| . . .
| 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? . . .
| Ludwik
Be careful:
Whichever effect produces the emf, q*VxB or a non-conservative E,
1) the emf will be in the same direction as the produced current
2) wherever static charges accumulate (and they typically will, in both
cases - due to bends, resistance, etc) they will generate a conservative E
whose field lines begin on positive and end on negative charges -
everywhere. In both cases, this conservative E co-exists with the "emf
force field" (q*VxB or non-conservative E).