Is the Magnetic Force Really Doing No Work?

[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\timesB]

Elementary work done :

dW= q[V\timesB].dr

=q[V\timesB].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]\timesB].[V₁+V₂]dt

=q[V₁\timesB].V₁dt + q[V₁\timesB].V₂dt + q[V₂\timesB].V₁dt + q[V₂\timesB].V₂dt
=0

q[V₁\timesB].V₁dt=0

q[V₂\timesB].V₂dt=0

q[V₁\timesB].V₂dt \neq0

q[V₂\timesB].V₁dt\neq0

dW= q[V₁\timesB].V₂dt + q[V₂\timesB].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/\muxB²xAxd

 

[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.[/color]

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/\muxB²xAxd

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.

 

[cabraham]

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.

I was not sidestepping any issue. Of course the dipole moments are important, I never denied that. The question I was addressing was how on Earth can a z-oriented B field lift an e- vertically. I was just pointing out that there is some component of B in the lateral x-y plane. The Lorentz force law involving v x B does not fail for individual e- as long as we account for thr lateral B component. That was my point.

I am well aware that orbiting e- & moving charges in general have an associated magnetic dipole moment. That has never been disputed. If you examine the moments involved you can ascertain the upward lift force. I was only demonstrating that the Lorentz law is still valid when viewing an individual e-. Some inferred that modern science cannot explain the upward lift, & that Lorentz' law was invalid, & I just addressed that issue to show that it is valid.

I don't have any beef w/ your magnetic moment discussion at all.

Claude

 

[gabbagabbahey]

The question I was addressing was how on Earth can a z-oriented B field lift an e- vertically. I was just pointing out that there is some component of B in the lateral x-y plane. The Lorentz force law involving v x B does not fail for individual e- as long as we account for thr lateral B component. That was my point.

And I feel this argument is misleading. In many cases, the horizontal components of the (external) field contributes very little to the lifting power of a magnet, and hence are often negligible (Like in the crane/car experiment). This doesn't mean that you can't apply the Lorentz force law and end up with a net vertical force from an entirely vertical field. The equation for the force on a dipole can be easily derived directly from the Lorentz force law. As can the equation for the force on a magnetizable material.

 

[UWStudent101]

Would anyone be able to give me an explanation of the magnetic forces regarding physically how they work? It is their physical property that I find fascinating rather than their equations and vector solutions. Any theories or explanations would be appreciated.

 

[quantum123]

Very interesting stuff. I happened to get an email notification and then start reading this thread. Thanks to you guys for the thought provoking discussion using various angles to look at this phenomenon, which is still rather mysterious to me.