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

[gabbagabbahey]

That's correct of course (and it's exactly what I explained).

Already explained in the other thread: the equations of classical mechanics dictate that the Lorentz force drives the motion. Surely it doesn't go to zero when the wires start to move, there is no law of nature according to which that would happen. But if you really think so, please give a reference in which such magical disappearance is derived or where that magnetic force disappearance law* is given.

*such a weird law should prescribe complete and instant magnetic force disappearance for a current loop, but none at all for an electron in a cyclotron!

The (magnetic part of the) Lorentz force doesn't need to be zero to not do any work, it only need to be perpendicular to the motion of the entities it is acting . In a current loop, it acts on little bits of moving charge/current, d\mathbf{F}_m = dq \mathbf{v} \times \mathbf{B}, and is always perpendicular to the motion \mathbf{v} of each little bit moving charge. Therefor it does no work.

 

[Dale]

Your theory that the motion creates an E field which does the work cannot be right.

It isn't my theory, it is Poynting's, and it follows directly from Maxwell's. If there is work done on matter then there must be E and j.

 

[Dale]

That's correct of course (and it's exactly what I explained).

Already explained in the other thread: the equations of classical mechanics dictate that the Lorentz force drives the motion. Surely it doesn't go to zero when the wires start to move, there is no law of nature according to which that would happen. But if you really think so, please give a reference in which such magical disappearance is derived or where that magnetic force disappearance law* is given.

*such a weird law should prescribe complete and instant magnetic force disappearance for a current loop, but none at all for an electron in a cyclotron!

Nonsense, I have never suggested anything like this, and it does not follow from Maxwells equations. However, you appear to be making the mistake of equating a force with work. There is indeed a magnetic force, but it does no work according to the laws of classical EM.

 

[harrylin]

Nonsense, I have never suggested anything like this, and it does not follow from Maxwells equations. However, you appear to be making the mistake of equating a force with work. [..]

I don't know how you can interpret "the Lorentz force displaces the wire in the direction of the force" as making the mistake of equating a force with work. Anyway, to clear up matters I just provided a simple case example in the new thread (applicable to both) and so it would clarify things a lot if you specify what your words mean when applied to that example.

 

[vanhees71]

I guess it's hopeless with this thread since we discuss in circles. One last try:

(1) Please learn to distinguish between force and work/power. The magnetic field, of course, causes forces on currents (including the magnetization currents, which in particular includes the force on permanent magnets, elementary dipole moments of particles in the classical limit, etc.), but it provides no direct transfer of energy from the magnetic field to the charges/magnetization sources.

(2) In a motor the power is provided by the induced magnetic field. Please check any texbook on electrical engineering about this.

(3) Last but not least the formula
P=\int \mathrm{d}^3 \vec{x} \vec{E} \cdot \vec{j}
is a quite simple mathematical consequence of Maxwell's equations, which are very well established from both first principles of relativistic (quantum) field theory and, even more important, to a very high accuracy from all empirical findings and observations of electromagnetic phenomena. Denying the validity of this equation necessarily means to deny this very successful description of em. phenomena, and for this there is really not the slightest justification from observations today!

 

[harrylin]

The (magnetic part of the) Lorentz force doesn't need to be zero to not do any work, it only need to be perpendicular to the motion of the entities it is acting . [..]

That is correct of course.

In a current loop, it acts on little bits of moving charge/current, d\mathbf{F}_m = dq \mathbf{v} \times \mathbf{B}, and is always perpendicular to the motion \mathbf{v} of each little bit moving charge. Therefor it does no work.

In contrast, textbooks explain how it acts on the wire, which leads to the contrary conclusion. In the new thread I now provided a specific example to illustrate this.

 

[harrylin]

I guess it's hopeless with this thread since we discuss in circles. One last try:

(1) Please learn to distinguish between force and work/power. [..]

Surely everyone here distinguishes between those.

(2) In a motor the power is provided by the induced magnetic field. Please check any texbook on electrical engineering about this.

I fully agree with that (and did so from the start).

(3) Last but not least the formula
P=\int \mathrm{d}^3 \vec{x} \vec{E} \cdot \vec{j}
is a quite simple mathematical consequence of Maxwell's equations, which are very well established from both first principles of relativistic (quantum) field theory and, even more important, to a very high accuracy from all empirical findings and observations of electromagnetic phenomena. Denying the validity of this equation necessarily means to deny this very successful description of em. phenomena, and for this there is really not the slightest justification from observations today!

I'm afraid that you did not clearly demonstrate how that equation answers the OP's questions. I hope to show with a simple example how the Lorentz force equation and the definition of the ampere allows to answer both questions in the new thread.

 

[Dale]

I don't know how you can interpret "the Lorentz force displaces the wire in the direction of the force" as making the mistake of equating a force with work.

Sorry, I guess I should have used a more specific quote. It was this part "the equations of classical mechanics dictate that the Lorentz force drives the motion. Surely it doesn't go to zero when the wires start to move, there is no law of nature according to which that would happen." There you seem to assume that because the magnetic force does no work it must drop to zero as the wire moves, which is why I think you are making the mistake of equating force with work.

Anyway, to clear up matters I just provided a simple case example in the new thread (applicable to both) and so it would clarify things a lot if you specify what your words mean when applied to that example.

Could you link to it? I am getting confused with the multiple threads.

 

[Dale]

In contrast, textbooks explain how it acts on the wire, which leads to the contrary conclusion.

The magnetic force acts on currents, not wires. To see that simply measure the force on a current without a wire and the force on a wire without a current.

 

[harrylin]

Sorry, I guess I should have used a more specific quote. It was this part "the equations of classical mechanics dictate that the Lorentz force drives the motion. Surely it doesn't go to zero when the wires start to move, there is no law of nature according to which that would happen." There you seem to assume that because the magnetic force does no work it must drop to zero as the wire moves, which is why I think you are making the mistake of equating force with work.

Obviously it did not contradict what I wrote just before; that would be illogical, to state it mildly! I'll expand that to make it hopefully clearer on that point:

W = F.d (Work equals force times displacement due to the force along the direction of force). The equations of classical mechanics dictate that the net Lorentz force drives the motion of the wires. Or, in other words: the wires are moved by the resulting Lorentz force - and this can only occur if the motion is at least partly along the direction of that force. Surely that force doesn't go to zero when the wires start to move along the direction of that driving force - there is no law of nature according to which that would happen. And of course, motion is not possible without displacement.

Could you link to it? I am getting confused with the multiple threads. [..]

Sure (sorry I thought that you are following that thread, there we also discuss your next point):
post #166 in
https://www.physicsforums.com/showthread.php?t=628896&page=11

 

[Dale]

W = F.d (Work equals force times displacement due to the force along the direction of force).

That is the definition of work that is appropriate for Newtonian mechanics. It is not generally applicable when you are using fields as fields don't move (d=0) and fields can do work on other fields without any forces (F=0). The general definition that is applicable for fields is: work is a transfer of energy other than heat.

http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1

By Poynting's theorem the energy transferred from EM fields to matter (work) is E.j.

 

[cabraham]

That is the definition of work that is appropriate for Newtonian mechanics. It is not generally applicable when you are using fields as fields don't move (d=0) and fields can do work on other fields without any forces (F=0). The general definition that is applicable for fields is: work is a transfer of energy other than heat.

http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1

By Poynting's theorem the energy transferred from EM fields to matter (work) is E.j.

But energy is transferred to & from the B field as well. Using your definitions, the power source did the work, as well as the E forces, as well as the B forces. Since energy transfer involves all of these quantities, you can claim that either or all of them did the work.

Once again, it may be simply a matter of how "work done" is particularly defined. You stated earlier that fields & their forces do not move so that F.d is inadequate to describe work. But if I have a ball in my hand, then release it, dropping to the floor it acquires KE, what did the work? I only let go of it, applied no force. I say the gravity field did the work, although said field did not move. To me W = mgh is perfectly applicable here (h = height, m - mass. g = accel due gravity). The gravity field did work on the ball.

I can't understand how this question is even controversial. We know not all about e/m fields, but we know enough to answer the OP question. The force on the loop results in a torque which spins the loop, doing work, W = Iω²/2. This force is Fm = qvXB. E.J does indeed transfer energy, but the E component of Lorentz force, Fe, acts in a direction to move charges around said loop, not to spin loop.

My sketches illustrate this. Please, not just Dale, but all those in disagreement with moi, point out where my sketch is wrong. I'm not asking a lot by doing that.

Claude

 

[Dale]

But energy is transferred to & from the B field as well. Using your definitions, the power source did the work, as well as the E forces, as well as the B forces. Since energy transfer involves all of these quantities, you can claim that either or all of them did the work.

I am fine with this. This is, IMO, correct and is essentially the reason why I say that B can do work indirectly. The work done on matter is E.j but B has energy and influences both E and j.

But if I have a ball in my hand, then release it, dropping to the floor it acquires KE, what did the work? I only let go of it, applied no force. I say the gravity field did the work, although said field did not move. To me W = mgh is perfectly applicable here (h = height, m - mass. g = accel due gravity). The gravity field did work on the ball.

Yes, energy was transferred from the gravity field to the ball. The point is that B doesn't transfer energy directly to matter, only through E.j.

E.J does indeed transfer energy, but the E component of Lorentz force, Fe, acts in a direction to move charges around said loop, not to spin loop.

My sketches illustrate this. Please, not just Dale, but all those in disagreement with moi, point out where my sketch is wrong.

Your sketches aren't wrong, I have said that previously multiple times. They are just sketches of force, not energy transfer. As you yourself admit, the energy transfer is given by E.j.

 

[harrylin]

That is the definition of work that is appropriate for Newtonian mechanics. [..]

The OP's questions are about mechanical work (moving wires, moving magnets). Throughout this thread I specified to which definition my answers related; sorry that I did not do so in every single post.

http://www.lightandmatter.com/html_books/lm/ch13/ch13.html#Section13.1

Yes, I think that I provided that link.
It also uses F.d. Where does it claim that this is wrong? (apart of energy loss which is not an issue here)

 

[Dale]

The OP's questions are about mechanical work (moving wires, moving magnets).

Yes, but mechanical work done by EM fields. So you need to use a general definition which can handle both the mechanical part and the fields part.

Yes, I think that I provided that link.
It also uses F.d. Where does it claim that this is wrong?

See section 13.6, and then in the summary he explicitly says "There are some situations in which the equation W=Fd is ambiguous or not true".

 

[harrylin]

Yes, but mechanical work done by EM fields. So you need to use a general definition which can handle both the mechanical part and the fields part.

Concerning that aspect, I have no issue with W=F.d (and neither has either article).

See section 13.6, and then in the summary he explicitly says "There are some situations in which the equation W=Fd is ambiguous or not true".

Ah yes: d relates to the displacement of the point(s) where F is acting over that displacement (for lazy onlookers: a pushing hand does work but a rigid bouncing wall does zero work). And of course heat loss isn't work. None of that is an issue for answering the OP's questions.

 

[Dale]

Concerning that aspect, I have no issue with W=F.d (and neither has either article).

I do (and you really are stretching credulity to claim that an article which says something is sometimes "ambiguous or not true" has "no issue" with it)

 

[harrylin]

I do (and you really are stretching credulity to claim that an article which says something is sometimes "ambiguous or not true" has "no issue" with it)

Here you mis-cited me.