Re: PHYS-L Digest - 25 Jun 2004 to 26 Jun 2004 (#2004-184)

[Leigh Palmer <palmer@SFU.CA>]

On 26-Jun-04, at 2:00 AM, Kenny Stephens wrote:

> This is my first posting to the list so please be tolerant.
>
> I'm currently reviewing some chapters for a "well-known" cal-based
> intro physics book. It uses a popular explanation (I've seen it
> elsewhere) for deriving the emf developed by a conductor moving with
> uniform velocity through a constant magnetic field. Just to be
> thorough, here's the setup:
>
> Two conducting rods are placed parallel to each other. Let's call them
> rails and say the left end of both rails are connected by a
> resistance, R. Another conducting rod of length L is placed across
> (perpendicular) to the rods and can slide freely along the rails. An
> external agent acts on the rod to give it a uniform velocity, v,
> parallel to the rails (and away from the resistance). A uniform
> magnetic field is applied perpendicular to the plane of the problem
> (let's say into the page). Assume a current flows through the circuit
> (through R, along one rail, up the rod and returns along the other
> rail) such that the charge carriers have a drift velocity v_d.
>
> The text says that each charge carrier, q, in the rod has the velocity
> v and since q moves in a magnetic field it experiences a Lorentz force
> F_M= qv cross B. The text then states that the work done by this force
> pushing the charges along the rod is F_M * L= qvBL. Since emf is
> energy per charge, the motional emf between the ends of the rod is E=
> vBL.
>
> Now this bugs the heck out me because magnetic forces are not supposed
> to do work. Using this explanation just sets the students up for
> confusion and puts me in a pickle to try to justify it.
>
> I prefer the explanation of calculating the changing flux, Phi_M= BLx,
> through the circuit where x is the position of the rod measured along
> the rails from the resistance. This gives the emf E= -dPhi_M/dt=
> BL(dx/dt)= BLv.
>
> After all this yacking, my reason for posting is to get a range of
> opinions of this text's derivation.

This is my first posting in quite a while. Please excuse its rough
form, but you need not be tolerant. I have been enjoying my retirement
too much, and a visit from Larry Woolf and his wife Wendy has convinced
to me that I ought to get back in the phys-l game. I have writer's
block on an essay I was writing, and working on a different problem
here may be helpful in breaking that block.

I certainly share your preference for an explanation of emf using
Faraday's law, the integral form of one of Maxwell's equations.
Maxwell's equations are to be shown, ultimately, to be all that is
necessary to treat problems in electrodynamics. It is important to show
the students that Faraday's law can be understood in terms of concepts
they have already learned to trust from mechanics, like force and work.
Demonstrating the consistency of these approaches is, I presume, the
reason your text's author has chosen to explain emf in this manner.

In my opinion the text's is an unnecessarily complicated explanation,
and a good example of why one should never tell students that magnetic
forces do no work. That statement is insufficiently qualified, and it
might leave a student under the impression that DC electric motors are
physically impossible. (It is easy to construct a comutatorless DC
motor, as Faraday did.)

It is unnecessary to assume a current is flowing in the circuit since
the emf is present whether a current flows or not. Indeed it is
unnecessary to introduce the circuit, since the emf exists as a
potential difference between the ends of the rod (in the rod's comoving
frame of reference) regardless of the presence of the rails and
resistor. These gratuitous complications can only confuse a student;
they will not help him learn. A better explanation would be to note
that the force acting on any charge q in the moving rod is qvB along
the rod, and that the work done on the charge in moving it through a
distance L is just qvBL. (I will leave the business of signs and sines
to the reader.)

At this point the circuit can be attached to the moving rod. Again, it
is unnecessary to introduce "charge carriers" and "drift velocity" -
they can only complicate the explanation unnecessarily. I should also
point out that "L" is now the distance between the rails, not the
length of the rod.

The modern interpretation of a rod moving with its length and velocity
mutually perpendicular with the applied magnetic field is to consider
the problem in the frame of reference in which the rod is fixed. In
that case the force on charges in the rod is interpreted as an
electrostatic force ascribable to an electric field directed along the
rod. It is the electric field that does the work in this frame, if that
makes one more comfortable. Such information should tempt the student
further into the study of relativity.

I hope there is some food for thought in the above.

Leigh