Does a Magnetic Field Do Work on Electrons in a Wire?

[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.

 

[nicksauce]

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

 

[pam]

A time varying magnetic field will lead to an electric field that does work.
An example is Faraday's law.

 

[Phrak]

An electric generator does work because the field changes?

 

[pam]

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

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

 

[pam]

An electric generator does work because the field changes?

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.

 

[Phrak]

Pam, thank you for answering.

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.

 

[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?