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

cabraham

Quote by vanhees71

Very nice! Now you yourself have shown that the work is done by the electric field, not the magnetic.

Of course, what you considered is the static case, i.e., the forces and torque at fixed loops, and that's why your electric field is only there to compensate for the loss due to resistance (producing heat through scatterings of the electrons providing the currents in the loops).

If you add the calculation in the paper, I've cited, for the dynamical case of the moving wires, you'll see that also the energy needed to set the loops in motion is provided by the electric field, and this shows that Maxwell's equations hold for this case as expected.

As the paper has also demonstrated, the same dynamics holds for the case when you substitute one of the loops by a permanent magnet, whose magnetism is due to the spins of the electrons and the quantum mechanical exchange force that directs the spins into macroscopic domains, which is the modern understanding of Weiss's model for ferromagnets.

Your cited paper calculation backs me up. It does not account for torque on the loop. That force is B, not E. You draw conclusions w/o any proof. Please draw a diagram & show the E force that spins the loop with torque. My diagram is consistent with the paper you cited & Maxwell's equations. How can E force spin the loop?

Paper you cited describes power density as "E dot J". Integrating over volume gives power. I've already affirmed that that is correct. To have current in the loop 2 types of work on the electrons are needed. We need to do work on the e- to transition it from valence to conduction band. Only E can do that. Second, when the e- loses energy due to lattice collisions, i.e. resistance, the E force restores this energy by doing work on the E.

If not for loop current, there would be no B force. So, E is all important & indispensable. Nobody is denying the important role played by E force. Without it, the motor does not operate. But the force spinning the loop is indeed B force, not E. Please show me the component of E in a direction radial to the loop. Which Maxwell equation applies here?

My diagram accounts for I, J, E, B, A, & velocity u. You keep citing that paper w/ the integral of E dot J. That integral proves that the work done by E is along the path of the current density J. But torque is normal to the current, where E dot J equals ZERO.

I recommend you draw a diagram for your own understanding. All you do is cite that integral, which clearly proves my case. BR.

Claude

DaleSpam

Quote by cabraham

I scanned & uploaded a diagram detailing the relevant force fields. Comments welcome.

It seems pretty well-done, and I only see 3 very minor mistakes, none of which substantially change the conclusions:

1) On page 1 you have the field B1 pointing in the wrong direction (or maybe the current I1 is in the wrong direction).
2) On page 2 it is not correct that [itex]E=-\frac{\partial}{\partial t} A[/itex]. Because the curl of the divergence of any scalar function is 0 you can add the divergence of an arbitrary scalar to E and still satisfy [itex]\nabla \times E = -\frac{\partial}{\partial t}(\nabla \times A)[/itex]. However, by assuming that everything is uncharged, I suspect that you can use Gauss' law and the remaining gauge freedom to set the divergence of the scalar function to 0.
3) On pages 3 and 4 it seems that, since both loops are in-plane Fm squeezes the loop without producing any torque or spin. It is an equilibrium position, however it is an unstable equilibrium and any deviation from being in plane will provide a torque. So, it is not critical.

So, overall I agree with the conclusion. The E-field provides the work and the B-field provides the torque.

DaleSpam

Quote by Miyz

Since when did I say the rotor is doing all the work by it's own?

I think that is exactly what you were trying to say in post 159. In that post you seem to be trying to say that increasing the input power increases the work done on the rotor by the B field because it increases the B field of the rotor. This only makes sense if it is the B field of the rotor which is doing the work on the rotor.

If you weren't even trying to say that then the fact that increasing the current increases the rotor's magnetic field is even more irrelevant to the work done.

Per cabraham's analysis, increasing B increases the torque, but it is still E.j which does the work.

cabraham

Quote by DaleSpam

It seems pretty well-done, and I only see 3 very minor mistakes, none of which substantially change the conclusions:

1) On page 1 you have the field B1 pointing in the wrong direction (or maybe the current I1 is in the wrong direction).
2) On page 2 it is not correct that [itex]E=-\frac{\partial}{\partial t} A[/itex]. Because the curl of the divergence of any scalar function is 0 you can add the divergence of an arbitrary scalar to E and still satisfy [itex]\nabla \times E = -\frac{\partial}{\partial t}(\nabla \times A)[/itex]. However, by assuming that everything is uncharged, I suspect that you can use Gauss' law and the remaining gauge freedom to set the divergence of the scalar function to 0.
3) On pages 3 and 4 it seems that, since both loops are in-plane Fm squeezes the loop without producing any torque or spin. It is an equilibrium position, however it is an unstable equilibrium and any deviation from being in plane will provide a torque. So, it is not critical.

So, overall I agree with the conclusion. The E-field provides the work and the B-field provides the torque.

1) Yes, I need to remember my right hand rule from the left. That negative sign threw me. You are correct.
2) Yes, I am aware that there is not a 1 for 1 equivalence, that uncharged de-energized conditions have to be assumed for my equation to be absolutely valid.
3) I did not do a great job drawing the loops. They are supposed to be oblique, but my lousy drawing skills ended up making them look co-planar. Based on the co-planar appearance, you are right, there would be zero torque, & a little motion either way results in non-zero torque.

Thanks for your feedback, we are in agreement. One point needs to be clarified however. I agree that E does provide the work when it comes to producing loop current, since work is done elevating valence e- into conduction, & restoring energy lost due to lattice collisions, which is resistance. E does this exclusively. We agree that B produces torque. But remember that torque times angle equals work. B does rotational work equal to torque times radian angle measure. Torque, however, would be 0 if current were 0. But current is nonzero due to E. So B does rotate the loop, but its torque would not exist w/o J, which would not exist w/o E. BR.

Claude

*Miyz*

Quote by DaleSpam

I think that is exactly what you were trying to say in post 159. In that post you seem to be trying to say that increasing the input power increases the work done on the rotor by the B field because it increases the B field of the rotor. This only makes sense if it is the B field of the rotor which is doing the work on the rotor.

If you weren't even trying to say that then the fact that increasing the current increases the rotor's magnetic field is even more irrelevant to the work done.

Per cabraham's analysis, increasing B increases the torque, but it is still E.j which does the work.

Umm, I think what lead you was my mistake of saying "within it" I apologize for that.

I like were this is going. Work is done by E.j and torque is by the B field. Good conclusion + agreement.

*Miyz*

Quote by cabraham

1) Yes, I need to remember my right hand rule from the left. That negative sign threw me. You are correct.
2) Yes, I am aware that there is not a 1 for 1 equivalence, that uncharged de-energized conditions have to be assumed for my equation to be absolutely valid.
3) I did not do a great job drawing the loops. They are supposed to be oblique, but my lousy drawing skills ended up making them look co-planar. Based on the co-planar appearance, you are right, there would be zero torque, & a little motion either way results in non-zero torque.

Thanks for your feedback, we are in agreement. One point needs to be clarified however. I agree that E does provide the work when it comes to producing loop current, since work is done elevating valence e- into conduction, & restoring energy lost due to lattice collisions, which is resistance. E does this exclusively. We agree that B produces torque. But remember that torque times angle equals work. B does rotational work equal to torque times radian angle measure. Torque, however, would be 0 if current were 0. But current is nonzero due to E. So B does rotate the loop, but its torque would not exist w/o J, which would not exist w/o E. BR.

Claude

Very nice point!

DaleSpam

Quote by cabraham

One point needs to be clarified however. I agree that E does provide the work when it comes to producing loop current, since work is done elevating valence e- into conduction, & restoring energy lost due to lattice collisions, which is resistance.

In a functioning motor E.j is greater than the energy dissipated by the resistance. It is equal to that plus the mechanical work.

Quote by cabraham

We agree that B produces torque.

Yes.

Quote by cabraham

But remember that torque times angle equals work. B does rotational work equal to torque times radian angle measure.

If that were true then energy would not be conserved. E.j is an amount of work done by E. That can be split into an amount of energy dissipated by the rotor's resistance plus some remaining amount. Now, you are saying that the B field does the mechanical work on the rotor, so what happens to the remaining amount of work done by E.j? It isn't increasing the thermal energy of the rotor, and according to you it is not doing mechanical work on the rotor, so where did it go? Also, if B does the work then the energy that B used to do the work must come from somewhere, so where did it come from?

DaleSpam

Quote by Miyz

Work is done by E.j and torque is by the B field. Good conclusion + agreement.

Excellent!

cabraham

Quote by DaleSpam

In a functioning motor E.j is greater than the energy dissipated by the resistance. It is equal to that plus the mechanical work.

Yes.

If that were true then energy would not be conserved. E.j is an amount of work done by E. That can be split into an amount of energy dissipated by the rotor's resistance plus some remaining amount. Now, you are saying that the B field does the mechanical work on the rotor, so what happens to the remaining amount of work done by E.j? It isn't increasing the thermal energy of the rotor, and according to you it is not doing mechanical work on the rotor, so where did it go? Also, if B does the work then the energy that B used to do the work must come from somewhere, so where did it come from?

Hmmm, 3 good questions. Here are 3 good answers.

1st bold: Agreed. E dot J is the work done by E. But why do you say that this work is split between rotor conduction thermal loss & mechanical energy? You're making a pure assumption. E dot J is the conduction loss, thermal, of the rotor. The current in the rotor is needed or else there is no B force to spin the rotor.

2nd bold: Where it went is into rotor loss, conduction current squared times resistance. That is all of it. "E dot J" cannot be what produces torque. Torque acts radially to the loop, whereas E dot J is tangential. Refer to my picture. I made it clear that E and J are in the wrong direction to produce torque. To produce torque we need a radial force, i.e. normal to current density J. E is along the J direction. Any component of E normal to J has ZERO dot product with J.

3rd: Agreed. The energy B used to do the work had to come from somewhere. We are in agreement thus far. Hopefully we are still in agreement when I say that the independent power source driving the motor (battery, wall outlet, car alternator, etc.) is replenishing the B energy.

No field, E, B, whatever, can supply energy long term. Just as the input power source replenishes the B field energy, it also replenishes E field energy as well. As the B magnetic poles align, energy is minimum, & energy maxes out when the poles are 90 degrees apart. But the input supply is providing current as well as voltage. The product times the power factor times the efficiency is the amount of power processed by the fields, B as well as E.

Like I said, E & B both do short term work. But the input power supply is doing the long term work. The energy from the supply is stored in B & E fields, transferred to charges & torque*angle, then said E & B field energy is replenished by the power source. Ultimately all the energy is provided by this input power source.

But fields such as B & E provide us with a means of focusing & controlling the energy & transfer. The winding length, number of turns, air gap, core shape, etc. allow us to modify the motor behavior based on the application. But in all cases the power source driving the input does all the work. BR.

Claude

Dadface

We can write:
V-E=IR (v=supply voltage,E=back emf,I=current,R=resistance)
From this we can write:
VI=EI+I^2R

VI=power supplied,I^2R resistive heating power losses and EI=mechanical power output.

(A more detailed treatment would consider the other energy losses due to friction etc)

Sorry if this is irrelevant to the discussion.

truesearch

Cabraham:
from post 171:We need to do work on the e- to transition it from valence to conduction band. Only E can do that.
Can you give me some idea of the width of the energy band between Valence and conduction bands in something like copper that you would use in any analysis?
From most recent post: But the input supply is providing current as well as voltage. The product times the power factor times the efficiency is the amount of power processed by the fields, B as well as E.
I am fimiliar with the concept of power factor in the analysis of AC circuits containing R, L and C but I have not met the idea applied to DC electric motors.... Can you amplify on this or quote a reference that I could access?
And.... how do you define efficiency in your analysis.
Looking forward to any explanations you feel able to give.

cabraham

Quote by truesearch

Cabraham:
from post 171:We need to do work on the e- to transition it from valence to conduction band. Only E can do that.
Can you give me some idea of the width of the energy band between Valence and conduction bands in something like copper that you would use in any analysis?
From most recent post: But the input supply is providing current as well as voltage. The product times the power factor times the efficiency is the amount of power processed by the fields, B as well as E.
I am fimiliar with the concept of power factor in the analysis of AC circuits containing R, L and C but I have not met the idea applied to DC electric motors.... Can you amplify on this or quote a reference that I could access?
And.... how do you define efficiency in your analysis.
Looking forward to any explanations you feel able to give.

For metals, the valence & conduction bands actually overlap. Some e- are already in conduction band. Those in valence that require a little work to move into conduction get this work from E, not B. For a good conductor, like Cu, E is small, since J = sigma*E. Since the bands overlap, many e- are already in conduction band & need no work from E. Those e- in valence need a small amount of work to transition up into conduction band.

I was referring to ac motors as far as power factor goes. For dc motors, power factor can still have meaning. For example, if the input is a pure dc voltage source, but the current is a square wave, I've heard "power factor" defined as "pi/4". The dc pedestal current times the dc voltage is the continuous average power we are familiar with. But the ac ripple current times the voltage integrates to zero real power. Thus the ripple component of current does not contribute to motor output mechanical power.

Again, I was implying ac motors whenever PF is computed. But with dc motors, or other types of load, the term "power factor" can still have meaning. It differs from the R-L-C definition of power factor. With R-L-C networks, PF is cos of phase angle between I & V. However, in switching power converters, a rectified waveform has a power factor involving fundamental line frequency & all harmonics. I.e. a full wave bridge rectifier outputs fundamental & harmonic frequencies, current & voltage.

Power factor is defined as real power over total power. Reactive power is due to products of I & V of differing frequency. Or it can be due to same frequency I & V with 90 degree phase difference. I hope I've answered your question. BR.

Claude

truesearch

You have 'answered' my questions but not completely satisfactorily....sorry.
'Most electrons are already in the conduction band and need no work done on them'...that is what I thought. How does this affect your presentation?
A lot of confusion regarding power factor !...you have 'heard it defined' as pi/4... definitions are not communicated by hearsay.... where is it published as pi/4
I did not realise that AC motors were implied in what you have produced...sorry.
Also, I am not certain what is meant by this phrase: ' & energy maxes out when the poles are 90 degrees apart'.....is this a recognised technical term?

vanhees71

To make it clear once again: I never denied that magnetic fields cause forces and also torques, but what has this to do with the fact that the magnetic field does not do work on charge and current distributions (including magnetization currents)! I think, in principle we agree now on this simple fact.

BTW: I don't like to draw diagrams but to use vector calculus since this is far more save.

Dadface

The original question enquired as to whether a magnetic field or force can work on a current carrying wire and the answer is yes it can.The answer is nicely illustrated in the labelled sketch(in the opening post) which shows a force(F=BIL) on each of the two opposite sides of the coil.F=BIL is found by using F=BeV and calculating the total force on all of the current carrying electrons in the length of conductor under consideration.In fact it is the total magnetic force.
If memory serves correctly F=BIl is used to define magnetic flux density which is used to define the Tesla which,in turn,is used to define the Ampere.
The motion of the electrons within the conductors is affected by the B field as is evidenced by the Hall voltage which is set up if there are constraints to the movement of the coil.
It should be remembered that the B field is not just that due to the permanent magnets or field coils only.There is also a major contribution to the field because of the current flowing through the coil.The resultant field is sometimes referred to as the "catapult field".

DaleSpam

Quote by cabraham

E dot J is the work done by E. But why do you say that this work is split between rotor conduction thermal loss & mechanical energy? You're making a pure assumption. E dot J is the conduction loss, thermal, of the rotor.
...
Where it went is into rotor loss, conduction current squared times resistance. That is all of it.

It looks like we have a disagreement of fact. I believe that E.j is greater than the Ohmic losses, you believe it is equal.

If you are correct on that fact, then I agree with your reasoning.

If I am correct on that fact, do you agree with my reasoning?

*Miyz*

Quote by cabraham

Hmmm, 3 good questions. Here are 3 good answers.

1st bold: Agreed. E dot J is the work done by E. But why do you say that this work is split between rotor conduction thermal loss & mechanical energy? You're making a pure assumption. E dot J is the conduction loss, thermal, of the rotor. The current in the rotor is needed or else there is no B force to spin the rotor.

2nd bold: Where it went is into rotor loss, conduction current squared times resistance. That is all of it. "E dot J" cannot be what produces torque. Torque acts radially to the loop, whereas E dot J is tangential. Refer to my picture. I made it clear that E and J are in the wrong direction to produce torque. To produce torque we need a radial force, i.e. normal to current density J. E is along the J direction. Any component of E normal to J has ZERO dot product with J.

3rd: Agreed. The energy B used to do the work had to come from somewhere. We are in agreement thus far. Hopefully we are still in agreement when I say that the independent power source driving the motor (battery, wall outlet, car alternator, etc.) is replenishing the B energy.

No field, E, B, whatever, can supply energy long term. Just as the input power source replenishes the B field energy, it also replenishes E field energy as well. As the B magnetic poles align, energy is minimum, & energy maxes out when the poles are 90 degrees apart. But the input supply is providing current as well as voltage. The product times the power factor times the efficiency is the amount of power processed by the fields, B as well as E.

Like I said, E & B both do short term work. But the input power supply is doing the long term work. The energy from the supply is stored in B & E fields, transferred to charges & torque*angle, then said E & B field energy is replenished by the power source. Ultimately all the energy is provided by this input power source.

But fields such as B & E provide us with a means of focusing & controlling the energy & transfer. The winding length, number of turns, air gap, core shape, etc. allow us to modify the motor behavior based on the application. But in all cases the power source driving the input does all the work. BR.

Claude

The problem to me this is another higher level of my education but you made sense to me there.

Quote by Dadface

The original question enquired as to whether a magnetic field or force can work on a current carrying wire and the answer is yes it can.The answer is nicely illustrated in the labelled sketch(in the opening post) which shows a force(F=BIL) on each of the two opposite sides of the coil.F=BIL is found by using F=BeV and calculating the total force on all of the current carrying electrons in the length of conductor under consideration.In fact it is the total magnetic force.
If memory serves correctly F=BIl is used to define magnetic flux density which is used to define the Tesla which,in turn,is used to define the Ampere.
The motion of the electrons within the conductors is affected by the B field as is evidenced by the Hall voltage which is set up if there are constraints to the movement of the coil.
It should be remembered that the B field is not just that due to the permanent magnets or field coils only.There is also a major contribution to the field because of the current flowing through the coil.The resultant field is sometimes referred to as the "catapult field".

Thanks Dadface, for that I think other then Darwin123, no one mentioned the simple digram contradict the idea of magnetic force/field not doing work on a loop... And this formula is like the main one to look at.

Unfortunately for my lack of experience with maxwell's equations I can't product a good argument based on what Dale + Van are saying... Thanks to Claude his showing the other side of things.

Again this is going more deeper and more interesting!

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Dadface Recognitions: Gold Member	We can write: V-E=IR (v=supply voltage,E=back emf,I=current,R=resistance) From this we can write: VI=EI+I^2R VI=power supplied,I^2R resistive heating power losses and EI=mechanical power output. (A more detailed treatment would consider the other energy losses due to friction etc) Sorry if this is irrelevant to the discussion.

vanhees71 Recognitions: Science Advisor	To make it clear once again: I never denied that magnetic fields cause forces and also torques, but what has this to do with the fact that the magnetic field does not do work on charge and current distributions (including magnetization currents)! I think, in principle we agree now on this simple fact. BTW: I don't like to draw diagrams but to use vector calculus since this is far more save.