Is the Magnetic Force Really Doing No Work?

[Studiot]

Other than the Lorentz force from the external magnetic field.

Which was the conclusion I thought we had all agreed.

The Lorenz force cannot act in an upwards direction and therefore cannot move the car upwards.
Therefore some other force does this.

 

[gabbagabbahey]

Which was the conclusion I thought we had all agreed.

The Lorenz force cannot act in an upwards direction and therefore cannot move the car upwards.
Therefore some other force does this.

Right. So, do you now agree that magnetic forces do no work?

 

[Studiot]

In short, the attempt to couple classical matter
to electromagnetism via the point particle hypothesis (4) and the Lorentz force law (5) fails
on both mathematical and physical grounds2.
Thus, it seems clear that the point particle charge-current (4) and Lorentz force law
(5) cannot be fundamental.

Thank you C.Iblis for that reference.
The above quote from p 3 of it sums up the futility of promoting a flawed explanation over a much simpler one.

The car is not a charged object, let alone a point object.
If you want to work in terms of yanking up spinning electrons, which in turn pull the protons with them, which in turn pull the neutrons as well, then you need an interaction theory that can at least direct the yank in the correct direction.

 

[gabbagabbahey]

Thank you C.Iblis for that reference.
The above quote from p 3 of it sums up the futility of promoting a flawed explanation over a much simpler one.

The car is not a charged object, let alone a point object.
If you want to work in terms of yanking up spinning electrons, which in turn pull the protons with them, which in turn pull the neutrons as well, then you need an interaction theory that can at least direct the yank in the correct direction.

Classically, intrinsic magnetic moments (which are the dominant contributor to ferromagnetism) can be modeled as a current loop in the limit that the size of loop is shrunk to zero. Are there problems with the classical model of Electrodynamics? Yes, absolutely! Infinite self-force on a charged point particle (like an electron), for one. That doesn't alter the fact that classically, magnetic forces do no work, or that all electrodynamic forces can be derived from the Lorentz force law and Maxwell's equations.

Moreover, if you agree with the model presented in this article, even at QFT level, magnetic forces do no work. According to their model, the work on a point particle with an instrinsic magnetic moment comes from the changing rest-mass energy of the particle.

 

[Studiot]

So, do you now agree that magnetic forces do no work?

I agree that
In my model (as in the diagram) the current in the coil creates a magnetic field which in turn creates a magnetic pole on face C of the upper slug, A.
This in turn induces a temporary opposite magnetic pole on face D of the lower slug.
The two poles attract in the normal way with a force F, sufficient to lift the weight, W, of the slug, B, a distance, D, doing work WD.
The strength of the magnetic field necessary to do this can be calculated from my equation and easily related to the coil current, if so desired.

I also agree that at a molecular or sub molecular level the magnetisation is caused by a complicated interaction of the coil field and the (charged) particles constituting the slugs but that the mathematics of considering things this way is profoundly more difficult.

As an analogy do I track and log all the velocities and masses of gas particles and the container surface contours using these to calculate the total impact forces upon the coantainer walls or do I just use Force = pressure time area?

 

[gabbagabbahey]

I agree that
In my model (as in the diagram) the current in the coil creates a magnetic field which in turn creates a magnetic pole on face C of the upper slug, A.
This in turn induces a temporary opposite magnetic pole on face D of the lower slug.
The two poles attract in the normal way with a force F, sufficient to lift the weight, W, of the slug, B, a distance, D, doing work WD.
The strength of the magnetic field necessary to do this can be calculated from my equation and easily related to the coil current, if so desired.

I also agree that at a molecular or sub molecular level the magnetisation is caused by a complicated interaction of the coil field and the (charged) particles constituting the slugs but that the mathematics of considering things this way is profoundly more difficult.

As an analogy do I track and log all the velocities and masses of gas particles and the container surface contours using these to calculate the total impact forces upon the coantainer walls or do I just use Force = pressure time area?

I agree with all of this too. However, it doesn't change the fact that classically, magnetic forces do no work. $F=\frac{B^2A}{2\mu_0}$ is the approximate force of attraction, and this net force does work. However, it is not a "magnetic force".

Do you still disagree?

 

[Studiot]

I have provided a purely magnetic analysis, generating a purely magnetic force, which does work WD in lifting the car.

You have yet to provide any vertical force resulting from your extensive analysis of the interaction of the magnetic field with the current loops or point charges.

The short answer is that I don't think classical analysis can cope, except in the simplistic terms I outllined.

 

[gabbagabbahey]

I have provided a purely magnetic analysis, generating a purely magnetic force, which does work WD in lifting the car.

No, you've provided what you claim is a purely magnetic force. You later agreed that other forces must be at play inside the car and the electromagnet.

The force you consider to be purely magnetic, is no such thing according to the laws of classical electrodynamics.

You have yet to provide any vertical force resulting from your extensive analysis of the interaction of the magnetic field with the current loops or point charges.

Sure I did. My derivation earlier showed a vertical force and was based directly off of the Lorentz force law. My derivation didn't explain all the nitty-gritty microscopic details, but it didn't need to. The result was straight from the Lorentz force law, and said law says magnetic forces do no work.

The short answer is that I don't think classical analysis can cope, except in the simplistic terms I outllined.

Sure it can! I can provide you with several respected references of texts and articles, where the authors claim that all electrodynamic phenomena can be described classically, by the Lorentz force law and maxwell's equations.

Of course, I can't prove their claims, since doing so would require analyzing every electrodynamic phenomena from first principles, which would take forever. However, you've yet to provide any references or proof of the claim you just made. And a proof of your claim would only require finding a single example of an EM phenomenon that can't be derived from the Lorentz force law, and showing that it can't be derived from the Lorentz force law.

 

[Studiot]

I have now received my copy of Griffiths and have had the opportunity to look at example 5.3 as advised.

In the face of all the aspersions cast about my ability to read I note very clearly one fact.

The magnetic B field in Griffiths' example 5.3 is perpendicular to the field described by the OP here and by my example.

In particular the field is horizontal. So as Griffiths states his Lorenz force is vertical.

We have all agreed that the field described in this thread is vertical.


So I await an explanation of how a vertical Lorenz force may be generated.

The force you consider to be purely magnetic, is no such thing according to the laws of classical electrodynamics.

Incidentally I don't consider any such thing. I merely state that there is a simple macroscopic classical theory that obtains correct working answers without pages of vector mechanics. This theory is known to be invalid at the sub atomic particle level but does provide useful answers.
Any better theory, to be an improvement, must be at least self as consistent and answer questions not resolvable by the simpler method.

A quick trawl through Griffiths suggests he does not present the older methods at all.

 

[gabbagabbahey]

At this point, I'm not sure what, if anything, will convince you.

The point of Example 5.3 was to show an example where it appears that a magnetic force does work, and then prove by explicit calculation that no work was done by the magnetic field. This is exactly analogous to the crane lifting a car example you seem to be dwelling on.

From what I can gather, your problem now is that in your crane/car example, the field is now parallel to the motion of the car (Unlike in ex. 5.3 where the field was perpendicular to the motion of the loop). I don't see why you would take issue with this. Would you be satisfied if I showed you that the force on a small current loop (magnetic dipole) in a non-uniform magnetic field is parallel to the field, and that that force can be derived from the Lorentz force law? Or would you then just say that it doesn't matter since a car isn't a single magnetic dipole?

 

[Studiot]

At this point, I'm not sure what, if anything, will convince you.

You don't need to convince me of anything. I know and agree there is a more modern and comprehensive explanation of magnetism. But I also know there are yet more comprehensive and modern theories. None are yet perfect.
If you read all of my posts you would surely have realized that my comment is simply that I don't see the point of replacing one inadequate explanation with an equally dubious one. Especially since the classical approach has easier mathematics.

I am now aware that Griffiths avoids all the inadequacies of the theory he presents. For instance he confines the Hall effect to a passing comment and completely avoids a detailed classical Lorenz-point charge - current loop explanation, perhaps because he knows it can lead to incorrect conclusions.

I simply observed that to obtain full explanations of the interaction of magnetic and electric fields within matter lattices requires abandoning such theory altogether and I am far from convinced that we are yet in a position to provide such a definitive theory, despite the great strides made in the last century.

The statement that the battery does the work is disingenuous to a point.

I would say it is the power station that charged the battery.

No I would say it was the dead plants that absobed the solar energy and created the oil/coal.

No I would say it was the sun that provided solar energy.

No I would say it was the gravity coalesing the gass clouds to form the sun and start the fusion fires.

No I would say it was the big bang that created the gas clouds...

That is all really dodging the point.

 

[Anamitra]

Consider a current carrying wire placed in a magnetic field such that the magnetic field is perpendicular to it.The electron experiences a magnetic force perpendicular to its direction of motion(and also perpendicular to the magnetic field).Though the wire may be a thin one the electron has enough space to move in a direction perpendicular to the length of the wire while the axial motion continues to take place.The motion (perpendicular to the length of the wire) invites an axial Lorentz force while the original axial motion continues to generate a force perpendiculat to the length of the conductor.In general the magnetic force has two components :
1) The axial component
2) The cross-axial component
The total work performed by these two forces(the algebric sum of the works) is indeed zero since the total magnetic force is a no work force . But the two works,the axial and the cross-axial works, are available separately in two different directions. We may use them individually for technological benefit though the sum of the works is zero.Theory does not put any restriction on such an utilization.

 

[cabraham]

https://www.physicsforums.com/showthr...37#post2413237

Once again, the above thread settled the issue. For those who won't read it, here is the long & short of it.

Two wires carrying current attract/repel in accordance with the magnetic field magnitudes & directions. The H field exerts force on the free electrons in the wire, yanking them in 1 direction. The protons in the conductor's lattice structure are tethered to the electrons via electric force. The H field displaces the electrons, then the protons tag along with the electrons like an obedient shadow, tethered by the E field.

Not in the above thread, but worth mentioning is that the neutrons get tethered by the strong nuclear force, herein referred to as SNF, so that they follow the protons.

An H field does not change the KE of an electron, only its momentum. But an H field can & does change the momentum & thus the direction of the e-. The p+ in the lattice is drawn towards the displaced e- by the E force. The neutron is dragged along with the p+ by the SNF.

But the H field is in complete control. A change in the magnitude of H results in a corresponding change in the e- displacement. The E & SNF provide tethering force & follow the new location of the displaced electron. H is in charge, & E & SNF tag along due to tethering.

So it is obvious that all 3 forces participate, & that H alone cannot lift the car. E & SNF provide tethering, so that the car moves upward. But the force due to H must include that for E & SNF. The E & SNF are not doing the work. E & SNF can store energy, but cannot supply energy indefinitely. The power source energizing the electromagnet is doing the work. The H field is the controlling parameter. The magnitude of H dictates the amount of upward force on the car. E & SNF are incurred due to built in tethering inherent in the atomic structure of metals.

I don't know how to make it clearer. H all alone can only displace a moving e-, but NOT the stationary lattice, including p+ & neutrons. But E & SNF do not do the work in lifting the car, the power source energizing the magnet does. Did I explain it well?

Claude

 

[Studiot]

Once again, the above thread settled the issue. For those who won't read it, here is the long & short of it.

Nothing has been settled.
The explanations here, like the link provided in the quote, lead to a dead end.

The plain fact is that non one has provided a calculation or explanation of what direction a point charge needs to be moving, in accordance with V x B, in order to experience a displacing force parallel to the B field.
This issue is clearly demonstrated in my post#22.

Saying the energy comes from the battery is not in dispute, but it is sidestepping the issue.

 

[diazona]

The plain fact is that non one has provided a calculation or explanation of what direction a point charge needs to be moving, in accordance with V x B, in order to experience a displacing force parallel to the B field.

And no one can, because there is no such direction. That should be obvious from the equation itself.

 

[cabraham]

Nothing has been settled.
The explanations here, like the link provided in the quote, lead to a dead end.

The plain fact is that non one has provided a calculation or explanation of what direction a point charge needs to be moving, in accordance with V x B, in order to experience a displacing force parallel to the B field.
This issue is clearly demonstrated in my post#22.

Saying the energy comes from the battery is not in dispute, but it is sidestepping the issue.

What does that have to do with anything? A cross product is normal to the plane of v & B. The cross of v & B cannot be parallel to B due to the nature of the cross product. Have you had courses in physics/mechanics/kinetics?

Seriously, the thread I linked to examined the question very thoroughly, & the conclusion was that no single force, H, E, or SN, can possibly do it all alone. The lift involves all 3 forces. Fields, however, store & deliver energy. A field, H or E, cannot indefinitely do work. It must be replenished to replace the energy delivered.

An E field for instance, can attract an e-. But the energy imparted to said e- is given up by the E field, which now has less energy than before. When the e- is drawn to the positively charged region, its negative charge reduces the net charge distribution, E field magnitude, & associated energy. The battery replaced the lost energy.

The redox chemical reaction in the battery is doing ALL the work, period. The fields, H, E, & SN, simply receive & deliver energy. Neither 1 is doing all of the lifting. Actually, the H field must exert enough force on the e- so that the p+ & n0 can be tethered along. Ultimately H is in control, but E & SNF assist since H cannot exert force on p+ or n0.

It's too easy. I cannot believe that this is even debateable. The battery chemical reaction provides all the needed work/energy. The fields interact per the above. Too easy, let's move on.

Claude

 

[Studiot]

What does that have to do with anything? A cross product is normal to the plane of v & B. The cross of v & B cannot be parallel to B due to the nature of the cross product. Have you had courses in physics/mechanics/kinetics?

Have you read post#22?

You keep recounting the same catechism without offering any proof or explanation of it.

The lift force against gravity is, by definition, vertical.
The B field produced by my simplified apparatus is vertical.

Do you deny that they are parallel?

Which is why I pose the question.

What direction must v, in the Lorenz formula, take if it is to be a correct explanation?

 

[cabraham]

Have you read post#22?

You keep recounting the same catechism without offering any proof or explanation of it.

The lift force against gravity is, by definition, vertical.
The B field produced by my simplified apparatus is vertical.

Do you deny that they are parallel?

Which is why I pose the question.

What direction must v, in the Lorenz formula, take if it is to be a correct explanation?

Yes I have read post #22. Let's call the vertical direction the "z" co-ordinate. The steel object being lifted is in the x-y co-ordinate plane. The H (or B) field lines emerge from the electromagnet pole downward in the z direction, wrapping around horizontally, & then upward in the z direction into the opposite pole. Thus there is a component of B in the x-y plane. Let's say that the horizontal component of of B is in the y direction, with the e- velocity in the x direction. The cross product vXB is thus oriented along the z direction.

You seem to be implying (I don't want to put words in your mouth so correct me if you mean otherwise) that the B field is pure vertical, which makes no sense. I apologize if you mean otherwise, but that is what I'm reading in your post #22. The B field cannot be pure vertical from the poles because they must wrap around, in a solenoidal manner. Hence there has to be a horizontal component. That is what results in the upward cross-product. Since v is in the x direction, with B in the y direction, we get a force in the z direction, + or - depending on polarity.

As I said, it's pretty simple. Have I overlooked something?

Claude

 

[Studiot]

The H (or B) field lines emerge from the electromagnet pole downward in the z direction,

Exactly.

So the closer the lifting magnet approaches the object the more the vertical component, relative to the horizontal.
And when they are an infinitesimal distance apart the flux is purely vertical and cannot exert an upward Lorenz force.

So the object falls off the magnet.

Of course experience, which suggests that the closer the magnet the stronger the attraction, does not concur with this theory.

 

[gabbagabbahey]

The plain fact is that non one has provided a calculation or explanation of what direction a point charge needs to be moving, in accordance with V x B, in order to experience a displacing force parallel to the B field.
This issue is clearly demonstrated in my post#22.

As many have said repeatedly, no-one can. The magnetic force on a moving point charge is always perpendicular to both the magnetic field and the point charge's motion..

However, the fact that the magnetic field is parallel (approximately, ignoring the fringing fields) to the motion of the car in your example should be of no concern. I've already shown a calculation where the Lorentz force gives rise to a net force of attraction, which is parallel to the magnetic field (and proportional to $B^2$), for a linearly magnetizable material. The car consists of many charges which interact with each other. The external magnetic field causes a Lorentz force on each moving charge, which is perpendicular to both the external field and the charge's motion. Those lorentz forces try to alter the motion of the charges, which changes the forces of interaction between them. The net effect of those interaction forces is an upward force on the car.

 

[cabraham]

Exactly.

So the closer the lifting magnet approaches the object the more the vertical component, relative to the horizontal.
And when they are an infinitesimal distance apart the flux is purely vertical and cannot exert an upward Lorenz force.

So the object falls off the magnet.

Of course experience, which suggests that the closer the magnet the stronger the attraction, does not concur with this theory.

Have you ever had your grade school teacher sprinkle iron filings on a sheet of paper, then place a magnet underneath? The lines are curving as soon as they exit each pole of the magnet. Many lines continue in the vertical z direction, but some curve laterally. There is always a vertical as well as horizontal component to the B field. Thus v x B always has a z component. That is why the object does not fall off. Also worth noting is that the lines nearest the pole are the strongest in magnitude, also having the greatest curvature. Thus the lateral component of B is quite substantial.

Of course our understanding of magnetism is incomplete. But the same can be said for all science. With time & study, we will learn more. Right now this is what we have. There is no inconsistency here. The v x B force explains this issue very well.

Once again, this issue is so settled, debate is pointles except for understanding. The science community got this right to a level where we can make measurements. When better equipment is produced, we can take down to a more basic level. Anyway, this has been interesting & fun. Thanks to all who participated in this thread.

Claude

 

[Studiot]

Cor ain't this exciting guv!

We now have two rival 'modern' classical explanations.

And to think engineers making electromagnets before the electron was discovered produced the formula I originally presented ( and Gabbagabbahey sort of derived) without the benefit of all these formulae.

All I have ever said is that the interaction of magnetic fields with 'particles' in lattices show some effects inconsistent with regarding them as point charges with mass (or point masses with charge). Some aspects of the origins of magnetism fall into this category.

 

[gabbagabbahey]

Have you ever had your grade school teacher sprinkle iron filings on a sheet of paper, then place a magnet underneath? The lines are curving as soon as they exit each pole of the magnet. Many lines continue in the vertical z direction, but some curve laterally. There is always a vertical as well as horizontal component to the B field. Thus v x B always has a z component. That is why the object does not fall off. Also worth noting is that the lines nearest the pole are the strongest in magnitude, also having the greatest curvature. Thus the lateral component of B is quite substantial.

Your argument doesn't seem to adress Studiot's main concern. For a magnet that is large in comparison to whatever it is lifting, the fringing fields are negligable when the object is far from the edges (i.e. more or less centered). Even for the car/crane example, a uniform verticle field is a decent rough approximation. More importantly, a purely vertical field can produce a net force of attraction that is vertical on an object containing many charges (but not on a single point chrge) due to the interactions between charges. This is born out by my previous calculation in which I assumed a vertical external field.

 

[Studiot]

There is always a vertical as well as horizontal component to the B field. Thus v x B always has a z component.

Are you quite sure this is what you meant?

That V x B always has a Z component because there is always a vertical component to the B field?

 

[Anamitra]

$$\times$$$$\neq$$

Magnetic force[from the Lorentz force equation] is given by:

F=q[V$$\times$$B]

Elementary work done :

dW= q[V$$\times$$B].dr

=q[V$$\times$$B].Vdt

=0

Let us break up the velocity into two parts

V=V₁ + V₂

We make sure that V1 , V2 and B do not lie in the same plane

Work performed is given by:

dW=q[[V₁+V₂]$$\times$$B].[V₁+V₂]dt

=q[V₁$$\times$$B].V₁dt + q[V₁$$\times$$B].V₂dt + q[V₂$$\times$$B].V₁dt + q[V₂$$\times$$B].V₂dt
=0

q[V₁$$\times$$B].V₁dt=0

q[V₂$$\times$$B].V₂dt=0

q[V₁$$\times$$B].V₂dt $$\neq$$0

q[V₂$$\times$$B].V₁dt$$\neq$$0

dW= q[V₁$$\times$$B].V₂dt + q[V₂$$\times$$B].V₁dt
=0
dW=dW₁+ dW₂
=0
But
dW₁ and dW₂ are in general individually not equal to zero.

A technologist may try to get the benefits of the individual works though the total work performed is zero!

 

[cabraham]

Your argument doesn't seem to adress Studiot's main concern. For a magnet that is large in comparison to whatever it is lifting, the fringing fields are negligable when the object is far from the edges (i.e. more or less centered). Even for the car/crane example, a uniform verticle field is a decent rough approximation. More importantly, a purely vertical field can produce a net force of attraction that is vertical on an object containing many charges (but not on a single point chrge) due to the interactions between charges. This is born out by my previous calculation in which I assumed a vertical external field.

Are you quite sure this is what you meant?

That V x B always has a Z component because there is always a vertical component to the B field?

1) No, the lateral component of the B field is not negligible. It is not "fringing", it is a solenoidal field. All B fields are. As I stated there is a lateral & vertical component to the B field, even when the object is flush against the pole of the magnet. If the lateral part was negligible, why would the iron filings on the paper exhibit curvature? The solution to this problem was given to all of us in middle school. Since then we've extended it, & refined it, but the answer lies in the curvature of the flux lines.

2)What I meant & said is that v x B has a z component because there is always a lateral B component. The lateral B component crossed with the lateral velocity component results in a z force proportional to the sine of the angle between v & B. That is just elementary vector calculus. The iron filings discussed above demonstrate the curvature of the flux lines, revealing the lateral (x,y) & vertical (z) components. This issue is so well closed that debate is pointless.

All science is based upon observation & measurement. All I've presented is built upon observation & measurement, & I am not aware of my omission of pertainent details. If I have erred by omission or otherwise, please enlighten me. The x-y-z issue has been not only discussed, but proven w/ scientific findings dating to the 19th century, still valid today. What more do you want? Best regards to all.

Claude

 

[Studiot]

One thing I notice is that no one has asked for any proof/derivation of the formula I gave.

In fact it comes out very nicely using the theorems of virtual work.

Let the car be suspended in equilibrium under the magnet.

Now consider a virtual displacement, d, in the direction of the lifting force, P.

By the principle of virual work

External work done by lifting force = Internal work done by magnetic field = internal energy loss/gain

Pd = 1/2HB = 1/2x1/$$\mu$$xB²xAxd

 

[Studiot]

The H (or B) field lines emerge from the electromagnet pole downward in the z direction, wrapping around horizontally, & then upward in the z direction into the opposite pole.

The first part of this statement appears to be at variance with your own later statements. It very clearly states that the flux emerges with only a vertical component, an infinitesimal distance below the surface of the magnet.

The second part is pure nonsense, which was carried on into later posts.

How can the flux curve upwards to the opposite pole, which lies below the magnet on the object being lifted?

The field is far from solenoidal.
Instead of insulting my schooldays I suggest you visit a manufacturer of electrical machinery and study the shape of the flux field in the gaps between the poles.

Why do you act as though you have the only answers, when even your links do not work?

 

[cabraham]

The first part of this statement appears to be at variance with your own later statements. It very clearly states that the flux emerges with only a vertical component, an infinitesimal distance below the surface of the magnet.

The second part is pure nonsense, which was carried on into later posts.

How can the flux curve upwards to the opposite pole, which lies below the magnet on the object being lifted?

The field is far from solenoidal.
Instead of insulting my schooldays I suggest you visit a manufacturer of electrical machinery and study the shape of the flux field in the gaps between the poles.

Why do you act as though you have the only answers, when even your links do not work?

Boy that was rude! I never claimed to have the only answers, but rather that the scientific community has had the right answer all along. You're saying that the whole science world has been wrong for over 100 yrs. & you have it right. You, sir, are the one that claims to have a monopoly on wisdom. I am adhering to the body of work of many others over the span of 2 centuries. Let's now examine the "infinitessimal distance below the pole" & the curvature of flux lines.

There is curvature right at the pole face & even in the interior of the bar magnet. It is universally known that a magnet is a collection of a great many tiny magnets called *domains*. Each domain has a N & S pole of its own with solenoidal (wrap-around) flux lines. At the pole face, each domain at the pole-air boundary has wrap-around flux lines. So the lateral component at the pole face is quite substantial.

This is the only view that agrees with observation. Considering the wrap-around nature of flux lines for each tiny domain, the center of the bar magnet exhibits the smallest lateral flux line component. That is likely why a magnet is weakest at its center.

Of course placing 2 bar magnets together, N against S poles results in a new magnet. The center is now the junction of the N & S poles of the individual magnets, & is the weakest in terms of force. The flux lines at this new center have much less wrap around when the magnets are joined due to the geometry than they had as individual separated magnets.

Cutting a single magnet in 2 results in 2 magnets. The original center of the 1st magnet was weak, but becomes quite strong after the cut. Why is that so? The answer lies in the wrap-around nature of flux lines emanating from the poles.

As far the the shape of the rotating machinery flux field in the gap between the poles is concerned, remember that we are dealing with 2 different magnets. The rotor has flux lines emerging from its magnetic pole as does the stator. In the rotor-stator air gap, we see a flux field formed by 2 discrete magnets. The electromagnet lifting a car is the topic under discussion. A single magnet has wrap around flux lines. If you don't know that, you should not be disputing anybody. Two magnets in proximity have a flux distribution in the gap as you mentioned, but each individual magnet has its own solenoidal field. All machinery texts will point that out. Your analogy is not an apple-apple comparison.

No other view fully agrees with observation. I do not claim a monopoly on wisdom, or access to inside info, or anything like that. Minds far greater than mine have already examined this issue & published the results for all to benefit from. Until better equipment can take measurements down to a finer level, what we currently have will have to do. Cheers.

Claude

 

[gabbagabbahey]

1) No, the lateral component of the B field is not negligible. It is not "fringing", it is a solenoidal field. All B fields are. As I stated there is a lateral & vertical component to the B field, even when the object is flush against the pole of the magnet. If the lateral part was negligible, why would the iron filings on the paper exhibit curvature? The solution to this problem was given to all of us in middle school. Since then we've extended it, & refined it, but the answer lies in the curvature of the flux lines.

My point is that even a purely vertical B-field can lift a magnetizable object upwards. The vertical component of the field is actually the primary contributor to the net upward force on the car when it is lifted by the crane's magnet. The easiest way to show that is to look at the force on a single dipole, $\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})$, the dot product indicates that the component of the field that is parallel to the dipole moment is responsible for the net force on it.

 

[gabbagabbahey]

The field is far from solenoidal.

Solenoidal, in this context, means that the divergence of the field is zero, which is in agreement with one of Maxwell's equations.

 

[Studiot]

Solenoidal, in this context, means that the divergence of the field is zero, which is in agreement with one of Maxwell's equations.

I stand corrected.

 

[cabraham]

My point is that even a purely vertical B-field can lift a magnetizable object upwards. The vertical component of the field is actually the primary contributor to the net upward force on the car when it is lifted by the crane's magnet. The easiest way to show that is to look at the force on a single dipole, $\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})$, the dot product indicates that the component of the field that is parallel to the dipole moment is responsible for the net force on it.

I realize that, but the argument that B fields can "do no work" is based on the force exerted on a single electron. That is where v x B comes into play. The force on a dipole is different. No dispute there. Hopefully you will concur that an e- being yanked upward towards the magnet is being acted on by a lateral component of B, not vertical.

A dipole will indeed be attracted to the magnet in accordance with the vertical (z) component of B. This is best described by the "A" vector, which is vector magnetic potential, & B = curl A. By definition, A is the vector corresponding to the magnitude & direction that a dipole is displaced when placed in the B field.

The question involved single charged particles. But if the magnetic forces involved are dipoles, such as the rotor & stator of a rotating machine, then the dipole approach is used. I believe that we are in agreement, so why argue? Cheers.

Claude

 

[gabbagabbahey]

I realize that, but the argument that B fields can "do no work" is based on the force exerted on a single electron. That is where v x B comes into play. The force on a dipole is different. No dispute there. Hopefully you will concur that an e- being yanked upward towards the magnet is being acted on by a lateral component of B, not vertical.

A dipole will indeed be attracted to the magnet in accordance with the vertical (z) component of B. This is best described by the "A" vector, which is vector magnetic potential, & B = curl A. By definition, A is the vector corresponding to the magnitude & direction that a dipole is displaced when placed in the B field.

The question involved single charged particles. But if the magnetic forces involved are dipoles, such as the rotor & stator of a rotating machine, then the dipole approach is used. I believe that we are in agreement, so why argue? Cheers.

Claude

I disagree with the part highlighted in red. The upward force on an electron, due to its orbital motion, in an external magnetic field must come from the horizontal component of the field. However, electrons (and atoms) also have an intrinsic dipole moment, and an entirely vertical field can still produce an upward force on them as a result. Close to a large (in comparison to the object being lifted) magnet, the vertical component of the field is much larger than its horizontal components (when you are more or less centered below one of its poles). Moreover, the intrinsic magnetic moments of atoms are often much larger than the dipole moments due to the orbital motion of their electrons. When you lift iron filings, or a car, using a magnetic field, the vertical component of the field is usually the dominant factor.

We are really talking about dipoles for the crane/car example of Studiot's (Intrinsic dipole moments (spin) are the dominant source of Ferromagnetism)--- or when lifting iron filings with a fridge magnet. As far as I can tell, Studiot's main issue at this point is how, according to the Lorentz force law, the force on a dipole can be parallel to the applied magnetic field. This is the issue I think your earlier argument was failing to address. I think you were sidestepping the issue by claiming that the horizontal components of the field were responsible.

 

[Born2bwire]

One thing I notice is that no one has asked for any proof/derivation of the formula I gave.

In fact it comes out very nicely using the theorems of virtual work.

Let the car be suspended in equilibrium under the magnet.

Now consider a virtual displacement, d, in the direction of the lifting force, P.

By the principle of virual work

External work done by lifting force = Internal work done by magnetic field = internal energy loss/gain

Pd = 1/2HB = 1/2x1/$$\mu$$xB²xAxd

Nobody has ever said that you cannot extract energy from the magnetic field. I have stated this multiple times previously. The caveat is that the direct mediator for transfering this force is always an electric field in respect to the frame of the charges. As such, it is perfectly consistent to be able to use a frame of reference where we only see magnetic fields to calculate the force from the energy since classical electromagnetics follows special relativity.