Re: Magnetic field and relative motion

From: xxein (xxein_at_bellsouth.net)
Date: 06/22/04 

Date: 21 Jun 2004 20:24:23 -0700

"Pmb" <someone@somewhere.com> wrote in message news:<pemdnchHRqaF0UvdRVn-tA@comcast.com>...
> "xxein" <xxein@bellsouth.net> wrote in message
> news:cce403e3.0406171624.3e7de408@posting.google.com...
> > "Pmb" <someone@somewhere.com> wrote in message
> news:<qvmdnfQKXbtLekzdRVn-tw@comcast.com>...
> > > "Pmb" <someone@somewhere.com> wrote in message
> > > news:ir2dnYsAgfel9kzdRVn-tA@comcast.com...
> > > >
> > > > "V.K.Tamhane" <vktamhane12@rediffmail.com> wrote
> > > >
> > > > > 1. Observer on the disc will notice magnetic field rotating along
> with
> > > > > the circuit and so seen from this frame, there is no emf generated
> in
> > > > > the circuit of stationary frame.
> > > >
> > > > It is incorrect to think of magnetic fields as rotating as you
> mentioned.
> > > > Doing so can lead to serious errors. You're also incorrectly assuning
> that
> > > > no E-field in one frame means no E-field in all other frames. That is
> also
> > > > incorrect.
> > > >
> > > >
> > > > > 1. When the magnet producing the field is rotated and if the field
> is
> > > > > fixed then there will not be an emf.
> > > > > 2. If the field is rotating along with the magnet then too, emf will
> > > > > be zero. Because equal and opposite emf will be produced in the
> > > > > lengths MB and CD.
> > > >
> > > > If the magnetic is rotated there will be no EMF but there will be an
> > > > electric field
> > >
> > > To be precise - an EMF is defined as the line integral of force per unit
> > > charge in a wire integrated around a complete circuit.
> > >
> > > Pmb
> >
> > xxein: And OWLS is defined as whatever we measure with TWLS? Just 'a
> > shot on the dark'.
>
> I dunno. I've never given much thought to that particular question in
> relativity. I will someday soon though.
>
> Pmb

xxein: Geez! You don't know what you are missing (like almost all of
any physical theory that is self-proclaimed to be consistant).

You surprise me. All this positive talk about this and that - and you
don't even know the rudiments needed to have even a semi-complete
physical theory?

It seems that you DO just shoot in the dark. But, hey, I can only
call it as I see it!

This is just a comment and you don't need to reply.

"shot on the dark" is my typo. I beg Peter Sellers to forgive me -
&#9835;&#9835;&#9835;&#9835;&#9835;&#9835;&#9835;&#9835;&#9835;.

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