# Magnetic Field and Work

1. Apr 3, 2008

### Phrak

How does a magnetic field do work on the electrons in a wire?

If F = q v x B, the magnetic field is always perpendicular to the velocity of the free charge. After a small time interval, dx = v dt, so that v and x are in the same direction.

For the magnetic field to do work on the charge, dW = F dot dx. But F and dx are perpendicular, aren't they?, so no work seems to be done.

2. Apr 3, 2008

### nicksauce

That's right. Magnetic forces never do any work.

3. Apr 3, 2008

### pam

A time varying magnetic field will lead to an electric field that does work.

4. Apr 3, 2008

### Phrak

An electric generator does work because the field changes?

5. Apr 4, 2008

### pam

A magnetic field acting on a permanent or electric magnetic can do work.

6. Apr 4, 2008

### pam

That is how an induction motor works.
Other motors use a permanent or electric magnet as an armature,
and a magnetic field does do work on the armature.

7. Apr 5, 2008

### Phrak

You threw me so far. I am now sure you are applying Faraday's law in integral form,
$$\int _{_\partial S} \overline{E} ds = -\int _{S} \partial_{t}\overline{B}dxdy$$ ,
it must surely be.

This makes some very good sense.

I could not possibly see how to apply Faraday's law in differential form,

$$\nabla\times \overline{E} = -\partial_{t}\overline{B}$$.

Unfortunately, I still cannot.

In the vicinity of the wire B is unchanging, we should presume. In such a manner $$\partial_{t}\overline{B}$$ should be zero as well, in this locally manner. But should this not make $$\nabla \times \overline{E}$$ zero as well?

Something I am missing.

8. Apr 5, 2008

### Phrak

Can this be understood in terms of a field that doesn't act at a distance?

Does a locally acting 4-vector potential, A and Dirac's equation of A acting on the phase of the free electrons in a conductor make some sense of it?