning the loop. Since Fm provides torque, & torque produces rotary motion, the torque times the angular displacement is the work done by Fm.

Fe is perpendicular to torque & loop motion, so by your own definition Fe cannot be doing work on the loop. Classical motor/generator texts affirm the Fm does the work spinning the loop, Fe does the work on charge carriers in the loop. The dipole formed by a loop carrying the induced current is due to Fe. To summarize, E & B, like Fe & Fm, are not the same entity, but then again, nor are they divorced. They work together. Wherever you find one, the other is right there next to the first, like Siamese twins.

I hope I have stated my case clearly. I will answer any questions if desired. BR.

Claude

 

[harrylin]

[..] the magnetic field can only push electrons and because it pushes them in such a way that the force is always perpendicular to motion, it cannot possibly do work on them or on anything else, because it doesn't affect anything else.

https://www.physicsforums.com/showthread.php?p=4053762#post4053683

 

[Q-reeus]

harrylin, do you know what a pulley is? Explain to be how a fixed pulley does any work without undergoing any displacement.

And I have a similar question in turn for you K^2. Back in #129 you gave a somewhat partial reply to my #124. Admittedly I had never considered precession or non-localization as relevant factors when arguing for zero electrical work being done on an intrinsic moment. However in the end I don't see it mattering. I realize you acknowledge an E field cannot take advantage of precession to do work, but you seem to imply the non-localization allows E.j type work - somehow. Fact is, magnetic response in ferromagnets is very far from the diamagnetism always predicted if real electrical work acts on perfectly conducting circulating currents. So please just explain how you arrive at, to quote from #129:

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.

What exactly are these induced currents? And how do you get them to equate to the mechanical power change when two fully magnetized magnets draw together? You are probably quite aware that the actual currents, as in translational motion of electrons - eddy currents, are exceedingly small in hard ferrites, and anyway even in reasonably good conductor ferromagnetic material will be a minor contribution to overall energy exchanges. Particularly if motions are slow, and we are free to make such motions arbitrarily slow. If these 'induced currents' are the fictitious Amperian currents arising from domain growth/reorientation, we are really back to arguing over whether E.j type work can even in principle be done on intrinsic moments. If you have some notion of 'smeared out' moment, please indicate how that squares with the fact of magnetic saturation in ferromagnetic material - especially sharp and dramatic in high susceptibility media like supermalloy.

 

[K^2]

Ah right - I meant rope, corrected now. :tongue2:

But when we are talking about B field acting on a segment of wire carrying the current, B field is playing role of pulley, not rope. It merely redirects the current, allowing the electric field applied from battery or whatever is pushing current through wire to apply a force perpendicular to the wire itself. The work here is done by applied electric field via the current with assistance of the Hall Effect. All B field does is bend the current. Absolutely no work is done by it in any sense of the word.

If you have some notion of 'smeared out' moment, please indicate how that squares with the fact of magnetic saturation in ferromagnetic material - especially sharp and dramatic in high susceptibility media like supermalloy.

Tell you what. Sit down and solve hydrogen atom in a magnetic field gradient. Now, compute expectation values for electric dipoles of each state. What happens to a net electric dipole of an atom placed in magnetic field if you have an un-paired electron in a d-orbital?

You don't need non-local eddie currents to form. You already have a nice little loop current around every single atom. And the B field gradient shifts the electron cloud. Naturally, the shift depends on orientation of electron spins. If average magnetic moment is zero, the average electric dipole will also be zero. If you have a ferromagnetic, then you'll end up with a rather strong electric dipole that will pull the magnet.

 

[Dadface]

In both parts of this twin thread one thing that a majority seem to agree upon is that the magnetic(Bqv) part of the Lorentz force cannot do work on an unconstrained moving charged particle.By "unconstrained" I mean that Bqv is the only force acting on the particle.If it is true that Bqv cannot do work on an unconstrained particle does it follow that Bqv cannot do any work at all whether it be on unconstrained particles or any sort of constrained particle(s) or systems?A lot of people here are giving the impression that it does follow and that Bqv can never do work.

Look at this in greater detail by considering real events.With such events there is usually more than one force acting on the particle such as the weight of the particle along with Bqv.When other forces are taken into account we see that the path of the particle is not circular and that depending on how the system is constucted the particle can be made to move with a component in the vertical direction.

With such movements there are changes in gravitational potential energy with work being done with or against the gravitational force.Bqv is instrumental in bringing about such energy changes and other potential energy changes such as those due to electrical forces.The fact that work can be done to change potential energy seems to have been largely overlooked here.

Now consider the following scenario.An observer is situated such that he observes a system setting up a B field,the system being observed to be at rest .A charged paricle is observed to enter the field and as a result experiences the Bqv force and follows a curved path.
Now consider the force which acts on the system that sets up the B field.According to Newton's third law forces occur in pairs which are.
1.Equal in size
2.Opposite in direction
3.Act on different bodies
4.Are forces of the same type

If these criteria are met we have a (changing)magnetic force acting on the system.Work is done on the system because it moves under the influence of the force,no matter how slight the movement.

 

[harrylin]

[..] B field is playing role of pulley, not rope.

Same question to you then: post #166. I'll give my analysis later (probably this weekend).

 

[K^2]

Same question to you then: post #166. I'll give my analysis later (probably this weekend).

I've just answered it. The work is being done by electric field that keeps these currents going. Again, the electric field that actually moves the atomic nuclei in the wires will be due to Hall Effect.

 

[Q-reeus]

Q-reeus: "If you have some notion of 'smeared out' moment, please indicate how that squares with the fact of magnetic saturation in ferromagnetic material - especially sharp and dramatic in high susceptibility media like supermalloy."

Tell you what. Sit down and solve hydrogen atom in a magnetic field. Now, compute expectation values for electric dipoles of each state. What happens to a net electric dipole of an atom placed in magnetic field if you have an un-paired electron in a d-orbital?

You tell me please - I'm no expert on such matters. But I get the impression you are claiming a strong magneto-electric dipole-dipole coupling, but this needs a lot of clarification. If it is an m-p coupling, are m and p coaxial, orthogonal, or what? Does this effect re hydrogen atom translate meaningfully and quantitatively to the case of ferromagnetic permanent magnets? And how does such have anything to do with a solenoidal E doing E.j type work on real currents? Because that's where it's supposed to be at - -dA/dt E acting on those Amperian 'currents' formally yields the minus of net mechanical power change. Where do induced electric dipole moments fit in there?

You don't need non-local eddie currents to form. You already have a nice little loop current around every single atom.

That is orbital contribution. I tried to get feedback on that before.

And the B field shifts the electron cloud. Naturally, the shift depends on orientation of electron spins. If average magnetic moment is zero, the average electric dipole will also be zero. If you have a ferromagnetic, then you'll end up with a rather strong electric dipole that will pull the magnet.

That last bit sounds really suss to me. How exactly does this induced dipole moment 'pull the magnet' so as to do the work we are talking about?

Lots of questions needing lots of answers - so fire away please!

 

[Dale]

"In physics, a force is said to do work when it acts on a body so that there is a displacement of the point of application, however small, in the direction of the force. Thus a force does work when there is movement under the action of the force."

Again, that is not the best definition of work to use, particularly for fields. Here is the usual definition:
http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1

Also already discussed in that thread, and here - the electrons are part of the wire. If I accused you of pushing my car, would you reply that you didn't act on my car but merely on the paint? :uhh:
Clearly if contact forces can act on a body then according to the definition here above which I now refer to for these discussions, you acted on my car, despite the fact that you only indirectly acted on most of the body.

There is a big difference though. The conduction electrons can easily move wrt the wire, the paint cannot. However, wheels can move wrt the car, like conduction electrons. So I like the analogy, but it would be better to talk about pushing on the wheels rather than the paint.

Consider a force which acts vertically downward on the frontmost point of the wheel and transfers some energy from an external motor to the wheel. There is also a frictional force acting horizontally on the bottommost point of the wheel, but no energy is transfered. The car moves horizontally, and yet the work is done by the vertical force acting on the wheel, not the horizontal force.

 

[Dale]

According to Newton's third law forces occur in pairs which are.
1.Equal in size
2.Opposite in direction
3.Act on different bodies
4.Are forces of the same type

If these criteria are met we have a (changing)magnetic force acting on the system.Work is done on the system because it moves under the influence of the force,no matter how slight the movement.

Newton's 3rd law is a little tricky in EM. Unless you consider the momentum of the fields it is easy to violate the 3rd law.

 

[K^2]

That is orbital contribution.

No, it is not an orbital contribution. The angular momentum here can be pointing either way. In fact, I'm pretty sure that for Iron it cancels perfectly, because it has even number of electrons, and each orbital has at least one electron in it. But even if angular momentum contributions cancel, the fact that electron spins in more orbitals point the same way will result in electron cloud getting distorted in B gradient in a way that creates an electric dipole. Simply because you have an S.B term in the Hamiltonian.

You are still trying to work with electrons as point objects with deterministic positions. If you want to understand how ferromagnetics work, you have to understand quantum mechanics. It's a purely quantum phenomenon. If you can't sit down and solve Hydrogen atom, you shouldn't be trying to understand this, you should be studying quantum mechanics.

How exactly does this induced dipole moment 'pull the magnet' so as to do the work we are talking about?

How does electric dipole apply a force on positively charged atomic nuclei? Are you really asking that?

 

[Darwin123]

And I have to think about the apparent disconnect between classical EM and the relativistic variant that lies in the effective redefinition of very many values such as E and B ;-) Does the Minkowski EM tensor preclude you from having separate fields where one (the former B field) deals only with the rotational aspects of EM and the other only the linear aspects of EM? Does the stepping from the transformational matrices of the Lorentz and Poincare groups into the tensor formulation force a reinterpretation of the distinction between rotation, boost and translation operations? (But this is certainly off-topic for this thread)

I think your reply is relevant to the OP's question. Many posts here make a distinction between electric fields and magnetic fields.
The controversy seems to concern the "fact" that a magnetic field can't do work on an electric charge, while an electric field can do work on an electric charge. However, relativity shows that the "percentage" of electric field energy and magnetic field energy varies with the observer. So your question is really whether the "fact" that a magnetic field can't do work on an electric charge is consistent with relativity. I would answer "yes", but I have to explain why.
The "fact" that a magnetic field can't do work on a charge is literally true only for one observer in an inertial frame (special relativity) or in a geodesic frame (general relativity). The OP wasn't asking about an accelerating frame or a frame that wasn't in fall. Relativistic transforms describe how different observers interpret the same series of events.
In anyone inertial frame, there is a big distinction between electric fields and magnetic fields. The diagrams shown by the OP and others were for an inertial frame by default. There was no description of a gravitational field, or accelerating magnets, or anything that implies more than one inertial frame. Because there is only one inertial frame presented in the problem, one must respect the sharp distinction between electric fields and magnetic fields.
The Lorentz force law is precisely satisfied only in an inertial frame. The fraction of force that is electric or magnetic may vary with different inertial frames. However, the Lorentz force law is precise and accurate in an inertial frame. The Lorentz force law and the work-energy theorem imply that the magnetic field can't be doing work on a free electric charge.
The questions which this thread started with really amount to asking whether the Lorentz force law is ever inaccurate in an inertial frame. I interpreted the OP's question about "classical electrodynamics" as being the self consistency of the Lorentz force law in an inertial frame. If there is any other interpretation, then maybe someone should explain it to me.
Analysis has shown that the Lorentz force law is consistent with relativity given relativistic modifications in mass. Therefore, a magnetic field can not do work on an electric charge in an inertial frame. If a magnetic field could do work on an electric field, the Lorentz force law would have to be modified.
Relativity does not say what the direction of the electric current is. The direction of the electric current is actually determined by constitutive relationships between the fields. Relativity does put constraints on what the constitutive relationships can be. However, changing the constitutive relationships can not change the basic fact. A magnetic field can not do work on an electric charge.
The key to this conundrum may be the direction of the electric current. I do believe the OP had a valid point many posts ago. A magnetic field can change the direction of an electric current without doing work. An electric field pointing in the direction of the new electric current can do work on the electric charges in this current. So a magnetic field can "enable" an electric field to do work. That is very different from the magnetic field doing work.

 

[Q-reeus]

Q-reeus: "That is orbital contribution."
No, it is not an orbital contribution. The angular momentum here can be pointing either way. In fact, I'm pretty sure that for Iron it cancels perfectly, because it has even number of electrons, and each orbital has at least one electron in it. But even if angular momentum contributions cancel, the fact that electron spins in more orbitals point the same way will result in electron cloud getting distorted in B gradient in a way that creates an electric dipole. Simply because you have an S.B term in the Hamiltonian.

You are still trying to work with electrons as point objects with deterministic positions.

No, just trying to make heads or tails of what you wrote earlier:

You don't need non-local eddie currents to form. You already have a nice little loop current around every single atom. And the B field gradient shifts the electron cloud. Naturally, the shift depends on orientation of electron spins. If average magnetic moment is zero, the average electric dipole will also be zero. If you have a ferromagnetic, then you'll end up with a rather strong electric dipole that will pull the magnet."

"nice little loop current around every atom" - well that read like orbital to me, yet you say not. If you meant spin/orbital combo, should have said so (like you have now per above). But then on to B gradient shifting electron cloud -> electric dipole. It's dawned on me now this is nothing more than saying it's the cushion/glue/mediator between intrinsic spins (overwhelmingly in ferromagnetic media) and the lattice. An irrelevancy wrt what I asked.

If you want to understand how ferromagnetics work, you have to understand quantum mechanics. It's a purely quantum phenomenon. If you can't sit down and solve Hydrogen atom, you shouldn't be trying to understand this, you should be studying quantum mechanics.

Does that apply to everyone else here, or just me? Presumably to everyone. But I don't think a course in QM is really needed to answer the basic OP question.

Q-resus: "How exactly does this induced dipole moment 'pull the magnet' so as to do the work we are talking about?"
How does electric dipole apply a force on positively charged atomic nuclei? Are you really asking that?

I think my previous comments above covers what you were really saying, and it's irrelevant to the issue I made plain. We all know there has to be such internal transferal of forces involved. The real question remains unanswered - where are electric forces performing all the E.j work - not simply acting as 'glue' between intrinsic moments (with a little orbital thrown in) subjected to *magnetic* gradient forces, and the lattice. Given your presumed QM prowess, thought you must know electric dipolar forces are irrotational in nature and incapable of providing an emf, and it's emf that is needed to do work on those Amperian 'currents'.

 

[K^2]

I'm not talking about work done on the currents. I'm talking about work done on the lattice. That work is done by electric dipole.

Yes, to compensate for that, induced EMF will do work on the currents. The energy does, ultimately, have to come from the field between magnets, after all. You cannot escape that.

 

[Q-reeus]

I'm not talking about work done on the currents. I'm talking about work done on the lattice. That work is done by electric dipole.

Yes, to compensate for that, induced EMF will do work on the currents. The energy does, ultimately, have to come from the field between magnets, after all. You cannot escape that.

We were probably talking past each other before and I just wasn't getting your meaning to various things, so sorry for any sense of impatience. I'm still not clear on what you understand by work done on the lattice above, but it can't mean stored elastic energy as Young's modulus or similar doesn't enter the discussion. If it simply means the mechanical work of magnet motion under the action of ∇(m.B) magnetic forces that are transmitted to lattice via electric dipole forces, ok I suppose it can be seen that way. The business of energy coming from the magnetic field seems straightforward, but I now retract partly my words in #117:
"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,".
The latter matter when worked through actually invalidates the former, except if one works from the artificial Gilbert model that assumes irrotational 'true magnetic dipoles' as sources. Not going to elaborate further on that here. No sign of any consensus emerging so at this point I'm ready to sign out. Have fun.

 

[harrylin]

I've just answered it. The work is being done by electric field that keeps these currents going. Again, the electric field that actually moves the atomic nuclei in the wires will be due to Hall Effect.

Sorry a phrase is not the specific case example calculation that I ask for in order to clarify what people mean with their phrases; in fact, it sounds again as if you are saying that a rope does no work! Please provide a calculation with equations.

 

[harrylin]

Again, that is not the best definition of work to use, particularly for fields. Here is the usual definition:
http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1

That's fine; as we discussed, the answer may depend on the definition that we relate to. It has to be referred to constantly.

There is a big difference though. The conduction electrons can easily move wrt the wire, the paint cannot. However, wheels can move wrt the car, like conduction electrons. So I like the analogy, but it would be better to talk about pushing on the wheels rather than the paint.

Consider a force which acts vertically downward on the frontmost point of the wheel and transfers some energy from an external motor to the wheel. There is also a frictional force acting horizontally on the bottommost point of the wheel, but no energy is transfered. The car moves horizontally, and yet the work is done by the vertical force acting on the wheel, not the horizontal force.

Thus you would say that that force does not do work on the car while I would say that it does. I won't discuss words anymore and it now appears that we have perhaps no disagreement about the physics.

 

[Miyz]

Your question was phrased generally, so I gave a general answer. If you are talking specifically about a case where there is no external E field, then yes, the only thing that can do work is the induced E field.

Thanks.
Just wanted to make sure that the B field's would induce the E field's that would eventually do the work :)

Really interesting how theses fields work.

Miyz,

 

[harrylin]

Thanks.
Just wanted to make sure that the B field's would induce the E field's that would eventually do the work :)

Really interesting how theses fields work.

Miyz,

It depends on your definition of "work"; a case example to discuss this in detail is in my post #166. However, this has not yet been discussed.

In the meantime, in a nut shell: if you pull a plow with a rope, then according to http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1 the tractor does work on the plow, while perhaps according to some people here the tractor does no work on the plow but only work on the rope which introduces E fields that eventually do the work.

 

[chingel]

It's like when the tractor does no work on the rope, no force through a distance in the direction it is moving, then how can it do work on the plow? The tractor can only affect the rope directly and if it applies no force through a distance in the direction it is moving to the rope, it cannot apply it to something on the other end of the rope.

It is not that the magnetic field does work on the electrons and then the electrons do work on the wire and people are arguing which is really doing the work, it is that the magnetic field does no work on the electrons in the first place! How can it then be doing work on the wire if it doesn't affect it at all, only the electrons?

 

[harrylin]

[..] the magnetic field does no work on the electrons in the first place! [..]

Then, same question to you: post #166
https://www.physicsforums.com/showthread.php?p=4053535#post4053535

 

[Dale]

I won't discuss words anymore and it now appears that we have perhaps no disagreement about the physics.

Then do you agree that the work done on matter in classical EM is always given by E.j?

 

[harrylin]

Then do you agree that the work done on matter in classical EM is always given by E.j?

Sorry, I haven't even considered that question. It could be a topic for this forum.

 

[Dale]

Sorry, I haven't even considered that question. It could be a topic for this forum.

It is a pretty central topic for this thread and is a key point in my post 3. I think that cabraham and I agree on the physics (agree on all forces and that the amount of work is equal to E.j), but not the semantics (work equal to vs. work done by). Until you can answer that question then I am not sure we agree on the physics.

 

[harrylin]

It is a pretty central topic for this thread and is a key point in my post 3. I think that cabraham and I agree on the physics (agree on all forces and that the amount of work is equal to E.j), but not the semantics (work equal to vs. work done by). Until you can answer that question then I am not sure we agree on the physics.

I find it of no relevance to this topic, as will become clearer from my coming analysis of my post #166.

 

[Dale]

I find it of no relevance to this topic, as will become clearer from my coming analysis of my post #166.

Please see eqts 1034-1036:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

It is of relevance to any topic involving work done on matter in classical EM, including this one.

 

[harrylin]

Please see eqts 1034-1036:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

It is of relevance to any topic involving work done on matter in classical EM, including this one.

"the rate of heat dissipation per unit volume in a conductor (the ohmic heating rate) is E.j"

The discussions in these threads is exactly not about energy that is lost in ohmic heating - that is at the detriment of mechanical work!

 

[Dale]

"the rate of heat dissipation per unit volume in a conductor (the ohmic heating rate) is E.j"

The discussions in these threads is exactly not about energy that is lost in ohmic heating - that is at the detriment of mechanical work!

While it is true that Ohmic work is given by E.j it is also true that all other forms of EM work are equal to E.j as well. This follows directly from the Lorentz force law. So E.j describes ALL work done by the classical EM fields on matter, including Ohmic and mechanical.

http://en.wikipedia.org/wiki/Poynting's_theorem#Poynting.27s_theorem

 

[harrylin]

While it is true that Ohmic work is given by E.j it is also true that all other forms of EM work are equal to E.j as well. This follows directly from the Lorentz force law. So E.j describes ALL work done by the classical EM fields on matter, including Ohmic and mechanical. [..]

[edited:] The total energy that comes out of a battery is E.j - that makes sense to me. I have no idea how (or why) you want to apply that to permanent magnets. Please show how you use it to calculate the mechanical work in the example of my post #166.

 

[cabraham]

While it is true that Ohmic work is given by E.j it is also true that all other forms of EM work are equal to E.j as well. This follows directly from the Lorentz force law. So E.j describes ALL work done by the classical EM fields on matter, including Ohmic and mechanical.

http://en.wikipedia.org/wiki/Poynting's_theorem#Poynting.27s_theorem

That is exactly what I've been saying since the 1st thread. E.J is what I think of as "the kitchen sink". Power density per E.J integrated over volume gives power. Part of said power is transferred into Ohmic heating, another part into force times distance resulting in mechanical work. To say that "E.J" accounts for the work lifting the other magnet is not incorrect because all energy must be accounted for. But E & J do not act in the direction of the motion, nor do the Lorentz force respective components. Fe is the Lorentz force due to E & it does not align with the motion & does no mechanical work lifting the magnet.

But the Fm force which lifts the magnet does the work, but the change in B field energy must be accounted for per E.J. As the lower magnet gains energy per mgh, the B field loses that amount. Without E.J, there would be no B field energy.

As I've stated before, E & B are not one & the same thing, but nor are they totally divorced. They are 2 sides of the same coin. Wherever one is doing work, the other may not be doing work on that same body, but it is transferring energy to the one that is doing the work.

E.J is indeed very important & cannot be overlooked. If your point is that w/o E.J, B cannot do anything, I reply with the following. --- Of course not! I've never disputed that E.J accounts for the mechanical power, plus thermal dissipation, plus any reactive power (motor), etc.

E & B under dynamic conditions (time varying) are like Siamese twins. Not one single person, but nor can they go their separate ways either. Wherever there is one, so there be the other.

As Master Yoda would put it:

"With time changing conditions, always both of them will you see. Without the other, neither one can ever be."

I don't argue with Master Yoda. He's way smarter than me.

Claude

 

[Dale]

[edited:] The total energy that comes out of a battery is E.j - that makes sense to me. I have no idea how (or why) you want to apply that to permanent magnets.

I want to apply it to permanent magnets because permanent magnets are governed by the laws of classical EM and E.j is a general result for the matter power density in ALL situations covered by classical EM.

 

[Dale]

E.J is indeed very important & cannot be overlooked. If your point is that w/o E.J, B cannot do anything, I reply with the following. --- Of course not! I've never disputed that E.J accounts for the mechanical power, plus thermal dissipation, plus any reactive power (motor), etc.

E & B under dynamic conditions (time varying) are like Siamese twins. Not one single person, but nor can they go their separate ways either. Wherever there is one, so there be the other.

I think we agree on the physics, just not the semantics. That is close enough to agreement to be satisfactory to me.

 

[Miyz]

As I've stated before, E & B are not one & the same thing, but nor are they totally divorced. They are 2 sides of the same coin. Wherever one is doing work, the other may not be doing work on that same body, but it is transferring energy to the one that is doing the work.

E.J is indeed very important & cannot be overlooked. If your point is that w/o E.J, B cannot do anything, I reply with the following. --- Of course not! I've never disputed that E.J accounts for the mechanical power, plus thermal dissipation, plus any reactive power (motor), etc.

E & B under dynamic conditions (time varying) are like Siamese twins. Not one single person, but nor can they go their separate ways either. Wherever there is one, so there be the other.

As I've stated before and now most of the member here in this thread agree upon.

I think we agree on the physics, just not the semantics. That is close enough to agreement to be satisfactory to me.

Glade to hear this.

The main OP question has been answered specifically multiple times and I thank all of you're efforts so far. It's much appreciated!

Regards,

Miyze.

 

[Miyz]

It depends on your definition of "work";

Transfer of energy = Ability to do work = Fd, the common definition?