Re: induced emf again

["John S. Denker" <jsd@MONMOUTH.COM>]

Ludwik Kowalski wrote:
> 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?

It depends on the direction of sliding and the
polarity of the change in B.

The rule:
        V equals flux dot
is applicable in each of these two cases and takes the same
form in each case.  (Feynman goes on to discuss cases where
the rule does not apply.)

> How can this be demonstrated?

Lots of ways.  Some better than others.

Trying to measure the voltage in a circuit made
entirely of low-impedance elements is not one of
the better ways.  Using N-1 low-impedance elements
plus one high-impedance voltmeter (or scope) seems
like a logical approach.

> 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.

Well, actually I was only imagining a single-trace
oscilloscope.

> One pair of leads (for trace 1) should be connected
> to the cross bar

It's not easy to this in a sensible way when the crossbar
is moving.  You need to do something rather tricky to
minimize and/or account for induced voltages in the scope
leads.  When I mentioned this yesterday I got scornful replies,
but it remains a nontrivial part of the physics.

> and another (for trace 2) should be
> connected to "the 4th leg" (the leg connecting two
> rails).

That connection I understand.

> In one case the polarities of two pulses will
> be the same and in another they will be different.

Not likely.

The EMF is just given by flux dot in both cases.  To be
explicit, the two cases are:
 -- change the flux by changing the field, leaving
the circuit unchanged, and
 -- change the flux by sliding the slider, leaving the
field unchanged.

The current, if any, under ordinary conditions will
flow in the direction motivated by the EMF.  A partial
exception could occur if you put a capacitor in the
circuit and manage to set up underdamped LC oscillations.
But the initial current and the average current will, even
in the LC case, be in the direction suggested by the EMF.

In an overdamped circuit (including purely resistive circuits,
and LR circuits without much C) you can accumulate charges
that reduce and/or steer the current, but they're not going
to reverse the current.  These charges don't even appear
until after the current has started to flow.

Why would anybody think otherwise?

(Nitpickers could contrive even more-bizarre exceptions by
putting batteries in the circuit, but let's not go there.)