John Mallincrodt's last post answered your question correctly, using your
numerical example:
I quote John:
"I'd expect to see a 7.5 V/m polarization field arise within the
rod in opposition to the driving vxB force. Over the 0.10 m
length of the rod, this would yield a 0.75 V "terminal voltage."
The "terminal voltage" may also be viewed as the result of a 1.0 V
vxB induced emf that has been reduced by a 0.25 V drop in the
"internal resistance" of the rod itself. due to the 0.25 A that
flow in the circuit. The 0.25 A that is responsible for this
internal loss is, of course, self-consistently related to the fact
that the 0.75 V terminal voltage drives current through a 3.0 ohm
external resistance." - John Mallinckrodt
As with other voltage sources, Terminal Voltage = EMF - I*Rint. As the
external load and current vary, the emf is constant at qVxB, the terminal
voltage and the E field in the rod are current dependent, exactly as in
John's example. To see this conceptually, your second model (a series of
dls) is perhaps more useful. Of course the Thevenin equivalent
conceptually separates the emf and Rint in order to make things
calculationally simple (and correct).
To answer directly: yes, qVxB = E holds only for zero current (or even
with current if the rod has zero resistance).
Note that in the external circuit, charges accumulate at the terminals of
a resistor so as to establish the necessary driving E field, resulting in
an electrostatic potential drop of IR in the direction of the current.
The internal (rod's) resistance experiences this effect in addition to the
effect of the emf, so that charges develop on the terminals which produce
the electrostatic potential difference equal to the Terminal Voltage =
EMF - I*Rint. (Here EMF = qVxB)
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Ludwik Kowalski" <kowalskiL@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Sunday, May 05, 2002 12:15 AM
Subject: Re: induced emf again
| 1) You are probably correct, Bob, about the "armature
| resistance." But let me say why I am questioning it.
| The derivation of the formula for motional emf
| treats the entire rod as one element. Are you saying
| that two forces, q*v*B and q*E, are equal only
| when the current is zero? If the equality of these two
| forces does not depend on the current then the DOP
| between the ends of the rod should remain constant.
|
| On the other hand, one can treat the rod as a set of
| many generators in series, each of the length dL. Each
| element contributes emf=v*B*dL and each elements
| sees other elements as a load. In that case the DOP
| between the ends of the rod should decrease when
| the current becomes larger. . . .