Re: indiced emf ....

[John Mallinckrodt <ajmallinckro@CSUPomona.Edu>]

On Thu, 19 Jun 1997, LUDWIK KOWALSKI wrote:

> ... But can we say that the situation is paradoxial? 

I hope not.  ;-)

> On one hand emf is zero on the other hand it is not zero. 

I think that if there is one thing we (or at least I) have concluded in
this thread, it is that emf is not very well defined.  For instance, it
appears that emf is often taken as the path integral of the tangential
component of the NONelectrostatic component of E or of (vxB or of E +
vxB.) This has no import in the case of closed paths, but it does for open
paths.  For instance in the case of the moving rod the observed E in the
rod (which is electrostatic in origin) is opposite the direction of the
emf.  We get what is usually taken to be the "motional emf" by integrating
only vxB or, equivalently, by integrating vxB + the nonelectrostatic
component of E (which is zero).  If we integrate E + vxB we will get zero
since the observed E = -vxB.

> We are talking about
> what happens when v<<c. Is it a classical (Galileo) relativistic effect.
> what happens when v<<c. I was under the impression that B transforms into
> a different B (and E), in any practical sense, only when v/c is not very
> small.

It takes a lotta v and a lotta E to make much B.  For instance, 100 kV/m
seen at .01 c looks only like a little more than 3 microTesla.

But not vice-versa; it's *easy* to make E outta B For instance 100 mT seen
at 10 m/s looks like 1 V/m.

This is why Faraday's law is *so* much easier to directly confirm
experimentally than Maxwell's displacement current correction term in
Ampere's law. 

John
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