Can a magnetic fields/forces do work on a current carrying wire?!

gabbagabbahey

Quote by cabraham

I say with firm conviction that E.J must be based on total E, not just inside wire.

The inner product between two vector fields is a scalar field; its value, just like the two vector fields, depends on position. If either of the two vector fields is zero at a specific location, their inner product will also be zero there.

Maxwell's equations, and the Lorentz Force Law both deal with vector fields. If you don't understand how to take the inner product of two vector fields, you don't fully understand Maxwell's equations.

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

Dotini

Quote by gabbagabbahey

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

Do the followers of Gauss-Ampere-Weber make up the remaining 0.1%?

Respectfully submitted,
Steve

gabbagabbahey

Quote by Dotini

Do the followers of Gauss-Ampere-Weber make up the remaining 0.1%?

Respectfully submitted,
Steve

I wouldn't count them as part of the physics community. I was thinking more along the lines of actual physicists who have just spent too much time staring at equations and now believe that all of physics can be described by tiny Leprechauns made of Ether throwing God particles at each-other in a giant game of dodge-ball.

/sarcasm

Dadface

Quote by gabbagabbahey

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

I don't think that a field of any type does work but a force does.When F=Bev is summed for all of the current carrying electrons in a wire at 90 degrees to a B field the total force is given by F=BIL.If free to move in a vacuum these electrons follow circular paths and no work is done.Within the confines of the wire,however,any resultant motion of the whole conductor is in the direction of the force.I think it's safe to say that movement of the wire results in work being done.But is it being suggested that F=BIL,which comes from summing F=Bev is not a magnetic force?

cabraham

Quote by gabbagabbahey

The inner product between two vector fields is a scalar field; its value, just like the two vector fields, depends on position. If either of the two vector fields is zero at a specific location, their inner product will also be zero there.

Maxwell's equations, and the Lorentz Force Law both deal with vector fields. If you don't understand how to take the inner product of two vector fields, you don't fully understand Maxwell's equations.

Now, as for whether Magnetic field/forces do work on any classical system, the answer is fundamentally no. This assumes only that you take Maxwell's equations and the Lorentz force law as the basis for classical electrodynamics (which 99.9% of the current physics community likely does).

In many cases, the work done on an object depends on the value of the magnetic field applied to it, but this does not mean that the magnetic field is directly doing the work (and it isn't!).

Good grief, you think I don't how to take the inner product of 2 vectors? Just who are you? Physicist, EE? How much education - Ph.D., MS, BS? You talk like I'm an 18 year old just out of HS. If I've erred, show me where & offer correction. You state your position that mag fields do no work, & offer nothing in terms of proof.

You are just 1 more talker who says "Claude you're wrong!" but then fails to follow through & show why I'm wrong. Prove your case, until you do all you offer is emoty word.

Clauide

gabbagabbahey

Quote by Dadface

I don't think that a field of any type does work but a force does.

Force is a field.

When F=Bev is summed for all of the current carrying electrons in a wire at 90 degrees to a B field the total force is given by F=BIL.If free to move in a vacuum these electrons follow circular paths and no work is done.Within the confines of the wire,however,any resultant motion of the whole conductor is in the direction of the force.

Yes, this is due to the fact that there are other internal forces at play inside the wire. In this case, the only forces that directly contribute to the work on the wire are the internal electric forces that maintain the current (these of course are created by a battery or other power source providing a potential difference along the wire. cf. Example 5.3 from Griffith's Introduction to Electrodynamics 3rd edition)

I think it's safe to say that movement of the wire results in work being done.But is it being suggested that F=BIL,which comes from summing F=Bev is not a magnetic force?

The general idea is that since all of Classical Electrodynamics (including all composite force laws such as F=IBL) can be derived from The Lorentz Force Law, Maxwell's equations and some assumptions about the composition of certain macroscopic bodies (like a bar magnet, etc), one can treat [itex]\mathbf{F}_{ \text{mag} } = q \mathbf{v} \times \mathbf{B}[/itex] (the magnetic part of the Lorentz force law) as being the fundamental magnetic force and [itex]\mathbf{F}_{ \text{e} } = q \mathbf{E}[/itex] as being the fundamental electric force. All other EM forces are treated as composites. So, when one says that magnetic forces do no work, one is unambiguosly referring to [itex]\mathbf{F}_{ \text{mag, net } } = \int dq \mathbf{v} \times \mathbf{B}[/itex] as the magnetic force on an object (where the integral is over the object's volume/charge distribution)

gabbagabbahey

Quote by cabraham

Good grief, you think I don't how to take the inner product of 2 vectors? Just who are you? Physicist, EE? How much education - Ph.D., MS, BS? You talk like I'm an 18 year old just out of HS. If I've erred, show me where & offer correction. You state your position that mag fields do no work, & offer nothing in terms of proof.

You are just 1 more talker who says "Claude you're wrong!" but then fails to follow through & show why I'm wrong. Prove your case, until you do all you offer is emoty word.

Clauide

I am only going by your previous posts.

Quote by cabraham

Outside the wire, I don't believe that the E.J value vanishes.

As Dale Spam has said, outside the wire the current density [itex]\mathbf{J}(\mathbf{r})[/itex] is zero, so the dot product of it with any vector field will also be zero there. Do you really need me to prove this to you?

cabraham

Quote by DaleSpam

True, but j is 0 outside of the wire, therefore the dot product is 0 outside of the wire.

Weren't you going to look at a textbook last night?

A field at a location far from the wire cannot deliver power to a wire. That energy must first be transported to the location of the wire, and then it can be delivered to the wire.

It is obviously nonsense to claim that the E field a light year away from a wire can do any work on the wire now. For the same reason that is nonsense, it is also nonsense that the field a foot away or a millimeter away can do any work on the wire now. Only the fields at the location of the matter at a given time can deliver power to the matter at that time.

I'll find the text book. But here is a flaw in your reasoning. If only the interior of the wire matters, E is zero inside a superconductor, herein SC. So E.J = 0, & you insist E.J = 0 outside the wire. So we have 0 power at the inpout hence 0 work done. But SC motors have been built & affirmed. They draw current from an input voltage source, store energy in fields, tranfer enrgy to output as rotor spins. Yet E.J = 0 for a SC.

How do you explain that one? Remember that the energy in a motor that transfers from stator to rotor is chiefly in the air gap. If the energy was all confined to the wire, how does it transfer? The stator L has a flux which is in the stator iron core, links to the rotor iron core via the air gap. This flux is energy, LI²/2. This energy does not appear in the E.J product inside the wire.

An example which is simpler to visualize is a 120 volt xfmr secondary driving a heater. The secondary winding resistance is 0.10 Ω, & the heater R is 11.9 Ω. The total is 12 Ω. Current is 10 amps. The power inside the secondary winding is 1.0V * 10A = 10W. If we only consider the inside of the sec winding, we get 10W, instead of 1200W total. Why is that?

The voltage across the sec terminals depends on path taken. Outside the Cu it is 119V, inside it is 1.0V. To get the total power you need to consider both. When both are accounted for we get 10 + 1190 = 1200W, the right answer. Once again we appear to agree on almost everything but for the inside/outside question. I will search for the text & post. BR.

Claude

gabbagabbahey

Quote by cabraham

I'll find the text book. But here is a flaw in your reasoning. If only the interior of the wire matters, E is zero inside a superconductor, herein SC.

The applied magnetic field will penetrate the superconductor up to the London penetration depth. The E-field will also be, in general, non-zero up to that depth.

DaleSpam

Quote by cabraham

I'll find the text book. But here is a flaw in your reasoning.

In this case it isn't reasoning, that is just what the equations say. E.j is 0 anywhere E or j is 0. That is simply what E.j means. If you have a problem with it you need to take it up with Maxwell and Lorentz, they aren't my equations.

Quote by cabraham

Remember that the energy in a motor that transfers from stator to rotor is chiefly in the air gap. If the energy was all confined to the wire, how does it transfer? The stator L has a flux which is in the stator iron core, links to the rotor iron core via the air gap. This flux is energy, LI²/2. This energy does not appear in the E.J product inside the wire.

If you believe that the energy can transfer from a point in the air gap directly to the wire without passing through the intervening space then do you also believe that a field a light year away can transfer energy to the wire now?

Quote by cabraham

I will search for the text & post.

Please do so.

andrien

Quote by cabraham

But E & B are from interacting loops. Maybe E¹, E², B¹, & B² are more appropriate.

E.J is no problem, & is consistent w/ what I've presented. You acknowledge B as producing torque, yet you deny its work contribution. For the 4thtime please draw a sketch showing the fields doing work. All work is ultimately done by input power source. We have 2 loops. Each loop has E & B. It is the interaction that makes motors run. An e/m have not ionizing an H atom is too simplistic dealing w/ motors. Please draw us a pic. Thanks.

Claude

you did not really answer that.but apart from that I am saying it once again that magnetic field can not do work on electric charges because of lorentz force law so F.v is zero.but on a magnetic dipole the force is ∇(m.B) and it is not perpendicular to any velocity so it can do work.electric charges no work but magnetic charges there can be.

gabbagabbahey

Quote by andrien

but on a magnetic dipole the force is ∇(m.B) and it is not perpendicular to any velocity so it can do work.

Classically, magnetic dipoles are treated on the same footing as any other current distribution.

Specifically, the force on an ideal dipole [itex]\mathbf{m}[/itex] is derived (typically as an example in a textbook or as an exercise for the student - c.f. Problem 6.4 from Griffith's Introduction to Electrodynamics 3rd edition ) from the Lorentz force law by treating the dipole as a limiting case of a current loop (with current [itex]I[/itex] running through it), as the vector area [itex]\mathbf{a} \equiv \oint \mathbf{ \hat{n} } da [/itex] scribed/enclosed (subtended?) by the loop is shrunk to zero in such a way that the product [itex]\mathbf{m}=I \mathbf{a} [/itex] remains unchanged. Thus [itex]\mathbf{F}=\nabla( \mathbf{m} \cdot \mathbf{B} )[/itex] is regarded as a composite force law in the sense that it can be derived from the Lorentz Force Law and Maxwell's equations, like every other EM force law, and not a pure magnetic force. The work on a dipole in an external magnetic field must once again come directly from the electric field that maintains the current, which is produced by an unknown (classically) energy source (Of course, permanent dipoles like those associated with an electron's spin are really a quantum phenomenon, but classically they are treated as described)

electric charges no work but magnetic charges there can be.

True Magnetic charges (monopoles) have never been observed, despite the repeated efforts of many experimentalists, so whether or not a magnetic field does work on them is purely speculation.

andrien

of course,but the result is of more general character then for any current loop.there is a mechanical energy associated with it amount to -m.B.the potential energy of the dipole from where the force formula comes.

vanhees71

Quote by andrien

you did not really answer that.but apart from that I am saying it once again that magnetic field can not do work on electric charges because of lorentz force law so F.v is zero.but on a magnetic dipole the force is ∇(m.B) and it is not perpendicular to any velocity so it can do work.electric charges no work but magnetic charges there can be.

Somewhere in this huge thread, I've shown that this is no exception to Maxwell's theory. The power density is again given by
[tex]\vec{E} \cdot \vec{j}=\vec{E} \cdot (c \vec{\nabla} \times \vec{M}).[/tex]

In fact you can derive the coupled set of equations of motion for fields and matter by using the fundamental conservation laws from space-time symmetry, i.e., for total energy, momentum, and angular momentum.

Dadface

Quote by gabbagabbahey

Force is a field.

( FROM DADFACE-No,work done is force times distance not field times distance.Perhaps I am being nit-picky here.)

Yes, this is due to the fact that there are other internal forces at play inside the wire. In this case, the only forces that directly contribute to the work on the wire are the internal electric forces that maintain the current (these of course are created by a battery or other power source providing a potential difference along the wire. cf. Example 5.3 from Griffith's Introduction to Electrodynamics 3rd edition)

(FROM DADFACE_Extending this, the B field of the magnets is one of the factors that determines how much input power there is and how much is converted to mechanical power and how much is converted to electrical heating power losses.As I have stated before the mechanical power is given by
EbI {Eb=back(counter) emf,I= current})

The general idea is that since all of Classical Electrodynamics (including all composite force laws such as F=IBL) can be derived from The Lorentz Force Law, Maxwell's equations and some assumptions about the composition of certain macroscopic bodies (like a bar magnet, etc), one can treat [itex]\mathbf{F}_{ \text{mag} } = q \mathbf{v} \times \mathbf{B}[/itex] (the magnetic part of the Lorentz force law) as being the fundamental magnetic force and [itex]\mathbf{F}_{ \text{e} } = q \mathbf{E}[/itex] as being the fundamental electric force. All other EM forces are treated as composites. So, when one says that magnetic forces do no work, one is unambiguosly referring to [itex]\mathbf{F}_{ \text{mag, net } } = \int dq \mathbf{v} \times \mathbf{B}[/itex] as the magnetic force on an object (where the integral is over the object's volume/charge distribution)

THE NOTES ABOVE IN BRACKETS ARE FROM MYSELF(Dadface).WHAT IS A BETTER WAY TO HIGHLIGHT THEM? THANKS.

The section starting with "the general idea" is the point that interests me most.It might be considered as trivial because it is to do with definitions.It all boils down to the question:

Is the force F as given by F=BIL a magnetic force?

I find your notes above confusing.Reference to a "composite force" seems to suggest that it is not a magnetic force,so what,if anything,should it be described as? In a previous post I suggested(if I remember correctly)that it could be described as an electromagnetic force.

The notes go on to express an integral in terms of Fmag.net and then describe this as as "the magnetic force on an object".So is it a magnetic force or not?(A quick google lead me to the hyperphysics text book which described it as a magnetic force)

If the wire contained just one moving electron BIl would equal Bqv, so if BIl is not regarded as the fundamental magnetic force it could be argued that nor is Bqv.For N electrons F=NBqv=BIl.The electrons may be in a wire which restricts their movements but the force on each one is still Bqv.

Dadface

Quote by vanhees71

Somewhere in this huge thread, I've shown that this is no exception to Maxwell's theory. The power density is again given by
[tex]\vec{E} \cdot \vec{j}=\vec{E} \cdot (c \vec{\nabla} \times \vec{M}).[/tex]

In fact you can derive the coupled set of equations of motion for fields and matter by using the fundamental conservation laws from space-time symmetry, i.e., for total energy, momentum, and angular momentum.

Hi vanhees.I would be interested to hear your answers to the following:

1.Can the force BIl do work?
2.Would you describe BIL as a magnetic force or otherwise?

Thanks if you can find the time to answer.

Per Oni

My take on the original op question is that a magnetic field can exert torque but do no work.

If a put a weight on a frictionless horizontal bar I can increase or decrease at will the torque or moment around each support without input of any effort/work/joules. So torque and work done are different concepts.

If I hold 2 equal poles of a permanent magnet close together and let go the 2 magnets will push each other away and therefore it looks like the magnetic fields are doing work. However in the process of flying apart there are always electrical fields involved, since there are travelling magnetic fields. The Poynting vector shows the flow of energy.

The question of HOW energy reaches the rotor of a motor is an entirely different problem.

From: http://arxiv.org/abs/1207.2173

Earlier studies of the Poynting vector and the rate of flow of energy considered only idealized geometries in which the Poynting vector was confined to the space within the circuit. But in more realistic cases the Poynting vector is nonzero outside as well as inside the circuit. An expression is obtained for the Poynting vector in terms of products of integrals, which are evaluated numerically to show the energy flow.

The above reference is not the only one I have come across in the past. As to how reliable the above source is I really don’t know.

Just to make again another point: LI2/2 does exist but has nothing to do with the output power of a motor. Asked for references are in short supply.

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andrien	of course,but the result is of more general character then for any current loop.there is a mechanical energy associated with it amount to -m.B.the potential energy of the dipole from where the force formula comes.