# Magnetic Fields and Work

1. Nov 28, 2009

### lugita15

I've been following the recent thread about work done by magnetic fields, and I'm more confused than ever. I have a few questions about the subject. I have phrased them as true/false statements.
Consider Statement A: If an object X moves in any direction which is not perpendicular to some force F, then the work done by F on X is nonzero.
Is Statement A true or false?
Consider the situation where there are two parallel current-carrying wires seperated by some distance d, each having a current of magnitude I and pointing in the same direction. For simplicity, let us assume that the currents are going from the bottom of the page to the top of the page, and that the first wire is to the left of the second wire.
Now consider Statement B: By the Biot-Savart Law, the magnetic field due to the second wire at any point on the first wire is directed out of the page.
Is Statement B true or false?
Now consider Statement C: Since the magnetic force on a current-carrying wire is proportional to $$\vec{I}\times\vec{B}$$, by the right hand rule the magnetic force on the first wire due to the magnetic field of the second wire is directed to the right (that is, towards the second wire).
Is Statement C true or false?
Now consider Statement D: As a result of the rightward force it is experiencing, wire 1 will move to the right, i.e. closer to wire 2.
Is Statement D true or false?
Now consider Statement E: The direction in which wire 1 is moving, which is to the right, and the direction of the magnetic force acting on it, which is also the right, are parallel, which obviously means they are not perpendicular.
Is Statement E true or false?
Finally consider Statement F: Since wire 1 is moving in a direction which is not perpendicular to the magnetic force acting on it, by Statement A the work done by the magnetic force acting on wire 1 is nonzero.
Is Statement F true or false?
If you think that Statment F is false please point to the earlier step in which I made a mistake.

Any help would be greatly appreciated.

2. Nov 28, 2009

### Longstreet

There is a missing statement that the current I is held constant. If you have two free interacting currents, say both $$I_{z}$$, they will move toward each other. But now the z component is different for each: $$I_{z}^{'} < I_{z}$$. Kinetic energy is not changed. What changes is the distribution of EM field. If you want to keep $$I_{z}$$ fixed you have to supply an emf in z direction. So, you'll find that the emf is what is doing the work to drive the wires together.

3. Nov 28, 2009

### lugita15

Yes, let us assume that the current in each wire does not change. Basically the assumption is that the distance between the wires is sufficiently great, and consequently the magnetic field strength is sufficiently low, that there is no significant change in the currents.
Out of the six statements I listed, which ones do you agree with and which one do you disagree with, and why?

4. Nov 29, 2009

### Longstreet

If I had to pick one it would be statement E. Once the wires are moving toward each other then jxB (or vxB) no longer points toward the other wire (to the right).

5. Nov 29, 2009

### lugita15

Obviously the direction of the magnetic force will change as a function of time. But the important thing is that, at least initially, the wire will move in a direction that is not perpendicular to the magnetic force acting on it. Therefore SOME amount of work, however small, must be done by the magnetic force due to the second wire.

6. Nov 29, 2009

### Bob S

Statement F is true.
Suppose the two wires are in the x y plane, and the currents are in the y direction, and wire 1 moves in the x direction.
The stored energy per unit length in the system is

W = (1/2)LI2,
where L is the inductance per unit length.

The force Fx= dW/dx = (1/2)d/dx( LI2) = (1/2) I2 dL/dx + LI dI/dx.

Since I is a constant, the second term is zero, so we have

Fx = (1/2) I2 dL/dx

So the motion of the wire is reducing the stored inductive energy.

The electrical energy was due to a voltage pulse V = I dL/dt + L dI/dt , where the second term is zero.

Statements A through E are also true.

Bob S

7. Nov 29, 2009

### Longstreet

So how do you explain jxB and j*E at the fundamental level.

8. Nov 30, 2009

### Bob S

j x B refers to the vector cross product of a current density and a static magnetic field, resulting in a force F on the current density. This can be Lorentz transformed to a force (density) j'*E +j' x B', where the force now comes from both electric and magnetic fields. The force can be completely transformed to electric fields if the particles in j are completely relativistic.
Bob S

9. Nov 30, 2009

### Hootenanny

Staff Emeritus
Bob S' point is that it depends on your reference frame, or point of view. Suppose you have a distribution of charges that is stationary relative to you. You see an electric field, but no magnetic field. Now if I'm moving relative to you, I see a magnetic field. The two are difference aspects of the same object rather than distinct objects in themselves.

This point is emphasised when we use the electromagnetic tensor to write Maxwell's equations

$$F_{\left[\alpha\beta,\gamma\right] = 0$$

$${F^{\alpha\beta}}_{;\beta} = \mu_0J^\alpha$$

Here, there is no distinction between electric and magnetic fields, we simply have the electromagnetic field.

Last edited: Nov 30, 2009
10. Nov 30, 2009

### Longstreet

I mean in terms of work. obviously jxB does no work because j*(jxB) = 0. So magnetic force does no work, in only changes momentum. Mechanical power transfer comes from J*E at a fundamental level so saying the power comes from magnetic force is misleading it comes from the electrical force. Transforming frames doesn't get around the fact because j*(jxB) is zero by definition in every frame.

Also, looking at the divergence of the poynting vector $$\nabla \cdot S = \nabla \cdot (E \times B) = B \cdot (\nabla \times E) - E \cdot (\nabla \times B)$$

you find

$$\frac{\partial u}{\partial t} + \nabla \cdot S + J \cdot E = 0$$

where u is electromagnetic field energy density.

So by this energy can only be exchanged between EM field and matter through the electric field and current. So I ask again, how do you say magnetic field does work, at the fundamental level.

11. Nov 30, 2009

### Longstreet

A square looks rectangular in another frame but its the same object. Just because they look different doesn't mean they are different. It's called lorentz covariance. You can describe an object and what it looks like in every frame.

And no I did not even write field potentials (phi and A I presume) in my post. And yes, but what is a) causing the force b) direction of force and c) what causes the work is being obfuscated. All you are saying is that work is done. Yes, well duh, if I may be frank. But not saying WHAT does the work, which is the initial question.

12. Nov 30, 2009

### Longstreet

How is it causing work when v*vxB = 0. just answer that.

13. Nov 30, 2009

### Bob S

Hi Drum-

The stored magnetic energy per unit length is

W= (1/2) LI2

where L is the inductance per unit length, also known as the stored magnetic energy per unit length.

L = (1/I2)∫B·H dVvolume

the force Fx is given by

Fx = dW/dx = (1/2)d(LI2)/dx = (1/2) I2 dL/dx + L dI/dt, (but dI/dt =0 at constant current),

where x is the separation between the two wires.

A reduction dx of the wire separation at constant current is producing a voltage pulse

V = L dI/dt + I dL/dt = I dL/dt

and removing stored magnetic energy between the two wires.

Bob S

Last edited: Nov 30, 2009
14. Nov 30, 2009

### Longstreet

x: displacement

dx/dt = v: displacement/time

So rate of work, also known as power, from the lorentz force: $$v \cdot (q v\times B)= 0$$.

It's zero.

15. Dec 1, 2009

### Hootenanny

Staff Emeritus
No it doesn't. If I'd have meant "two different things" I would have said that.
But I am saying that they are the same thing. Whilst this may be a matter of taste with no 'practical' significance, I think it's a discussion worth having. As I said in my previous post, in relativity there is no difference in magnetic and electric fields, it is simply a matter of reference frames. If a choice of reference frames determines whether we observe a magnetic field or an electric field, then it is obvious that the distinction between electric and magnetic fields is an artificial one, imposed by us, rather than one imposed by the underlying laws. In other words, there is no difference between electric and magnetic fields.
The difference here is that an apple remains an apple irrespective of your relative velocity. If the apple changed to an orange if you move relative to it, then we would have a situation which was analogous to EM fields. If this would be the case, it wouldn't make any sense to have apples and oranges since whether you observe one or the other depends on how you are travelling relative to them. Therefore, we would be forced to conclude that apples and oranges are in fact the same objects and we would work with apple-oranges.

16. Dec 1, 2009

### Bob S

Your statement paraphrases Humpty Dumpty in Alice in Wonderland:
‘When I use a word,’ Humpty Dumpty said, in a rather scornful tone, ‘it means just what I choose it to mean, neither more nor less.’

‘The question is,’ said Alice, ‘whether you can make words mean so many different things.’

‘The question is,’ said Humpty Dumpty, ‘which is to be master – that’s all.’

Bob S

17. Dec 1, 2009

### Hootenanny

Staff Emeritus
The magnetic field resulting from a current is nothing else than an application of length contraction applied to the distance between the positive ions in the wire.

A single current carrying wire: http://galileo.phys.virginia.edu/classes/252/rel_el_mag.html

The result for two generalises in an obvious manner for two wires.

As for my motivation for making this point, it is simply that I think it a nice result. By combining special relativity and the electric field, we can arrive at electromagnetism. Do you not agree that this is a nice result?

18. Dec 1, 2009

### Longstreet

Those are the equation of motion for charged particles and power delivered to them. Why would they go back to original positions? Work has to do with change of ENERGY, not position. The magnetic force is not a conservative force, for no other reason that it does no work in the first place. The magnetic field cannot change the mechanical energy of the system. That is what work is.

The magnetic force is NOT in the direction of displacement. You are being confused by saying that current is in one direction, and then saying that it's moving in a different direction. It's redundant to say a current is moving. Current IS movement (of charges).

Compelled by the fact that electromagnetism makes no sense unless you do this. Take the magnetic force $$qv\times B$$. Now change your reference frame x' = x - vt. v' = v - v = 0. So $$qv'\times B = 0$$. The particle no longer has an acceleration of any kind. Physics has apparently CHANGED just by switch reference frames, which makes no sense.

19. Dec 1, 2009

### Longstreet

The shape of the magnetic field lines has nothing to do with this. I will try to explain this without vector algebra. Your picture is only valid for a static situation which by definition has no work associated with it. To understand what happens when the wire actually moves, you need to look at the forces internally.

If we assume electrons are the source of current, then in wire 1 they will be moving up and to the left. This causes a force up and to the right, toward wire 2. No work has been done yet though because nothing has moved. Now, if the wire starts moving up and to the right , then the electrons are not moving in the same direction they were. They have a net motion more directly upward, slightly toward wire 2. But they are not moving directly toward wire 2 either. Also, because of this motion toward wire 2, there is an additional force down and to the right. So the net force is no longer pointed directly toward wire 2 either. It turns out that since the electrons are being displaced more to the up, and the force is more to the right, that force and direction are perpendicular still. No work has been done by magnetic force even when moving.

And no we do not agree. You cannot have a *complete* description of the electromagnetic field with only electric, or only magnetic fields. They are two properties of the same field. Properties are different, yes, but they are still part of the same entity. The maxwell equations even have what is called duality. They are completely symmetric and you can interchange electric and magnetic fields and electric and magnetic charges or have combinations of them and you will get the same equations. This has been well understood classically far 100 years. If you are really interested in understanding it there are many sources etc available.

Last edited: Dec 1, 2009
20. Dec 1, 2009

### Nirgal

The question is ambiguous. From a certain philosophical point of view you could say that the Electric and Magnetic fields are different, in the sense that the different phenomena you observe can be understood by invoking the idea of an Electric field or a Magnetic field. Again, as iterated many times by previous posts, observation is what distinguishes the two.
The Electric and Magnetic fields are just conceptualizations to understand problems. They do not exist. What actually exists is the Electromagnetic field. Call it what you will, the point is that Electrical phenomena and Magnetic phenomena all come from the same source. That "source", the Electromagnetic field, is a real thing that exists at every point in space. Electricity cannot exist without Magnetism and vice versa. This means the two phenomena are two sides of the same coin.

Think about light. As I said before, the Electromagnetic field exists at every point in space, when you move a charge then you have changed the local electric field in your region. But regions far from you still experience the "old" electric field. You can think of a discontinuity in the field that spreads out from the charge at the speed of light, allowing regions far from the charge to experience the new structure of the field. This is a simple description of an Electromagnetic wave. But wave propagation cannot occur with out the both fields, because a time changing magnetic field induces a time changing electric field and vice versa. The E-field and the B-field create each other in a kind of self-referential tango.

You might be tempted to say "wait a minute, all you have are two fields interacting to create light!" But you arrive at paradoxes in other areas of physics when you assume independence of one another. Really think about their relative nature. If you ride with an electron you will not experience the magnetic field due to the current. Suddenly you stop riding with it and you experience a magnetic field. Where did it come from?

I think the root of your confusion is philosophical. How humans classify things. What words to give phenomena.

21. Dec 1, 2009

### Born2bwire

DRUM, you need to brush up on the basics of electrodynamics. First, work is defined as the application of force over a distance, the integral quantity
$$W = \int \mathbf{F}\cdot d\mathbf{\ell}$$
When the path of displacement is not along the force vector, then the work done is zero. It is obvious from the Lorentz force that the contribution from the magnetic field gives rise to zero work. The displacement is along the velocity vector yet the cross product of the velocity and magnetic fields means that the force vector will be normal to the velocity vector. This can be seen most easily in cyclotron motion. If I have a charge moving at some velocity v in a constant magnetic field applied perpendicular to the velocity, then the charge will be induced to move in a circular orbit. And just like a mass on a string being revolved in a circular orbit, the force on the charge is always pointing towards the center of the orbit, causing the work done over any length of path to be zero.

Maxwell's equations satisfy special relativity, indeed they helped to directly precipitate the derivation of the theory. As such, we are often at home working with electromagnetic fields in the Minkowski space through the use of four-vectors, covariants, tensors and such. In this respect we see that electric and magnetic fields are purely a matter of reference frames. They are equivalent to each other in the respect that they always exist together (I just mentioned Jefimenko's equations in another thread that gives one insight into this) and that by manipulating the reference frames one can induce the magnetic field from a electric field. The calculation of the magnetic field from a line of current using Lorentz transformations on the electric field of the charges in their frame is a common undergraduate physics problem in electromagnetics. I am sure you can find the appropriate derivations in say Jackson or Griffiths.

22. Dec 2, 2009

### Born2bwire

Displacement is not along the force vector. The force vector is the direction of acceleration. As I mentioned previously, if you take the electron in a cyclotron you can see most clearly that the force vector is normal to the path of displacement at any point. And again, we can see that the magnetic field's contribution to the force does no work because of this. We can even see this explicitly with the wires, the movement of the charges are perpendicular to the force for our given static initial condition. In this case, it is apparent that the work at the initial condition is zero since the force is being applied normal to the path of displacement of the charges at that moment.

And no, there is nothing where classical electrodynamics can explain a phenomenon where relativistic electrodynamics cannot. First, classical electrodynamics satisfies special relativity already, there is no modification needed to the theory for them to agree. As such, careful examination of classical electrodynamics will give rise to the Lorentz transformations (as they did originally for Lorentz when he derived them) and you can then apply them to show the equivalence between the electric and magnetic fields. Once again I can only implore you to pick up an undergraduate electrodynamics textbook and familiarize yourself with relativistic electrodynamics. However, your confusion of acceleration with the path of displacement also suggests that you may want to also work out a few simple trajectory problems to refamiliarize yourself with the basic mechanics of the problem. Force only points along the path of displacement for special conditions, the primary one being when the current velocity of the mass is zero.

23. Dec 2, 2009

### Born2bwire

Oh no, I wouldn't make such a prediction off-hand. First off there are two things to discuss here. The first is that the Lorentz force is not acting on the wire, it is acting on the charges confined to the wire. In this case, the wire is made up of a lattice of atoms which we can freely strip the valence electrons off of in reaction to fields and forces leaving behind positively charged, but largely immobily tied to the lattice, ions. Now the Lorentz force acting on the moving electrons displaces them and the displaced electrons pull the wire's lattice with them from their own Coulombic attraction (via electric fields), in a very brief hand-wavy kind of explanation. So it is not meaningful to talk about the Lorentz force working on the wire, it is only working (if at all) on the moving charges. So we have to take into account that the charges are moving normal to the direction of force acting on them.

The second point is that once we allow time to progress, we no longer have a magnetostatic problem. The movement of the wires now means that we have time varying magnetic fields, which invariably means we also have time varying electric fields. So now we have Lorentz forces from electric and magnetic fields. And with electric fields we now have Lorentz forces capable of performing work on the charges.

So does the Lorentz force point along the path of the wire's displacement? Most likely I am sure though it isn't a simple problem but I would expect one could easily use Lagrangian physics, assuming a plasma or electron gas in place of the wires. But that doesn't matter, the Lorentz force isn't acting on the wire, it is acting on the charges in the wire.

EDIT: In response to your edit, I am not really interested in the OP as I am sure that the previous posters were able to address that. In addition, this is a thread that seems to pop up constantly and so there are a multitude of previous threads that have already explained this. I am more concerned about what I feel are misconceptions in your posts about the physics of the forces and interactions and your comments in response to Hootenanny's and Bob S' posts.

24. Dec 2, 2009

### Born2bwire

The point of contention is stating that the magnetic fields are doing the work. There is nothing wrong with the magnetic fields being a source of a Lorentz force, however, they act on the charges in a manner that does not expend work. However, in this case the actual trajectory of the system, over which we would calculate the work done, causes an electrodynamic problem, not a magnetostatic problem. So one of the problems here is assuming that we only have a force that arises from the magnetic field. So assuming a force arising from the magnetic fields is fine, but it is a different problem when we start talking about the work done in the dynamic system.

And again, the displacement here is not along the direction of the force. The force is not acting on the wire, it is acting on the charges that make up the currents in the wire which have an initial non-zero velocity which makes the path of displacement not the same as the direction of force. The mention of special relativity allows us to look at the problem in a different manner. We can work the problem using Lorentz transformations and Coulombic attractions, in which case there are no magnetic fields.

25. Dec 2, 2009

### Hootenanny

Staff Emeritus
Born2bwire has already addressed the majority of your comments, so there may be some overlap here. However, I would like to respond to the comments you directed at me.
Yes, the electro-magnetic field. Why do you think we call it the electro-magnetic field?
You are probably used to seeing Maxwell's equations as a set of four differential vector equations. However, as I said in one of my earlier posts, we can write Maxwell's equations in covariant form as a set of two tensor equations. In this form there is no distinction between magnetic fields and electric fields. Since you seem to require a reference for every statement I make, see here: http://en.wikipedia.org/wiki/Covariant_formulation_of_classical_electromagnetism. Although, I note a distinct lack of references on your part.
That is precisely my point, we can just have electric fields. It is possible to make a coordinate transformation such that the magnetic field disappears, leaving us only with the electric field. See for example, Grant & Philips; Electromagnetism, or Griffiths as Born2bwire suggests. Indeed, most undergraduate physics texts on electromagnetism will at least mention this.
As opposed to your well written and logically consistent arguments?
Yes. See the texts I mentioned above. I was trying to be helpful by providing a website, since I assumed that you wouldn't have access to any physics texts.