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Re: induced emf again



Viewing the effect of a distributed internal resistance (Rint):
Let me try to approach this from an energy standpoint. If the rod were
lossless (superconductor) the work done by the qVxB emf force would all go
into lifting the carrier charges up a potential hill equal to the emf :
Terminal V = EMF. This is accomplished by accumulating terminal charges
which establish an internal electrostatic field E =VxB to balance the qVxB
force and "store" the work it does. The carriers would then fall down
this hill and give this energy to the devices in the external circuit.

If some energy (I^2*Rint) is internally lost to heat as the carriers are
driven through the rod, that means that there is in the rod a frictional
force opposing the VxB force, leaving less work for the E field to do.
The electrostatic E field which builds up will then be accordingly
lessened, the accumulated charge will be lessened, and the Terminal V will
be lessened. Less energy is stored in the lifted carriers and less energy
goes into the external devices.

An equivalent viewpoint is that terminal charges accumulate to accomplish
two effects:
1) to generate an E field to balance the qVxB force (this E field opposes
the current inside the rod), and
2) to generate an E field to balance the resistive, frictional force (this
E field is in the direction of the internal current), as it does in
external resistors.

1) produces a PD=EMF, a rise in the direction of the internal current ; 2)
produces a PD=IRint, a drop in the direction of the internal current. The
resulting Terminal V = EMF - IRint, a net rise in the direction of the
internal current.

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor