# Why do the magnetic force does no work?

1. May 3, 2010

### Taturana

Why? Is there an explanation to that?

I know that it's because the magnetic force is perpendicular and so on, but then why the magnetic force works that way?

Thank you

2. May 3, 2010

### LostConjugate

Well a magnetic force can move a metal object over a distance which is the definition of work in physics. Energy is conserved though as it takes just as much energy to move that piece of metal a distance from the force. In other words it is a conservative force like gravity.

3. May 3, 2010

### Taturana

But a conservative force is a force such that the work done by this force is independent of it's path (or something like that).

And I read in a book that the magnetic force does NO work, the work done by that is always 0.

Is that book wrong then?

4. May 3, 2010

### LostConjugate

Exactly, if your starting position is with the metal attached to the magnet, no work is done in the total system of say human and magnet.

If you look at the magnet alone though it is doing work to make up for the work the human or machine moving the metal around the path is doing. They are both doing work, but energy is conserved and work is conserved in the entire system, though the human will have burnt a lot of calories after a day's hard "work" and the magnet none...

5. May 3, 2010

### n.r.g.

I think the o.p. Is referring to the right hand rule. Remember work is accomplished by the total effect of the deviation of the electron field. When you use magnetism as force for work you "aim " the field to effect the most electrons from their orbit thus 90 degrees works best. someone correct me if I missed please. This will help explain it all.http://hyperphysics.phy-astr.gsu.edu/Hbase/hframe.html

6. May 3, 2010

### Born2bwire

The magnetic force is not doing the work, this shown trivially by examination of the Lorentz force. The work is actually being done by electric fields. What happens in the case of magnets is that when the objects start to move, the magnetic field undergoes a Lorentz transformation. This causes an electric field to appear in the frame of the object under attraction/repulsion. It is the electric fields that appear from the transformed magnetic fields that do the work. The fact that it is a conservative field has no bearing on whether or not work is done. It only means that the net work is zero in any given closed path. You can still do work over a non-closed path. The electric field is a conservative field too but we consider it to be able to do work while the magnetic field cannot do work.

I just mentioned this in the other thread, but Griffiths' text is a good reference on this, particularly sections 5.1.2-5.1.3 and example 5.3.

So just to reiterate, the magnetic field does not do any work. We can still transfer energy out of the field though. However, this is done by electric fields. So you can think of electric fields as being the mediator for transferring magnetic energy to kinetic energy in charges. The reason why the magnetic fields do not do any work is that the force, given by the Lorentz force, is defined to be:
$$\mathbf{F} = q\mathbf{v}\times\mathbf{B}$$
So the force is always normal to the velocity and thus is always normal to the path of displacement. Thus, when you try to calculate the work:
$$W = \int_a^b \mathbf{F}\cdot d\mathbf{\ell}$$
the dot product between the force and the path of displacement is always zero.

7. May 4, 2010

### LostConjugate

Interesting.

So... what causes the metal to "start to move" from a standing position? There is no Lorentz transformation if both objects are at rest.

8. May 4, 2010

### Born2bwire

If we take a look at a cyclotron, we can easily see that the magnetic field applies a centripetal force that makes a charge move in a circular orbit. Even though the charge is deflected, no work is expended. So the magnetic field can still cause a displacement.

Doesn't actually answer your question but I do not want to imply that magnetic forces do not move charges directly at all. However, there is an inherently more subtle problem here. The magnetic field only exerts a force on moving charges. So in any situation were there is a magnetic force present, then the objects that experience the force are in motion and thus require a Lorentz transformation. So while the metal's bulk is stationary, the microscopic magnetic dipoles in the ferrite that the magnetic fields exert a force on are loop currents, moving charges.

9. May 4, 2010

### LostConjugate

That is fascinating. So the free electrons in motion within the material are deviated from their course perpendicular to the field. What causes them to start moving in the direction of the field? Or did I miss something?

10. May 5, 2010

### Born2bwire

The electrons are already in motion technically. There is the random thermal motion of electrons, since we are talking about a metal here, in the conduction band. But this isn't what what really contributes to the magnetic properties, the conduction electrons are responsible for conduction currents (which are in response to both magnetic and electric fields, like in a wave where the electric field gives the primary movement to the charges and then the magnetic field gives a force on the moving charges). The actual physics get a bit muddy here because this is still a classical description. Classically, we view a material as being filled with microscopic loop currents. What these loop currents are do not have a true physical representation. A naive way to think of them would be that they are the electrons orbiting an atom, but this of course falls apart when we think about the true nature of the atom in terms of quantum mechanics. However, in quantum mechanics, the electrons do retain a magnetic moment so it is not that bad of an assumption. Modeling it as a loop current is not correct but the fact that they can have a magnetic moment is.

So the point is that these loop current have magnetic moments. Normally the moments are randomly oriented and thus cancel each other out. The permeability of a material is a measure of how much these moments will align in response to an applied magnetic field. That is, if we place a magnetic field throughout the material, the microscopic moments will find it energetically advantageous to align with the field, thus they experience a torque. Upon aligning, all these microscopic moments will be in the same direction and thus generate a macroscopically significant magnetic field. This is the magnetization field and is included in the B field of a problem. A ferrous material is special because it has large regions of the material where the moments are already commonly aligned. However, there are many many regions in the material and they too are randomly aligned. So these small regions again do not contribute to a net magnetic field. However, the application of an external magnetic field will align these regions and you get a strong magnetization. In addition, when we turn off the applied field, these domains will remain aligned, giving the ferrite a permanent (or lasting) magnetization. Thus, we now have a magnet.

Simply put then, the magnets are operating on the magnetic moments of the individual atoms which are created by the orbiting electrons. Part of this is by the electron spin in what is known as paramagnetism. Another magnetic moment is produced by the actual movement of the electrons around the atom, this is diamagnetism I believe. Ferromagnetism is like paramagnetism, but as I mentioned above, the moments do not need an applied field to line up, they will do so naturally.

11. May 5, 2010

### LostConjugate

Thanks for putting all that together for me Born!

What is to prevent a non-ferrous material from eventually lining up with the magnetic field. After all, Aluminum has plenty of unpaired electrons that each have a magnetic moment, and it is very light, yet you can't pick it up with a magnet.