Re: induced emf again

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

"John S. Denker" wrote:

> Ludwik Kowalski wrote:
> > 1) We have copper wire whose diameter is 5 mm.
> > The length of the U-shaped rail will be 5+5+5=15 cm;
> > the effective length of the cross bar will be 5 cm.
>
> OK.
>
> > This means that the total R will be 0.0173 ohms.
>
> Maybe.  You need to worry about additional resistance
> at the point where the slider meets the rail.  I suggest
> MEASURING the resistance... or doing an experiment
> where the resistance is unimportant (see below).
>
> > 2) Suppose B=0.5 T (a neodymium magnet) and that
> > dt=0.25 second. With the d(AREA)=25 cm^2 the
> > emf will be 0.01 volts. The current will be 1.15 A.
> >
> > 3) The voltage across the bar will 25% of the total
>
> Again, if the contact resistance can be kept small.
>
> > or 2.5 mV; its polarity can be measured with the
> > oscilloscope.
>
> OK.  That should be easy.
>
> > Note that the voltage on the cross
> > bar would be negligible is q 100 ohm galvanometer
> > was inserted into the loop.
>
> So don't do that.  If you want to measure the current
> without disturbing the circuit, measure the voltage drop
> across the "4th leg" i.e. the stationary piece that joins
> the two rails, i.e. the leg that isn't the slider and isn't
> either rail. Use Ohm's law to infer the current.  This
> is a standard way to measure current.  Just as you want
> voltmeters to have a large input impedance, you want
> ammeters to have a low input impedance, and this
> scheme certainly achieves that.
>
> On the other hand, I don't have a clear idea of what
> the point of this experiment is.  The Subject: line
> says something about EMF.  EMF is a voltage, not a
> current, so why are we using ammeters (galvanometers)
> at all?????  Why not just insert a high-impedance
> voltmeter in series with the 4th leg?  That gives a
> direct, uncomplicated measurement of the EMF.  The
> result is insensitive to the resistance of the
> rails, slider, et cetera.
>
> If this uncomplicated experiment doesn't answer the
> question, please explain in more detail what the
> question is.

I agree that measuring I would not be essential. But
knowing the direction of I is essential. This can be
done with the quick compass needle. The contact
resistance will have the effect on the magnitude of I
but not on its direction.

My understanding of Feynman is that the direction
of I in the current loop and the direction of E lines in
the cross bar should be the same in one case (stationary
cross bar, changing B) and opposite in another case
(sliding cross bar, constant B). Is this correct? How
can this be demonstrated?

My first suggestion was not good because "the armature
resistance in the generator" was ignored. The second
suggestion, as indicated this morning by Bob, also has
an error.

John's idea, if I understand it correctly, is to keep
everything as described yesterday but measure two
DOPs, for example, with a dual trace oscilloscope.
One pair of leads (for trace 1) should be connected
to the cross bar and another (for trace 2) should be
connected to "the 4th leg" (the leg connecting two
rails). In one case the polarities of two pulses will
be the same and in another they will be different.
Is this an acceptable method to demonstrate  the
idea of "two distinct phenomena" behind the so-
called "flux rule", as explained by Feynman?
Ludwik Kowalski