How Does Faraday's Law Explain Zero Work by Magnetic Forces in Motional EMFs?

[Aubin]

Hello,

I am currently trying to really come to grips with electromagnetic induction and Faraday's law.

The text I'm using is Young and Freedman's University Physics, 12th edition.

How can the emf for a conducting loop moving in a magnetic field be given by the closed loop integral of (v cross B) dot dl where dl is an infinitesimal loop element, when (the way I understand it) that essentially represents work per unit charge done by the MAGNETIC force? And of course we know that the magnetic force does zero work. This is where I'm confused. Also, when we do work to keep such a loop moving, we move it against a magnetic force, how come (as my book says) the magnetic force does no (negative) work as we move this wire?

Thanks

Aubin

 

[Philip Wood]

What an interesting question!

As you say, the magnetic force never does any work on the particle, because:

dW = q(vxB).ds = q(vxB).vdt = 0

Here, ds is an element of the particle’s path.

Now let's consider the particles constrained to move along a conductor. Let the velocity through space in our frame of reference of an element dl of the conductor be v_cond, and let the velocity, dl/dt of the particles relative to the conductor be v_rel.

Then our formula becomes

dW = q{(v_cond+v_rel)xB}.(v_cond+v_rel)dt = 0

The terms involving just v_cond in the brackets either side of the dot clearly amount to zero, as do the terms with just v_rel. So we are left with the heterogeneous terms

q{v_condxB}.v_reldt + q{v_relxB}.v_conddt = 0

that is

q{v_condxB}.dl + q{v_relxB}.v_conddt = 0

Note that here we're using dl as originally defined - as a directed element of the conductor, rather than as a displacement of the particle through space (for which we're using ds).

Now, in the second term, q{v_relxB} is the ‘BIL’ motor effect force contributed by q on the element of the conductor, so - {v_relxB} is the external force needed to keep the element moving at constant speed, so - q{v_relxB}.v_conddt is the work an external agency has to do to keep the wire moving at constant speed. So:

Work done by external agency = - q{v_relxB}.v_conddt
=q {v_condxB}.dl
= work done pushing q along the conductor

Hope you like this. It shows that the work done pushing q through the conductor does not come from the magnetic force, but from another, external force.

 

[Aubin]

Yes, that makes sense, thank you very much. (Took a while to digest because my vector algebra was a bit rusty from last term). This also shows very clearly why the work done must equal the energy dissipated in the circuit doesn't it?

 

[vanhees71]

A very nice pedagogical paper on the subject is

P. J. Scanlon, R. N. Henriksen, J. R. Allen, Approaches to Electromagnetic Induction, Am. J. Phys. 37, 698 (1969)