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Re: induced emf again

1) Bob is right again. To see this I am looking at
Feynman's Figure 17.1 (vol 2). In this illustration
the lower end of the cross bar would be positive
and the induced current would be clockwise.

To produce a clockwise current with v=0 the magnetic
field must decrease, for example, by pulling the S-pole
of a magnet toward me (away from the plane of the
loop). Note that I am referring at the same Figure 17.2
and applying Lenz's rule. The lower end of the cross
bar is again positive.

2) This realization brings back the initial question. What
can one do, experimentally, to show that Feynman is
correct? His arguments are based on gedankenning,
that is on theoretical considerations. If two effects are
different then there should exist an experimental way
of distinguishing between them. Any suggestion?

3) Note that in Figure 17.2 Feynman shows an easy
to perform experiment in which the emf is induced
while the dFLUX/dt is zero. But I am having difficulty
in penetrating his Figure 17.3 experiment. Assuming
that the "slightly curved edges" of the copper plates
are circular I made a paper model of them. Wires
connected to the galvanometer (in my paper model)
touch the plates at their centers. I keep one semicircle
fixed and rock the other. This convinces me that the
area of the loop is indeed changing so that the
dFLUX/dt is not zero.

But what evidence is used to justify that "there is no
emf" in this case? The galvanometer in Figure 17.3,
by the way, shows that the current is flowing. This
would indicate the the emf is not zero. What am I
missing in this case?

4) It occurred to me that it would be desirable, in the
experiment suggested yesterday (see below), to
increase the resistance of the cross bar, for example,
by using three times thinner wire. In that case the
voltage on the cross bar could be made nearly as
large as the emf (rather then only 25% of emf).
Ludwik Kowalski

Bob Sciamanda wrote:

I don't think so, Ludwik.
With the currents in the same direction, the driving emfs will also be in
the same direction. In the moving bar (stationary magnet) case you measure
the PD created by the terminal charges - this will be a rise in the
direction of VxB and the emf. If instead you move the magnet (staionary
wires) so as to get a current in that same direction, you will measure an
emf in that same direction - it will now be due to a CurlE=-dB/dt effect and
will be distributed around the loop but the direction will still be in the
direction of the current! I don't see what you are proving.

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: Monday, May 06, 2002 9:38 PM
Subject: Re: induced emf again

In the previous message I suggested an experiment to
test Feynman's idea that the so-called "flux rule" covers
two distinct phenomena. I wrote:

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 experiment should
be performed twice: (a) when the rod is sliding in a
constant magnetic field and (b) when the rod is
stationary but the magnetic field is changing. In
the first case the direction of E inside the rod will be
opposite to the direction of conventional current, in
the second the two directions will coincide.

It turns out that using a galvanometer (to determine
the direction of the current) makes determinations
of the direction of E (in the cross bar) very difficult.
This has to do with the galvanometer's resistance
of about 100 ohms. The direction of I can be
determined with a compass needle.

Here are some practical considerations:

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. This
means that the total R will be 0.0173 ohms.

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
or 2.5 mV; its polarity can be measured with the
oscilloscope. Note that the voltage on the cross
bar would be negligible is q 100 ohm galvanometer
was inserted into the loop. The measured polarity
(for a fixed direction of electric current) is expected
to be different for two ways of changing the flux
(sliding the bar versus moving the magnet). Any comments?
Ludwik Kowalski