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Re: induced emf again



On Sat, 4 May 2002, Ludwik Kowalski wrote:

Suppose the rod whose length is 0.1 m is moved at v=10 m/s in
a uniform field of 1T. In this case the emf=1 volt. Suppose
the rod resistance is 1 ohms while the resistance if the
"rail," on which it slides, is 3 ohms. The expected i1=333 mA
is larger than i2=250 mA which would flow if the emf was
distributed ("working" against the total resistance of 4 ohms
rather than against 3 ohms only).

Bob has properly compared this situation to that of a battery or
any other "source of emf." Perhaps to add some specifics and to
relate his answer to the terms of my previous contribution at

<http://www.csupomona.edu/~ajm/special/rodnrail.pdf>,

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 mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm