# Magnetic force does no work?

1. Apr 15, 2008

### quantum123

Anyone who has studied electromagnetism knows that the magnetic force on a moving electrically charged particle such as an electron in a magnetic field is always perpendicular to its direction of motion. Since the magnetic force on it is perpendicular to its motion, no work is done on the electron by the magnetic force.
However, we all know that we can use magnetic force to raise cars to a certain height, thus doing work on it.
How do you resolve this contradiction?

2. Apr 15, 2008

### HallsofIvy

What contradiction? A car is not a "moving electrically charged particle".

3. Apr 15, 2008

### quantum123

Then why do you think a metal car is attracted to a powerful electromagnet?

A car is a piece of metal consisting of travelling electrons - electrically charged particles that experience magnetic forces.

In fact 2 wires if carrying current travelling in the same direction will experience an attractive force. Just imagine one of the wire carries a weight and is pulled upwards to the other one on top, gaining gravitational potential energy.(the electrons too gain that energy) What is the origin of this pull? Isn't it magnetic forces? Try drawing a free body force diagram on an electron.

So if magnetic forces do no work, how did the wire gain gravitational potential energy?

Furthermore, the wire is initially at rest, and then it started to move and gain kinetic energy. So if magnetic forces do no work, what is doing the work?

Last edited: Apr 15, 2008
4. Apr 15, 2008

### quantum123

I think I got the answer.

The work on the electrons are done by the electrostatic force / field in the conductor.

5. May 1, 2010

### rjllacubtan

I think it has the same analogy with the gravitational force. There is no net work produce by the gravitational force but the change of potential energy causes the gravitational force to do work. The car is attracted to the electromagnet when the current is supplied and we are confident that it is work. It is because we change the potential energy. The same as the electric motor which convert the electrical energy to mechanical by changing the magnetic field. The bottom line is that the change in magnetic field does the work. I do not have an equation but you might discover it.

6. May 1, 2010

### Redbelly98

Staff Emeritus
Welcome to Physics Forums.
That sounds like a contradictory statement. At any rate, gravitational forces do perform work if the mass changes elevation.

This is correct.

7. May 2, 2010

### Studiot

I really don't see the issue.

Work is done on any material body, charged or not, for any displacement in the direction of the force.
It makes no difference what velocity the body may have in any other direction or whether it is charged or not. Neither of these properties of themselves bring about any work.

So if a magnetic force acts on an electron and displaces it in a certain direction a distance d then the work done = force times the d.

Alternatively
Since force = mass times acceleration work = mass (of electron) x acceleration imparted x d

It is true that one way to calculate the force is the vector cross product V X B = VBsin(angle between them) and it is true that this force will be perpendicular to the plane containing both V and B and therefore perpendicular to V, but so what? It is not V that does the work.

8. May 2, 2010

### diazona

That is a true but useless statement. A magnetic force never displaces an electron by any distance d.
Actually that is a dot product,
$$W \sim m_e \vec{a}\cdot\vec{d}$$
Since acceleration (caused by magnetic force) and displacement are perpendicular, the value of the dot product is zero.
Okay, first of all, the statement "It is not V that does the work" makes no sense. Velocity never does work, only a force can do work; specifically, only a force acting parallel to (or antiparallel to) the displacement. Since the magnetic force is perpendicular to the velocity, as I said above, the value of the dot product is zero so that force does no work.

9. May 2, 2010

### Studiot

Can't see why you say this.

10. May 2, 2010

### rjllacubtan

I'm sorry! What I mean of no "net" work done by the gravitational force is that the gravitational force just converts the potential to kinetic energy or vice versa but if you remove the gravity on the system, there is a net work there. If we place an iron near an electromagnet without a current yet, there is no potential energy and no work done but if we supply current on the electromagnet, magnetic field is produced and the iron is attracted to the electromagnet and there is work there. Why? Because you convert the electrical energy to kinetic energy.

The real issue here is that the velocity is perpendicular to the magnetic field that is why the only responsible of the work done is the electrostatic force. This is the Lorentz force law. Now, if we place two permanent magnet close to each other, who is responsible for the attraction or repulsion, electrostatic force or magnetic force?

It is very confusing because we are not sure who is responsible of the work done and therefore I blame it to the potential energy.

11. May 2, 2010

### Studiot

Let us go back to the example given by quantum.

Consider a scrapyard with a crane electromagnet 6 feet above a vehicle.

The current is off and nothing happens. Both the electromagent and the vehicle are uncharged electrically.

Now switch the current on and slowly increase it. Both the eelctromagent and the vehicle remain uncharged.
However at the point where the magnetic attraction exceeds the force of gravity each and every particle of the the vehicle will be displaced upwards, including the electrons.
Work is done by the difference between the force of gravity and the force exerted by the electromagnet. No other forces are acting. Since there is no charge there are no electrostatic forces acting.

Now I see no point in trying to break this macroscopic theory down to particle level.
If you do that you have to explain why the protons and neutrons are also displaced upwards. They are not charges 'in motion' in the same way as electrons, so why should they be subject to the Lorentz force?
The truth is you need a different model at particle level.

12. May 2, 2010

### gabbagabbahey

The Lorentz force law for a point-like particle with charge $q$ subjected to electric and magnetic fields $\textbf{E}$ and $\textbf{B}$ is $\textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B})$. Classically, all currents and magnetizations (excluding the intrinsic magnetic moment (spin) of protons and electrons, which have no satisfactory classical explanation) are made up of moving point-like charged particles and hence are subject to the Lorentz force law.

Therefor, the acceleration imparted on a charged particle of mass $m$ by electromagnetic fields is given by

$$\textbf{a}=\frac{\textbf{F}}{m}=\frac{q}{m}(\textbf{E}+\textbf{v}\times\textbf{B})$$

Taking the dot product with the particles velocity, and using the identity $\textbf{v}\cdot(\textbf{v}\times\textbf{B})=\textbf{B}\cdot(\textbf{v}\times\textbf{v})=0$ proves diazona's statement.

No, work is defined as $W_b-W_a=\int_a^b\textbf{F}\cdot d\textbf{r}$. When you take away the gravitational field, there is no force to do any work, and hence none is done.

Electric forces are responsible. Keep in mind that a permanent magnet can be modeled as a collection of current loops, and each current loop contains accelerating charges (if their velocity wasn't changing, their motion wouldn't loop back on itself). Maxwell's equations tell you that accelerating charges produce time-varying magnetic and electric fields. The electric fields are what actually does the work, and in accordance with Faraday's Law, you shouldn't be surprised that they are closely related to $\textbf{B}$. When you add up all the fields and forces, you get that the net force of attraction of a permanent magnet, in an external magnetic field (possibly caused by another permanent magnet) is $\textbf{F}=\mathbf{\nabla}(\textbf{m}\cdot\textbf{B})$, and so it appears that the net magnetic field is what
is doing work. However, it is the intermediary electric fields which always do the work.

I suggest, as an exercise, you use the Lorentz force law to calculate the force on a magnetic dipole $\textbf{m}$ in an external magnetic field $\textbf{B}$, by modeling it as a circular current loop and show that you get the expected result.

Again, there are electric fields present which are responsible for doing the work. In this case, the electric field responsible is coming from the battery (or whatever power source is used) that maintains the current in the electromagnet. See for example, Griffiths' Introduction to Electrodynamics, example 5.3 (3rd edition).

Again, to make thing clear, I will take a quote from that same text:

You really see no point in trying to understand how what happens at the microscopic level results in what is observed at the macroscopic level? If so, this would explain why you haven't taken the time to properly understand classical electrodynamics.

All charged particles are subject to the Lorentz force. All of classical electrodynamics can be explained by the Lorentz force law and Maxwell's equations.

You certainly need a different model at the quantum level, but classically, the Lorentz force law works just fine; you simply do not understand it completely.

Last edited: May 2, 2010
13. May 2, 2010

### Studiot

Are you seriously suggesting there is an electric field from a battery, sufficiently powerful to reach out across 6 feet and lift a car?
If this were indeed the case I should be able to interpose an electrostatic screen, that does not interfere with the magnetic flux and determine whether the connection is magnetic or electrostatic, since such a screen would prevent an electrostatic force operating, but not a magnetic one.

So what is the classic Lorentz force on a stationary charge?

14. May 2, 2010

### gabbagabbahey

Sure, with a big enough battery. In practice, the cranes are powered by fossil fuel, which is converted to electricity. In order for electricity to flow in a normal conductor/wire, there must be some potential difference/drop and hence an electric field. It doesn't really matter whether the field comes from a battery, or a rotating magnet powered by internal combustion.

Simple; $\textbf{F}=q(\textbf{E}+\textbf{v}\times\textbf{B})=q\textbf{E}$ when $v=0$.

15. May 2, 2010

### Studiot

OK so do you maintain that the only force acting upwards on a charged particle within the material of the car is electric?

16. May 2, 2010

### gabbagabbahey

Anyways, to answer, no. But that does not mean that the magnetic force is doing work. Some of the moving charges in the car will be moving perpendicular to the magnetic field of the crane. That field will then create an upward force which results in work being done on the car. However, in order to maintain that field in the first place, an electric force must be doing work in the crane's magnet.

On the surface, it appears that a magnetic field is doing work, but on closer inspection it is an electric field that is responsible.

17. May 2, 2010

### Studiot

And exactly half of them are not moving at all (except upwards when the magnet is operational).

18. May 2, 2010

### Born2bwire

What's the point of all this?

It comes down very simply to just the Lorentz force that was explained earlier. All magnetic forces act normal to the direction of displacement of a charge. This means that magnetic fields can do no work. That's it. Classical electrodynamics does not have another source of force. Anytime that we have a situation that appears to have magnetic fields doing work needs to be looked at more carefully. Usually this is a very complicated process and isn't worth doing. In the end, it often comes down to the fact that a magnetic field in one frame can have an electric field in a moving frame via the Lorentz transformations. So even if we have a pure magnetic field acting on a ferromagnetic material, as soon as the ferrite starts to move it experiences an electric field. So the initial force that acts on the ferrite may be purely magnetic, it isn't in a direction that can expend work. Once the ferrite is displaced, it gets worked upon by the transformed electric field.

Again, no matter how you pose the situation, you still have to find a way to take the Lorentz force and get work out of the magnetic fields.

19. May 2, 2010

### rjllacubtan

so magnetic force is not a force because it cannot produce work. Is that what you mean? it's only a representation? Is it the electrostatic is always responsible for the work?

20. May 3, 2010

### Born2bwire

It is a force, it's just not a force that does work. This is perfectly fine because a force is a change in momentum. In the case of the magnetic force it is like a centriputal force. If I have a planet orbiting in a perfect, stable circle around a star, then no work is being done. This is because the force of gravity is always pulling the planet towards the star. In this case, this force will point along the radius of the orbit which is always perpendicular to the circle of the orbit. Thus, the dot product between the force and path of displacement is always zero. This is expected though, if we have a friction free system and balance the speed of the orbit and the force of gravity, the system should be stable. If it is stable and without any means of expending energy (no friction) then we should not be doing any work on the planet, else its orbit would change. We can have the same thing happen in a cyclotron. If we have have permeate a volume with a static magnetic field and have a charge move into the field in the plane normal to the field, then the magnetic force will cause the charge to move in a circular orbit. The size of the orbit is dependent upon the strength of the applied magnetic field, the speed of the charge, and the charge's mass and charge.

The eletrostatic force can do work, but it is not the only force. By electrostatic I assume you mean the simple Coulomb's force. The Coulomb force is derived using the Coulombic field of a point source with the Lorentz force, qE. But we can have a time-varying electric field and still have a force, qE, that does work. So the force due to the electric field is the one that is responsible for directly transferring energy from the fields to the charge. It is the direct mediator. The magnetic field can still store energy as well as the electric field. It is just that for the energy in the magnetic field to be transferred to a charge, an electric field must arise to do so. This electric field may not even exist in the observer's frame but it must exist in the rest frame of the charge.

For example, if we have two infinite parallel wires of current, a common example of the Lorentz transformations is to show that while in the rest frame we only see the magnetic fields from the currents, in the frame of one of the moving electrons we see an electric and magnetic field. In the electron's frame, the electric field points in a direction that allows it to do work. So while we may not see any electric fields in the lab from from the view point of the stationary wires, the charges in the wires see an electric field.

This is all a very overly technical discussion of the problem though. If we wanted to know the amount of work done when the wires move or repel from each other, the easiest thing to do is to calculate the total magnetic fields before and after they move and find the change in energy. We would not want to actually work out the work done on the charges by changing frames, finding the trajectory of the charges and then finding the work done. Since there is no energy loss when we do these transformations, we can still work with the magnetic fields and find the correct answer.

Last edited: May 3, 2010
21. May 3, 2010

### cabraham

The above post discussed this issue in detail, and we reached a consensus. At the particle level, an H field can change an electrons momentum, but not its kinetic energy. The force due to H is normal to velocity, so that H changes the direction of the e-, but not the KE. An E field can change both.

E & H are so inter-related that it is difficult to say which is respnsible for lifting the car since they both act in unison. Read the post I linked above, and then we can discuss it further. BR.

Claude

22. May 3, 2010

### Studiot

At least you ask for clarification of my point. Thank you for that courtesy.

My point is that I don't think quantum was looking for a highly mathematical answer, which has its own difficulties to explain.

If we consider

F=q(E+vXB) as sugested we must answer the following.

If we propose an electostatic field or force to reach and interact with the electrons of the vehicle's material, we must allow it to reach the other particles (protons and neutrons).

Since these particles make exactly the same movement as the electrons how is this accomplished?
Both particles are some 1700 times more massive than the electron.
Both particles are at rest in the system under consideration.
The proton has a positive charge.
The neutron has no charge.

Incidentally I was wrong in saying the exactly half the particles are at rest, since I forgot to count the neutrons.

The above formula is also pretty difficult to employ since we must have a figure for velocity and electric field strength in order to make calculations.

Can anyone derive the following simple, useful and true formula from the above?

$$F = \frac{{{B^2}A}}{{2{\mu _0}}}$$

I have modelled the crane and car as two slugs of ferrous metal of identical cross section area, A and separated by a distance, d. F is the force of attraction and B is the flux density produced by current I in the coil wound round the upper slug to model the lifting crane.

Further for practical calculations the approach based on magnetic poles and the magnetic inverse square law has much to commend it.

$$F \propto \frac{{{m_1}{m_2}}}{{\mu {d^2}}}$$

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23. May 3, 2010

### Born2bwire

Again, you are getting far too wrapped up in this specific problem. Like I stated before, showing how exactly the work is propagated by the electric fields can be a very tricky problem because it usually requires us to use different frames of reference than the lab frame (usually accelerating frames too).

The fact to focus on is that classical electrodynamics uses the Lorentz force to give rise to all forces. You need to figure out how to pull out an energy dissipative force from the magnetic contribution.

To do this, keep in mind what the macroscopic properties of the material mean on the microscopic level. For example, a magnetic moment is actually a loop current, so the actual force is operating on an accelerating charge. In the rest frame, which includes the protons of the atoms, there is only a magnetic field so they feel no force. However, in the frame of the electrons that are orbiting around the atoms and moving about the conuction band, they see an electric field from the transformation of the magnetic field. This electric field acts on the electrons and displaces them. The displaced electrons then exert a Coulombic attraction on the unmoved protrons and pulls them along with them. The protons will pull the neutrons with them due to the various forces that hold the nucleus together.

Take a look at Griffiths, particularly what he has to say in section 5.1.2-5.1.3 and example 5.3. Example 5.3 talks about an equivalent situation of the electromagnet and ferrite.

24. May 4, 2010

### Studiot

Why do responders keep claiming either
'the only force acting is electrostatic'
'no other forces are acting except electrostatic ones'

BUT

Introduce other forces, eg the nuclear force, to complete their argument.

I requested a derivation of a practical formula, such as the one I presented, from this much vaunted Lorenz vector formula, if I actually needed to calculate the force between the car and the crane I could do.

I asked for an opinion of what would happen I interposed an electrostatic screen to prevent the transmission of any electrostatic force between the crane and the car.

None of these questions have been answered.

25. May 5, 2010

### Born2bwire

As I said earlier, I am not claiming that it is done by electrostatic forces alone. This is obviously a time-varying system and thus the electric fields involved are not static. If you look at say Jefimenko's equations you can obviously see that there is always an electrostatic contribution to the electric fields but in addition time-varying sources also contribute to any observed electric field.

I'm am not willing to take the time to try an answer the other questions. As I stated previously, it is pointless to delve into an overly complicated problem. In the end you still need to show how you can apply a magnetic field that gives rise to a force that does work. The proof is in the basic physics not in some convoluted thought experiment. It simply comes down to the fact that the Lorentz force does not allow a magnetic force to do work, PERIOD. Unless you can find a legitimate way of breaking this or come up with another force relationship between charges and fields in classical electrodynamics that is not described by the Lorentz force then this is just a losing battle. Again, take a look at Griffith's sections section 5.1.2-5.1.3 and example 5.3. Purcell's chapter 5 may be interesting too. While he does not discuss magnetic fields and work, he approaches magnetic fields differently from Griffiths. Griffiths just introduces both the electric and magnetic fields and only in the later chapters does he show the Lorentzian relationship between the two. Purcell introduces the electric field and then, in chapter 5, dicusses how the magnetic force comes about simply from electric fields and special relativity. In fact, I don't think he mentions the magnetic field in the chapter except for the Lorentz force but he clearly solves problems that do have magnetic fields purely from the electric field standpoint. So it is a good demonstration of how the magnetic force is an electric field force transformed under special relativity.