Does magnetic field do any work

In summary, a charged particle moving through a magnetic field feels a force that is perpendicular to the motion and the magnetic field. This force does no work on the particle.
  • #1
AlbertEinstein
113
1
Sorry, but while going through the article on magnetism in wikipedia I read the following which i couldn't not catch it.

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:
F= q V x B
where is the electric charge of the particle is the velocity vector of the particle is the magnetic field.

Because this is a cross product, the force is perpendicular to both the motion of the particle and the magnetic field. It follows that the magnetic force does no work on the particle; it may change the direction of the particle's movement, but it cannot cause it to speed up or slow down.

This might give you pause: Simple bar magnets seem to be entirely able to pick up small metal objects, which certainly seems to require that they do work on those objects. As David Griffiths points out in his textbook Introduction to Electrodynamics, this law is absolute -the magnetic field doesn't do any work. However, quite like the normal force of an inclined plane, which also can't do work, the magnetic field can redirect the efforts of existing forces, and then those forces can indeed do work in the relevant direction.

I am not able to understand the underlined lines. I do not know about which "existing forces" is he talking about; and the same applies to "those forces".I will really appreciate the help.Thanks in advance.
 
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  • #2
Hi Albert,

I think I know what he's talking about. An exemple will be best to illustrate the idea.

Consider a charged particle voyaging in an electrostatic field at constant speed v following a trajectory of electrostatic equilibrium* (i.e. E=0, along its path but E[itex]\neq[/itex]0 just a little off its path). No work is done on the particle since E=0. But now we turn on a uniform magnetic field in a direction perpendicular to the velocity of the particle. As a result, the particle changes its course, which bring it in a region of non-vanishing E and work is being done on it.

I like this exemple because at first no work is done. Then the magnetic force is applied, which in itself, does no work on the particle, but is the cause of the resulting electrical work done on the particle by the surrounding E field.

The much simpler exemple of a charged particle moving in a constant electric field and constant magnetic field could have been evoqued just as well but I find it less elengant, because there is work done by the electric field in the begining. The the idea remains that the magnetic field cause the particle trajectory to change and thus affect the work done on it by the electrical force.

*A particle voyaging in the middle of a quadrupole lens for instance.
 
  • #3
First, although static magnetic fields can't do work, time varying magnetic fields can. Second, whenever work is done under the influence of a static magnetic field, some external source will have to supply the energy require (as opposed to an electric field that can do work on its own).

Let's analyze the example you gave, about the bar magnetic picking up staples. We'll start with a simpler example: two dipole moments - current loops - pointing in the same way. Both have mass, but one is fixed and the other is directly under the the first one. Let's say the magnetic forces are greater than gravity so the fixed loop pulls the other up. So, obviously, someone did some work against the gravity, but where did the energy come from? the loop itself. While the loop is ascends, the current in both decreases, due to mutual inductance, so the loops serve as some sort of battery.
When you move a magnet near a staple, the staple feels a changing magnetic field that does work and changes the staple's internal energy - it makes many of the current loops in the metal point in the same direction as those in the magnet. After that both magnet and staple behave as magnetic dipoles. (actually some quantum effects such as ferromagnetism make this a bit more complicated, but the basic idea is similar)

Bottom line, time dependent magnetic field can do work, and static magnetic fields can take energy from other sources and use it to move things.
 
  • #4
Thanks, quasar987 and pseudovector.
But again I do not understand one thing said by pseudo
While the loop is ascends, the current in both decreases, due to mutual inductance, so the loops serve as some sort of battery.
Afterall, why the current in both will decrease if the second loop ascends(actually I am confused betwn mutual induction and self induction).I'll be grateful if you explain it once more.
thanks once again.
 
  • #5
AlbertEinstein said:
Sorry, but while going through the article on magnetism in wikipedia I read the following which i couldn't not catch it.

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:
F= q V x B
where is the electric charge of the particle is the velocity vector of the particle is the magnetic field.

Because this is a cross product, the force is perpendicular to both the motion of the particle and the magnetic field. It follows that the magnetic force does no work on the particle; it may change the direction of the particle's movement, but it cannot cause it to speed up or slow down.

This might give you pause: Simple bar magnets seem to be entirely able to pick up small metal objects, which certainly seems to require that they do work on those objects. As David Griffiths points out in his textbook Introduction to Electrodynamics, this law is absolute -the magnetic field doesn't do any work. However, quite like the normal force of an inclined plane, which also can't do work, the magnetic field can redirect the efforts of existing forces, and then those forces can indeed do work in the relevant direction.

I am not able to understand the underlined lines. I do not know about which "existing forces" is he talking about; and the same applies to "those forces".I will really appreciate the help.Thanks in advance.
It's not clear which part of this is from the wiki and which are your own comments.

Nevertheless, you might find this useful: https://www.physicsforums.com/showpost.php?p=997210&postcount=14
 
  • #6
I notice that all these posts ignore ferromagnetism, which changes everything.
 
  • #7
Meir Achuz said:
I notice that all these posts ignore ferromagnetism, which changes everything.
Don't leaving us hanging...please explain.
 
  • #8
A permanent magnet does work when it attracts an iron nail.
 
  • #9
Meir Achuz said:
A permanent magnet does work when it attracts an iron nail.
Very often it will look like magnetic force really does do work and often very difficult to find out what really is responsible for performing the work, but it is never the magnetic force.
The fundamental equation [itex]q(\vec v \times \vec B)[/itex] tells you magnetic forces will never ever do work, whether they are time varying or not.
Look up how a magnetic field exerts forces on dipole moments and see if the real work being done is by the magnetic field.
 
  • #10
The myth of free energy from permanent magnets is difficult to dispell. Just keep in mind it takes energy to create a permanent magnet in the first place. Potential energy is not the same thing as 'work'. A boulder rolling downhill looks like free energy too, until you consider the amount of energy it took to place the boulder there to begin with.
 
  • #11
ok, let me try to explain this. As Galileo stated, the magnetic field does no work on elements of charge or current, but, as I interpret it, megnetic fields CAN do work.

Let us take two rings of current. One ring of current has current flowing in the clockwise direction and we will say it lies in the x-y plane and the other ring has current flowing in the counterclockwise direction and it is lying in a plane parallel to the x-y plane but translated along the z axis by a certain distance. Both rings have the same diameter. The rings will produce a magnetic field with a magnetic dipole moment given by the direction of the surface defining the loop, which is defined by the direction of the current in the loop. So, both magnetic moments due to both current rings are pointing in the -z direction. Now, keep in mind that the B-field due to the second ring (which is NOT the same thing as the magnetic moment and, unlike the moment, the B-field varies) will exert forces on the the invividual current elements that are not in the direction of the motion you observe between the rings (they will attract), but these forces add up to a net force which causes the attraction between the rings. It is important to realize here that the moment is not the same as the B-field. If the B-field was the same as the moment then there would be no net force as the forces would all be pulling radially outward on the ring, and assuming the ring is rather strong, these forces would balance.

So, the magnetic field from one current ring DOES do work on another current ring, but as Galileo pointed out, a B-field doesn't do work on moving charges flowing in the current loops. However, the sum of the mechanical forces on all of the moving charges in the loop causes a net force to be exerted.

Keep in mind that the field due to a ferromagnetic material (permanent magnet) is caused by current also...it is caused by currents called eddy currents flowing through the magnetic material and the analysis of the two current loops can be easily applied to the a couple of bar magnets.

Does this mean that you get free energy? Well, in the case of the current loops you need to apply an EMF to get the current flowing in the loops in the first place, and as the loops are brought close together there is a net change in flux through each loop, so by faraday's law there is an EMF induced in the loop that opposes the current in the loop, so constant work must be done by the EMF source, even in a superconductor to maintain the force of attraction.

In the case of the bar magnet, as chronos pointed out, it take energy to take iron (or any ferromagnetic material...a ferromagnetic material is just something with a permanent dipole moment) and make it into a magnet. You must have an initial changing magnetic flux through the iron to produce that moment. You might ask why the permanent magnet maintain its moment and thus its eddy current? It's simply because of faraday's law! If the eddy current was reduced for some reason there would be a reduced B-field through the magnet, and therefore by faraday's law there would be an EMF to counter this reduced eddy current. There is generally no net reduction in eddy current over time. Also, with bar magnets, if you let them attract at some speed the by faraday's law the eddy current is reduced, but then you pull them back apart at some speed and the eddy current is restored! However, you can reduce the eddy current and thus the magnetic moment by having the magnets attract very rapidly, but then pull them away very slowly. If you do this enough you can kill the magnet's dipole moment.

Also, I'd like to note that many people feel that the lorentz force is not accounted for in maxwell's equations. when you look at the differential forms of the equations (in maxwellian form, not hertzian form) Faraday's law DOESN'T take into account the lorentz force because certain parts of the integral form of the equation are ignored when the differential form is derived, but it is taken into account if you look at the integral form of faraday's law.
 
  • #12
leright said:
Keep in mind that the field due to a ferromagnetic material (permanent magnet) is caused by current also...it is caused by currents called eddy currents flowing through the magnetic material and the analysis of the two current loops can be easily applied to the a couple of bar magnets.
leright, this is...lewrong! (bad joke, sorry)

Ferromagnetism requires no currents whatsoever and eddy currents are not what you think they are.
 
  • #13
Gokul43201 said:
leright, this is...lewrong! (bad joke, sorry)

Ferromagnetism requires no currents whatsoever and eddy currents are not what you think they are.

All magnetic fields are generated by currents, and permanent magnets are no exception. Except in ferromagnets the "current" is produced by a bunch of nuclei and electrons with quantum spins all aligned in one particular direction, which results in a net "current".

Sure a ferromagnet doesn't behave identically to a ring of current, but the ferromagnet can be modeled in quite the same fashion.
 
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  • #14
leright said:
All magnetic fields are generated by currents, and permanent magnets are no exception. Except in ferromagnets the "current" is produced by a bunch of nuclei and electrons with quantum spins all aligned in one particular direction, which results in a net "current".
The electron magnetic moment, which is responsible for ferromagnetism, is a relativistic QM effect that is not related to any current or "current".
 
  • #15
Meir Achuz said:
The electron magnetic moment, which is responsible for ferromagnetism, is a relativistic QM effect that is not related to any current or "current".

This is an old topic but it seems to be not closed yet. I've seen repeated claims that magnetic field can do work under some conditions.

1) Electron spin and magnetic momen are indeed quantum (and also relativistic) in nature. Yet, there is quantum-mechanical expectation of current associated with a single spin. Derivation is given, for example, in Landau, v.3.

2) Whenever work is done on a spin, it is the work of electric field on the current mentioned in point 1. This, for example, happens if a spin flips in an external H-field by emitting or absorbing a photon.

3) Since the Lorentz force applies to every elementary particle that makes up a macroscopic object, the magnetic field does no work on each particle separately nor on the system as a whole.

4) Whatever work is apparently done can be attributed to the work of electric fields that the particles of a macroscopic object create on the atomic scale, even if the average is zero. This is always done at the expense of the internal energy of the object (as in the example of two current loops which was offered in a previous post).

5) Magnetic field can do no work without exceptions, either it is constant or time- and spatially-varying, in both classical and quantum theories.
 
  • #16
Chronos said:
The myth of free energy from permanent magnets is difficult to dispell. Just keep in mind it takes energy to create a permanent magnet in the first place. Potential energy is not the same thing as 'work'. A boulder rolling downhill looks like free energy too, until you consider the amount of energy it took to place the boulder there to begin with.

How does it take energy to create a permanent magnet? In a ferromagnetic material, ferromagnetism is the ground-state. Domain boundaries which allow different parts of the material to have different magnetisms, and thus, allow the whole chunk of material to have less than its maximum magnetic moment, require energy to create.

The classical picture of having permanent currents in the magnet which are reduced as the magnet does work seems untrue. The "currents" (electron spins) cannot decrease, and it's energetically favorable to align them and have a ferromagnet. It costs energy to break their alignments, and I'm fairly sure that below the Curie temperature, this results in a net increase in free energy. Of course you can think of picking things up with a magnet at zero temperature, there the free energy of the magnet is minimized by maximizing the magnetic moment of the material. So the magnet doing work can't be compensated by decreasing the magnetism in the material, it seems.
 
  • #17
Ah, I think maybe I understand now. When a magnet picks up an iron nail, the energy to do work on the iron comes from within the iron itself. The spins inside the iron can align with the magnetic field, which lowers the energy inside the nail. That's where the energy to do work comes from.
 
  • #18
Suppose I had a pair of coreless electromagnets. And I used one to pick up the other. Would work be done? By which one?
 
  • #19
TVP45 said:
Suppose I had a pair of coreless electromagnets. And I used one to pick up the other. Would work be done? By which one?

Work is always done by a force. The classical expression for a point particle is dW/dt=vF. You can't say that an object such as an electromagnet does work.

In your example, work is done on each elementary particle that has moved from its original equilibrium position to a new one. If you are interested in the work of the magnetic force which is given by the Lorentz formula, it is exactly zero. If you are thinking about the work of some phenomenological force which is not described by the Lorentz formula, it may not be zero. You can then attribute this phenomenological force to the existence of magnetic field. This doesn't mean that magnetic field does work.

One difficulty is that the macroscopic objects can not be understood in purely classical electromagnetic theory without addition of some phenomenological forces that are needed to hold the object together. As is well known, a system of classical charges can not be in stable equilibrium. Therefore, one must assume, sometimes implicitely, the existence of phenomenological forces which are not electromagnetic in nature.
 
  • #20
kanato said:
How does it take energy to create a permanent magnet? In a ferromagnetic material, ferromagnetism is the ground-state. Domain boundaries which allow different parts of the material to have different magnetisms, and thus, allow the whole chunk of material to have less than its maximum magnetic moment, require energy to create.

You must consider the permanent magnet and whatever it picks up and the EM field that will be radiated in the process as a system. The permanent ferromagnet isolated from anything else may be in its QM ground state. But the magnet+nail system is not. Its original state is not the lowest energy state. When the nail is attracted to the magnet, the energy of the system is decreased to a lower value.

Disregard gravity for a while. Assume that this has happened in vacuum. When the nail flew towards the magnet, it acquired some kinetic energy. Then it hit the magnet. The kinetic energy must be dissipated into healt (lattice vibrations). The thermal energy will then be dissipated due to the black body radiation, however weak.
 
  • #21
burashka said:
Work is always done by a force. The classical expression for a point particle is dW/dt=vF. You can't say that an object such as an electromagnet does work.

In your example, work is done on each elementary particle that has moved from its original equilibrium position to a new one. If you are interested in the work of the magnetic force which is given by the Lorentz formula, it is exactly zero. If you are thinking about the work of some phenomenological force which is not described by the Lorentz formula, it may not be zero. You can then attribute this phenomenological force to the existence of magnetic field. This doesn't mean that magnetic field does work.

One difficulty is that the macroscopic objects can not be understood in purely classical electromagnetic theory without addition of some phenomenological forces that are needed to hold the object together. As is well known, a system of classical charges can not be in stable equilibrium. Therefore, one must assume, sometimes implicitely, the existence of phenomenological forces which are not electromagnetic in nature.

Thanks for the reply. Note that I was careful to stay away from permanent magnets. I'm afraid I don't know what a phenomenological force is; can you give me a reference or some examples?

Anyway, my next set of questions would be:
1) Can a coil of wire have a magnetic field?
2) Does it require work to create that field if it exists?
3) Is any part of such work stored as potential energy?
4) If so, is any part of that potential energy available to do work?
 
  • #22
burashka said:
You must consider the permanent magnet and whatever it picks up and the EM field that will be radiated in the process as a system. The permanent ferromagnet isolated from anything else may be in its QM ground state. But the magnet+nail system is not. Its original state is not the lowest energy state. When the nail is attracted to the magnet, the energy of the system is decreased to a lower value.

That's true, but we don't have to consider the nail hitting the magnet. Consider the initial system to be the magnet and nail before the nail begins to move, and the final system to be the magnet and nail where the nail has picked up some kinetic energy. In this case I can set the final time to be well enough before the collision that the internal quantum systems of the two systems are isolated enough from each other to be independent, and the only interaction between them is through a classical B field.

In this case, the internal free energy of the magnet can't decrease; it's already at its lowest value. (I suppose there may be some feedback from the B field caused by spins aligning in the iron nail, but I would imagine that's small compared to the other energy scales in the problem, because the nail doesn't pick up the same magnetization as the magnet.) The iron nail has picked up classical kinetic energy, and the only place that could really come from is from the breakdown of domain walls in the unmangetized iron allowing it to form a net magnetic moment.

burashka said:
Disregard gravity for a while. Assume that this has happened in vacuum. When the nail flew towards the magnet, it acquired some kinetic energy. Then it hit the magnet. The kinetic energy must be dissipated into healt (lattice vibrations). The thermal energy will then be dissipated due to the black body radiation, however weak.

It's true, although I don't see how that changes anything. You don't have to consider the collision and what happens afterward for my argument.

TVP45 said:
Suppose I had a pair of coreless electromagnets. And I used one to pick up the other. Would work be done? By which one?

It's a good question. You can simplify it a bit; consider two parallel current carrying wires. There will be a force between them which is perpendicular to the B field: http://hyperphysics.phy-astr.gsu.edu/Hbase/magnetic/wirfor.html#c1

I guess you can think of it as the B field causes the charge carriers to "bump" into the edge of the conductor, pushing it in the direction of the force. My intuition tells me that this will result in an increase in the resistance of the wire, so that in order to maintain the same amount of current, the power source has to work harder. Or, if the power source is a constant voltage source, then the current would decrease. Either way, energy for the work done comes from the battery.
 
  • #23
TVP45 said:
Thanks for the reply. Note that I was careful to stay away from permanent magnets. I'm afraid I don't know what a phenomenological force is; can you give me a reference or some examples?


Well, what is the force that holds electrons inside a piece of metal? You know that there is a finite exit work needed to extract an electron. However, if we simply assume that electrons simply move in a positively-charged background, this system would not be stable. A tiny external field would ionize the system. The problem can be solved by either invoking quantum mechanics or by assuming that there is a phenomenological potential barrier at the metal surface that keeps the electrons inside.

Another example, is the harmonic restoring force in the Lorentz' model of dielectrics.

Anyway, my next set of questions would be:
1) Can a coil of wire have a magnetic field?

Sure

2) Does it require work to create that field if it exists?

Yes

3) Is any part of such work stored as potential energy?

Generally, yes, but the notion of potential energy does not easily apply to EM fields. Let me say that the whatever work was done to start the current was (i) partially dissipated to heat, radiation, etc.; (ii) partially stored as kinetic energy of of moving charges and (iii) partially stored as the energy of EM field that the current creates around itself. The magnetic field can have energy density; it just can't do any work (I hope, this doesn't sound like a contradiction).

4) If so, is any part of that potential energy available to do work?

Within the constraints of the second law of thermodynamics, yes.
 
  • #24
kanato said:
That's true, but we don't have to consider the nail hitting the magnet. Consider the initial system to be the magnet and nail before the nail begins to move, and the final system to be the magnet and nail where the nail has picked up some kinetic energy. In this case I can set the final time to be well enough before the collision that the internal quantum systems of the two systems are isolated enough from each other to be independent, and the only interaction between them is through a classical B field.

How can these two sysyems be isolated frome ach other if they interact electromagnetically and exchange energy?

kanato said:
In this case, the internal free energy of the magnet can't decrease; it's already at its lowest value. (I suppose there may be some feedback from the B field caused by spins aligning in the iron nail, but I would imagine that's small compared to the other energy scales in the problem, because the nail doesn't pick up the same magnetization as the magnet.) The iron nail has picked up classical kinetic energy, and the only place that could really come from is from the breakdown of domain walls in the unmangetized iron allowing it to form a net magnetic moment.

But the energy will decrease later, so it can not be at its lowest value

The bottom line is that the total energy of a closed system is always conserved. There are however no truly closed system because of radiation and similar stuff. So you need to consider the initial energy (before the motion started) and the final energy (after the collision and everything settling down). Obviously, the second will be less that the first and the difference will be carried away by radiation to infinity.

Another aspect is the following: you can write a Lagrangian and a Hamiltonian of a system of moving charges only to the order [tex]\beta^2[/tex], where [tex]\beta=v/c[/tex]. This is significant because radiation is an effect which is third order in [tex]\beta[/tex]. That means that in our energy balance analysis we can only consider the initial and final states of the system when the is no motion and no radiation.
 
  • #25
AlbertEinstein said:
Sorry, but while going through the article on magnetism in wikipedia I read the following which i couldn't not catch it.

When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:
F= q V x B
where is the electric charge of the particle is the velocity vector of the particle is the magnetic field.

Because this is a cross product, the force is perpendicular to both the motion of the particle and the magnetic field. It follows that the magnetic force does no work on the particle; it may change the direction of the particle's movement, but it cannot cause it to speed up or slow down.

This might give you pause: Simple bar magnets seem to be entirely able to pick up small metal objects, which certainly seems to require that they do work on those objects. As David Griffiths points out in his textbook Introduction to Electrodynamics, this law is absolute -the magnetic field doesn't do any work. However, quite like the normal force of an inclined plane, which also can't do work, the magnetic field can redirect the efforts of existing forces, and then those forces can indeed do work in the relevant direction.

I am not able to understand the underlined lines. I do not know about which "existing forces" is he talking about; and the same applies to "those forces".I will really appreciate the help.Thanks in advance.

The fields cannot do work because the charge does not accelerate in the direction of the field, and will not accelerate at all if it is stationary. You fail to realize that moving charges also produce magnetic fields (such as those that produce electromagnets):

550px-Electromagnetism.svg.png


The net effect of the Lorentz force that you mentioned from the magnetic fields produced by the first moving charge on nearby charges moving in the same direction is to draw them closer. Two aligned magnets that have their opposite poles in proximity also have internal domain currents moving in the same direction, and so they are drawn closer. I hope that this answers your question.
 
  • #26
shadowpuppet said:
The fields cannot do work because the charge does not accelerate in the direction of the field, and will not accelerate at all if it is stationary. You fail to realize that moving charges also produce magnetic fields (such as those that produce electromagnets):

550px-Electromagnetism.svg.png


The net effect of the Lorentz force that you mentioned from the magnetic fields produced by the first moving charge on nearby charges moving in the same direction is to draw them closer. Two aligned magnets that have their opposite poles in proximity also have internal domain currents moving in the same direction, and so they are drawn closer. I hope that this answers your question.

The fields can not do work under any cirumstances. Only forces can do work.

In electromagnetic theory, the force acting on an electric charge is given by the Lorentz formula.

The Lorentz formula has two terms: one with electric field and the other with magnetic field.

The term with the magnetic field is always pointinmg perpendicularly to the particle velocity.

The work (per unit time) is given by a dot product of the force and velocity, Since the second term in the Lorentz formula is perpendicular to the velocity, it drops out from this dot product.

That pretty much sums it all. Everythin else is more or less irrelevant.

It is however easy to get confused if you either assume that objects such as electromagnets can do work (no, only forces can) or that energy can do work (no, only force can) or that if work is being done in the presence of magnetic field and even requires such presence, then somehow the magnetic field is doing work (no, fields can not do work, only forces can, and the Lorentz force due to the magnetic field does no work by definition).
 
  • #27
Ignoring the work aspect of a magnetic field, I would assume that if a magnetic field is producing a force on a charged particle, then an equal and opposite force is applied to what ever object that is generating the magnetic field?
 
  • #28
Jeff Reid said:
Ignoring the work aspect of a magnetic field, I would assume that if a magnetic field is producing a force on a charged particle, then an equal and opposite force is applied to what ever object that is generating the magnetic field?

The Newton's third law holds in electromagnetic theory only approximately, when all velocities are small.
I am quite sure that the third law breaks down in the third order in [tex]\beta=v/c[/tex]. Note that v is the velocity of elementary particles inside the object, not the macroscopic velocity of the objects center of mass. In the third order, radiation appears and the third law doesn't really hold.

But it is probably the case that the third law breaks down already in the second order, as soon as you take the retardation into account. So action=reaction is only the property of static, instanteneous fields.
 
  • #29
burashka said:
The fields can not do work under any cirumstances. Only forces can do work.

In electromagnetic theory, the force acting on an electric charge is given by the Lorentz formula.

The Lorentz formula has two terms: one with electric field and the other with magnetic field.

The term with the magnetic field is always pointinmg perpendicularly to the particle velocity.

The work (per unit time) is given by a dot product of the force and velocity, Since the second term in the Lorentz formula is perpendicular to the velocity, it drops out from this dot product.

That pretty much sums it all. Everythin else is more or less irrelevant.

It is however easy to get confused if you either assume that objects such as electromagnets can do work (no, only forces can) or that energy can do work (no, only force can) or that if work is being done in the presence of magnetic field and even requires such presence, then somehow the magnetic field is doing work (no, fields can not do work, only forces can, and the Lorentz force due to the magnetic field does no work by definition).

I never said that all fields can do work, just as not all forces can do work (such as the electroweak force and centrifugal force). FYI I was explaining why magnets attract each other.
 
  • #30
burashka said:
Whatever work is apparently done can be attributed to the work of electric fields that the particles of a macroscopic object create on the atomic scale, even if the average is zero. This is always done at the expense of the internal energy of the object (as in the example of two current loops which was offered in a previous post).

This is incorrect. If magnetism was not considered a separate influence in electrodynamics then James Clerk Maxwell wouldn't have incorporated it into A Treatise on Electricity and Magnetism.

kanato said:
The classical picture of having permanent currents in the magnet which are reduced as the magnet does work seems untrue.

Eddy currents are practically negligible in comparison to the source currents that cause them, otherwise they would be effectively canceled out by a regression of currents that they themselves cause both in themselves and in the source.

burashka said:
Therefore, one must assume, sometimes implicitely, the existence of phenomenological forces which are not electromagnetic in nature.

You are making the subject seem harder than it is.

TVP45 said:
Thanks for the reply. Note that I was careful to stay away from permanent magnets. I'm afraid I don't know what a phenomenological force is; can you give me a reference or some examples?

Anyway, my next set of questions would be:
1) Can a coil of wire have a magnetic field?
2) Does it require work to create that field if it exists?
3) Is any part of such work stored as potential energy?
4) If so, is any part of that potential energy available to do work?

1. Yes, it is caused by the net contribution of all of the individual coils, usually one direction is reinforced on the inside which depends on the direction of the current and the other is reinforced on the outside, as is evident when iron filings are spread over a bar magnet.

2. It requires work to move the charges which are themselves subject to both electric and magnetic fields, but it is more common to use the electric potential to induce current because this can be controlled chemically and magnetic fields consequentially manifest.

3. Yes, the potential energy is equal to the inductance of the coil measured in Henrys multiplied with the square of the current in the coil measured in Amperes. The inductance is [itex]L = \frac{{{\mu}_0}{N^2}R}{2}[/itex] where [itex]N[/itex] is the number of turns and [itex]R[/itex] is the radius.

4. If the current is reduced an electromotive force equal to the inductance times the negative derivative of the current with respect to time will try to replace the current until the energy has been dissipated from the system in a finite time as an exponential decay.
 
  • #31
burashka said:
kanato said:
That's true, but we don't have to consider the nail hitting the magnet. Consider the initial system to be the magnet and nail before the nail begins to move, and the final system to be the magnet and nail where the nail has picked up some kinetic energy. In this case I can set the final time to be well enough before the collision that the internal quantum systems of the two systems are isolated enough from each other to be independent, and the only interaction between them is through a classical B field.

How can these two sysyems be isolated frome ach other if they interact electromagnetically and exchange energy?

I meant that they are far enough apart that the quantum state of one system doesn't affect the other, except through the classical B field that is generated.

burashka said:
kanato said:
In this case, the internal free energy of the magnet can't decrease; it's already at its lowest value. (I suppose there may be some feedback from the B field caused by spins aligning in the iron nail, but I would imagine that's small compared to the other energy scales in the problem, because the nail doesn't pick up the same magnetization as the magnet.) The iron nail has picked up classical kinetic energy, and the only place that could really come from is from the breakdown of domain walls in the unmangetized iron allowing it to form a net magnetic moment.

But the energy will decrease later, so it can not be at its lowest value

What energy will decrease later? The total energy? But below you say it's conserved?

I'm dividing the system up into several energy systems:
[tex]E_{int}^{mag} + E_{int}^{nail} + T_{nail} = \mathrm{const}[/tex]

where [tex]E_{int}[/tex] is the internal energy of the magnet or the nail, determined by the quantum mechanical structure of the material, and T is the macroscopic kinetic energy of the nail. The total energy is conserved. My argument was that [tex]E_{int}^{mag}[/tex] is already at its minimum value due to its Ferromagnetic order, but [tex]E_{int}^{nail}[/tex] can go down if the spins within the nail align. This allows the kinetic energy to increase.

burashka said:
The bottom line is that the total energy of a closed system is always conserved. There are however no truly closed system because of radiation and similar stuff. So you need to consider the initial energy (before the motion started) and the final energy (after the collision and everything settling down). Obviously, the second will be less that the first and the difference will be carried away by radiation to infinity.

You don't have to consider the final state to be after the collision. You can if you want, but energy is conserved <i>at all times</i> so I can pick any initial and final times that I want.

When the magnets attract each other, there is an increase in kinetic energy. I'm interested in understanding when where that energy comes from. Once they collide, yes that kinetic energy goes somewhere (thermal energy, perhaps into disordering the spins too) but I don't care about that; it's not fundamentally different from any other collision.

burashka said:
Another aspect is the following: you can write a Lagrangian and a Hamiltonian of a system of moving charges only to the order [tex]\beta^2[/tex], where [tex]\beta=v/c[/tex]. This is significant because radiation is an effect which is third order in [tex]\beta[/tex]. That means that in our energy balance analysis we can only consider the initial and final states of the system when the is no motion and no radiation.

Can you elaborate on this? The expansion comes from expanding [tex]\gamma = (1-\beta^2)^{-1/2}[/tex] in a Taylor series, right? This can be carried out to arbitrary order right? What barrier is there to preventing doing this in the Lagrangian formalism?

Are you talking about radiation from accelerating charges? Because IIRC the radiation for positive and negative charges has a pi phase shift, so accelerating a neutral system (like the iron nail) doesn't cause a significant amount of radiation because of destructive interference of the waves.
 
  • #32
kanato said:
I meant that they are far enough apart that the quantum state of one system doesn't affect the other, except through the classical B field that is generated.



What energy will decrease later? The total energy? But below you say it's conserved?

I'm dividing the system up into several energy systems:
[tex]E_{int}^{mag} + E_{int}^{nail} + T_{nail} = \mathrm{const}[/tex]

where [tex]E_{int}[/tex] is the internal energy of the magnet or the nail, determined by the quantum mechanical structure of the material, and T is the macroscopic kinetic energy of the nail. The total energy is conserved. My argument was that [tex]E_{int}^{mag}[/tex] is already at its minimum value due to its Ferromagnetic order, but [tex]E_{int}^{nail}[/tex] can go down if the spins within the nail align. This allows the kinetic energy to increase.



You don't have to consider the final state to be after the collision. You can if you want, but energy is conserved <i>at all times</i> so I can pick any initial and final times that I want.

When the magnets attract each other, there is an increase in kinetic energy. I'm interested in understanding when where that energy comes from. Once they collide, yes that kinetic energy goes somewhere (thermal energy, perhaps into disordering the spins too) but I don't care about that; it's not fundamentally different from any other collision.



Can you elaborate on this? The expansion comes from expanding [tex]\gamma = (1-\beta^2)^{-1/2}[/tex] in a Taylor series, right? This can be carried out to arbitrary order right? What barrier is there to preventing doing this in the Lagrangian formalism?

Well, assume you have a system of N classical charged particles. You then want to write a Lagrangian for this system. In the zeroth approximation

[tex] L = \sum_i m_iv_i^2/2 - \sum_{i>j} q_i q_j /r_{ij} [/tex]

But this completely neglects retardation.
At the next step, we can try to write the lagrange function for every particle in a given field and then compute that field as a function of all particle positions. We have

[tex] L_i = -mc^2\sqrt{1 - v_i^2/c^2} - q_i\varphi + (q_i/c){\bf A} \cdot {\bf v}_i[/tex]

Here [tex]\varphi,{\bf A}[\tex] are the scalar and vector potentials. It is not difficult to see that if we do not make any approximations, the resultant sum of all L_i is not the proper Largange function of the system. In particular, it would not result in proper energy conservation laws. This is explained by the fact that the field in electromagnetic theory itself has a number of independent "degrees of freedom" which must be included into the Lagrangian. Only then the total energy is conserved.

It is possible, however, to expand the Lagrangian to the second order in [tex]v_i/c[\tex]. This amounts to neglecting radiation.


Are you talking about radiation from accelerating charges? Because IIRC the radiation for positive and negative charges has a pi phase shift, so accelerating a neutral system (like the iron nail) doesn't cause a significant amount of radiation because of destructive interference of the waves.

A truly neutral system of course can not radiate. But the magnetized object, when moves through space, would create time-varying fields and this leads to radiation. You can not view a magnetized object as electrically neutral. It has currents running in it and these currents are made of charged particles.
 
  • #33
kanato said:
I meant that they are far enough apart that the quantum state of one system doesn't affect the other, except through the classical B field that is generated.

But you can not through this interaction away since it is important. So the magnet and the nail are not isolated. When you assume that they are, you neglect the very effect you want to consider.


kanato said:
What energy will decrease later? The total energy? But below you say it's conserved?

Yes, the total energy is conserved, including the energy of radiation that will eventually go to infinity. The energy of the atoms that make up the magnet and the nails will be decreased.
Or do you suppose that a system can never go from an excited state to a ground state because the total energy must be conserved?
 
  • #34
You don't have to consider the final state to be after the collision. You can if you want, but energy is conserved <i>at all times</i> so I can pick any initial and final times that I want.

When the magnets attract each other, there is an increase in kinetic energy. I'm interested in understanding when where that energy comes from. Once they collide, yes that kinetic energy goes somewhere (thermal energy, perhaps into disordering the spins too) but I don't care about that; it's not fundamentally different from any other collision.

I am not sure I understand. The energy doesn't have to come from "somewhere". It is simply conserved at all moments in time. But you have to account for all energy, including the energy of EM fields. In the example of a magnet and a nail, the magnet makes an H field around itself and that field has energy (and also mass according to E=mc^2). When the nail starts moving, it acquires some kinetic energy but the EM fields arond the objects also change and the total energy is conserved. It is however, not easy to see in detail how this happens. So it is better to consider the end-points, when everything is static.
 
  • #35
burashka said:
I am not sure I understand. The energy doesn't have to come from "somewhere". It is simply conserved at all moments in time.

The kinetic energy has to come from "somewhere." In order for the kinetic energy of the iron to go up, the energy of some other subsystem has to go down. That's what conservation of energy means. That's what I'm saying I want to understand, what other energy subsystem goes down? I don't understand how is this not clear to you.

burashka said:
But you have to account for all energy, including the energy of EM fields. In the example of a magnet and a nail, the magnet makes an H field around itself and that field has energy (and also mass according to E=mc^2). When the nail starts moving, it acquires some kinetic energy but the EM fields arond the objects also change and the total energy is conserved. It is however, not easy to see in detail how this happens. So it is better to consider the end-points, when everything is static.

You don't have to account for all the energy, only the energy subsystems that change significantly. I have a very hard time believing that there is significant radiation moving the weakly magnetized iron. The speeds here are such that v <<<< c, so any effect that is [tex]\beta^3[/tex] should be quite negligible. So I'm assuming that it's negligible.

That lagrangian you wrote down is not the related to the system I'm describing. You have charges in there, but there is nothing charged here. The magnet and the iron are both neutral.
 

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