# Work done by magnetic field

1. Aug 14, 2014

### reasonhunt

when a magnetic field exert force on a moving charge the work done by it is 0.But bar magnet do work.And i read that magnetic field is a non conservative field so work done by it in a closed path is not 0.But in case of a moving charge it is 0 and magnetic field also store energy so it have the ability to do work.but i am confused in one case magnetic field is unable to do work and in other it is able.
Also magnetic scalar potential is not defined.
So i have doubt whether magnetic field is conservative or non-conservative
and also is it necessary to have magnetic moment for magnetic field to do work
And i also want to know the physical significance of magnetic moment

2. Aug 14, 2014

### vanhees71

As always the answer is in Maxwell's equations and in the Lorentz force law. First of all you have
$$\vec{\nabla} \times \vec{B}= \frac{1}{c} \partial_t \vec{E}+\frac{1}{c} \vec{j}.$$
This means if the right-hand side is not 0, the magnetic field cannot be a potential field, because it's curl doesn't vanish.

Further the force on a particle is
$$\vec{F}_{\text{mag}}=\frac{q}{c} \vec{v} \times \vec{B}.$$
The work done (no matter whether the force is conservative or not) is given by
$$\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}=0.$$
So the magnetic field does not do work.

Some time ago we had a monster discussion on this topic with a lot of confusion created by not sticking to the fundamental laws. I hope this time this won't happen again!

3. Aug 14, 2014

### cabraham

The magnetic field itself does no work as it is normal to the displacement. But the magnetic component of Lorentz force does work, i.e. Fm = uXB, where u is the velocity vector.

I remember that long discussion we had. Those who claimed Fm does no work could not produce one picture or proof of any kind. They just assert themselves and expect whatever they say to stick. A bar magnet lifting a slab of iron is an example of work done by Fm.

Claude

4. Aug 14, 2014

### Staff: Mentor

Note that Fm is perpendicular to u. No work done.

In Introduction to Electrodynamics (4th Edition), Griffiths devotes a chapter to this very misconception. Check it out.

5. Aug 14, 2014

### vanhees71

@cabraham:
NO! I gave the proof in my answer. I gave it already within this earlier thread. If you still don't believe the mathematics and Maxwell's Laws, I can't help you. We even discussed more complicated issues concerning how a electric motor works etc. The master equations for energy-momentum conservation in electrodynamics can really be found in any sufficiently complete textbook on electrodynamics.

6. Aug 14, 2014

### Delta²

Simple question is who/what does the work in the case of a bar magnet (or even an electromagnet) attracting a piece of iron? Cause that piece gains kinetic energy...

7. Aug 14, 2014

### cabraham

But u is the electron velocity not the velocity of the moving mass. Fm does no work on the electron but does work on the loop in which the electron is conducting. A rotor in an induction motor is a prime example.

The loop is in the x-y plane, current is counter-clockwise. B field is in y direction tangent to electron. Electron velocity u is in x direction. Fm = uXB must lie in z direction. Fm is normal to electron velocity u, and cannot do work on electron. Likewise, B is normal to u. So far no work can be done. But Fm acts in the z direction on the loop, producing torque and spinning said loop. The angular displacement θ, times the torque is the work done by Fm.

If Fm does no work on the loop, then which force is doing the work spinning said loop?

The above thread includes a hand drawn picture illustrating the forces in action. I will elaborate if needed. Best regards.

Claude

8. Aug 14, 2014

### reasonhunt

@vanhees71
According to you magnetic field does no work then how it is able to store energy because to consume energy it is necessary to do work so how the energy stored in field can be used

i'm unable to clear my doubt regarding this matter.

9. Aug 14, 2014

### reasonhunt

And the case given by you is in the case of a moving charge but when a closed loop carrying current is placed in the magnetic field it experiences torque and we are also able to calculate the work done by the magnetic field in rotating the closed loop and that's why we associate potential energy with the system.

10. Aug 14, 2014

### milesyoung

From Matter and Interactions, 3rd Edition, Chabay and Sherwood, ISBN-13: 978-0470503478;
Excerpt from page 1448-1449 (PDF) in "Chapter 21: Magnetic Force":

It should be absolutely perfectly clear from the Lorentz force law that the magnetic force does not do any work. An electric force ultimately does the work, as it must.

11. Aug 14, 2014

### Jano L.

Yes, the magnetic force $\mathbf F_m$ does have torque and does impart angular momentum to the loop in general.

It is mechanical work on the loop alright, but it is not done by $\mathbf F_m$. Remember, you defined $\mathbf F_m$ above by saying that it acts on the electron in the loop. Proper way to calculate work of this force is to multiply it by displacement of the electron. Using displacement of the element of the rest of the loop (the lattice + nonconducting electrons) makes no sense for this force, because this element body has a different velocity than the electron inside it.

If a force acts on the electron, it cannot act on an element of the rest of the loop (ERL). One force cannot act on two distinguished bodies - if two distinct bodies experience force, even of the same magnitude and direction, there are two distinct forces. Total force acting on the ERL without the conducting electrons deserves its own symbol, say $\Delta \mathbf F_w$.

This force $\Delta \mathbf F_w$ and displacement of ERL $\mathbf u$ is what should be used to calculate net mechanical work done on the rest of the loop.

From experiments, we know the force $\Delta \mathbf F_w$ can be approximately expressed as

$$\Delta \mathbf F_w \approx \Delta V\,\mathbf j \times \mathbf B_{ext},$$
where $\mathbf B_{ext}$ is external magnetic field and $\mathbf j$ is current density of the conduction electrons in the wire of the loop.

This force does have magnetic field in its approximate expression, but it is not a magnetic force in the sense of the formula

$$\int \mathbf j_r \times \mathbf B_{ext}\,dV,$$
because the current density due to the rest of the loop $\mathbf j_r$ is very different from the density of conduction electrons $\mathbf j$ used in expression for $\Delta F_w$.

Therefore $\Delta \mathbf F_w$ have to be sums of electric forces or forces of non-electromagnetic nature. They are due to conduction electrons pushing or pulling on the rest of the loop due to their modified motion in the external magnetic field. These electrons exert electric and possibly non-electromagnetic forces on the rest of the loop and these can do work. Thus $\Delta \mathbf F_w$ are sums of internal forces. Internal forces do all the mechanical work in this case. Energy for this is transferred from the energy of magnetic field and the source of voltage.

Mechanical work done on the ERL is

$$\Delta \mathbf F_w \cdot \mathbf u$$
where $\mathbf u$ is velocity of the element, not of the electrons inside it, which is parallel to $\mathbf j$ and would give zero work. That's why the actual work can be non-zero; the force $\Delta \mathbf F_w$ acting on the element is not perpendicular to its velocity $\mathbf u$.

12. Aug 14, 2014

### milesyoung

Look, I'm not going to spend 20+ pages chasing your red herrings.

After all this, if you're still not convinced, then I doubt me or anyone else is going to change your mind.

13. Aug 14, 2014

### pleco

http://en.wikipedia.org/wiki/Ampere's_force_law

Just like depicted force vector acting on wire 1 points towards wire 2, so does force vector acting on wire 2 points to wire 1, and thus two wires attract. As always the movement is in the direction of the force, in this case it's Lorentz force. It is important to distinguish field lines and force lines. While electric field and force lines coincide, magnetic field lines and force lines are perpendicular.

Magnetic force only acts between magnetic fields. It acts on protons, electrons and even neutrons, because all of them have their spin magnetic moments, whether they move or not.

It's mostly free electrons in wires that we say interact with external magnetic fields because protons, neutrons and other electrons usually prefer to stick to their molecular arrangement since those forces are greater for them.

All these electric and magnetic fields superimpose and on average neutralize, in an average material. But when free electrons are drifting additional magnetic fields are produced and then there is an excess of magnetic field with the same orientation relative to only one particular direction of the drift.

They calculated magnetic force, renamed it to electric force and asserted they are equal, out of the blue. Magnetic and electric force can hardly be equal, but if they were equal then we could equally say for either is the one which does the work.

14. Aug 14, 2014

### Pythagorean

An analogy I came up with during my undergrad is a ramp. A ramp does no work, but it translates horizontal work to vertical motion, and vertical (as done by gravity) to horizontal motion.

Obviously the coupling between magnetic and electrical fields is more complicated than that of gravity and ramps, but it illustrates how you can get forces on an object from the ramp without the ramp doing work.

15. Aug 14, 2014

### Staff: Mentor

That is also my hope.

16. Aug 14, 2014

### pleco

This looks to me like a weird semantic argument like when people say the same word but mean different things, only this time they say different words but still think they are talking about the same thing.

Can magnetic fields do work? Of course they can not, no field can. Forces do work, not fields. They are different things, with different units. They don't even compare. Work is not field, but force times displacement.

17. Aug 14, 2014

### Staff: Mentor

The key theorem regarding energy conservation in EM is called Poynting's theorem. It says:

$$-\frac{du}{dt}=\nabla \cdot S + J \cdot E$$

Where u is the energy density of the field, $u=\frac{1}{2}(E\cdot D + B \cdot H)$ and S is the flux of the EM energy $S=E\times H$.

So Poynting's theorem says that the EM energy in the field at a point can only decrease (-du/dt) if the EM energy flows out to other parts of the field (∇.S) or does work on matter (J.E). The work done on matter is always J.E, but the magnetic field can decrease as part of u decreasing and the magnetic field is always involved in moving EM energy from one place to another as part of S. So the work done is always through the E field, but the energy required can come from the magnetic field.

18. Aug 14, 2014

### pleco

I found these two threads and I see what the problem is, in the case of two parallel wires.

There are two forces, magnetic force acting on drifting electrons and electric force between electrons and protons.

Magnetic force displaces drifting electrons in the direction of the Lorentz force, therefore magnetic force does that work on electrons.

Sure electrons are responsible for dragging protons along, but whether electric force does any work as well depends on whether there is any displacement between electrons and protons.

There is probably some relative displacement between electrons and protons as magnetic force is trying to yank or blow away electrons from protons. That is _away from protons. So if there is any displacement at all it will be in the opposite direction of the electric force and in the direction of magnetic force, thus magnetic force would be the one doing the work again.

Protons basically just add to resistance, that is mass, but we don't need protons at all. We can bend electron beams with magnetic fields, so then without a doubt we can see magnetic force can indeed do work.

19. Aug 14, 2014

### Staff: Mentor

See Poyntings theorem. Work done by EM is always J.E. Bending an electron beam is not a counter example since no work is done.

20. Aug 14, 2014

### pleco

There is a displacement of electrons in the direction of the magnetic force whether it is an electron beam in a magnetic field or a current carrying wire, so why do you say no work is done?