|
| 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? . . .
|Ludwik
Ludwik,
You are reading too much into this "reversed" internal E field of the
accumulated charges, in the moving bar case.
Suppose you connected an external circuit, with resistance, to the
secondary of an ideal transformer. Charges would accumulate at the
secondary terminals, just as across your moving bar. This happens because
the EMF cannot directly act on the external carriers - the electrostatic
field of the accumulated charges is needed to do this, as in the moving
bar scenario.
On the other hand, if you had a magical conducting loop which could grow
in radius. In the presence of a static magnetic field, this would generate
a VxB force throughout the loop. Furthermore, this force would be
tangential and would act on all carriers around the loop. There would be
no tendency for charges to accumulate and we would largely mimic a
corresponding dB/dt case. Think on it!