Re: induced emf again

[Ludwik Kowalski <kowalskiL@MAIL.MONTCLAIR.EDU>]

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.

The dilemma can be resolved experimentally. If the
internal resistance of the rod does have an effect on
the induced current then the current in the rail
produced by a rod made from lead, for example,
should be smaller than the current produced by the
rod made from copper. I am assuming that B, v and
sizes of two rods are identical. I believe that many
experiments of that kind have been performed by
electrical engineers.

2) The first experiment described by Feynman (on
page 17.3  of volume 2) is nice and not hard to do.
But it neither confirms nor contradicts our model
according to which the direction of E lines and the
direction of conventional current are opposite in
the part of the disk next to the N pole.

The experiment I suggested this morning was not
good because it was based on a presumably wrong
assumption that the DOP between the rod terminals
does not depend on the current. Here is another
suggestion. Use a voltmeter to find out about the
direction of E lines inside the rod, use a
galvanometer to determine the direction of the current.
The wires leading to the voltmeter should be parallel
to the magnetic field lines.
Ludwik Kowalski

Bob Sciamanda wrote:
> The rod resistance acts the same as the armature resistance
> in a generator (in fact, that is what it is). In both cases it
> must be included as the internal resistance of an emf
> device - as you would do with a battery.

John Denker wrote:
> ... You could build the apparatus in Feynman's figure
> 17-2 and figure 17-3.  Use a pull-string around the
> shaft in figure 17-2 to spin the disk. ...

Ludwik Kowalski wrote:
... Consider a metal rod of length L sliding with the speed
v along the rigid U-shaped wire frame perpendicular to
the uniform magnetic field B. Do the electric field lines
INSIDE THE ROD have the same direction as the
conventional current? My answer, especially after the
discussion about the induced emf we had last winter,
would be "yes." But now I think it would be a wrong
answer. I now think that the electric field inside the loop
is conservative, all electric lines begin and terminate on
static charges. In other words, the sliding rod behaves
as if it were a battery whose emf is B*v*L.

The origin of my misconception is probably rooted in
the rule according to which "the way in which the flux
changes is not at all important, it can be a stationary B
but changing area, or it can be a constant area but
changing B." But Faynman argues that two ways of
changing the flux result in two different phenomena.
Right or wrong? I wish I could refer to an experimental
verification of this theoretical claim.