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Re: induced emf again



1) What still bothers me is this. The emf=B*v*L can
also be derived from Faraday's law of induction.

d(flux)/dt = B*d(area)/dt=B*L*dx/dt = B*L*v

This derivation does not imply any gradients of
charge density; it contradicts the existence of such
gradients. The other derivation, also in the laboratory
frame, implies the existence of charge density
gradients. Both can not be correct.

2) Does anybody know of an experiment in which
voltage along a rod moving in the B field (for example,
airplane wings) was actually measured? I am referring
to a rod which is NOT sliding along a rigid wire frame
perpendicular to B lines. (The wires connecting the rod
to the voltmeter would have to be parallel to B lines in
order to keep the flux = 0; I suppose.)
Ludwik

John Mallinckrodt wrote:

I have written up a brief treatment of this problem that
some may find helpful. Please see:

<http://www.csupomona.edu/~ajm/special/rodnrail.pdf>

John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm

On Tue, 30 Apr 2002, kowalskil wrote:

I am puzzled by motional emf (v*B*L) produced
when a metal rod of length L slides with the speed
v along the rigid wire frame perpendicular to the
uniform magnetic field B. The v*B*L formula
can be derived in two ways: (a) from Faraday's
law of induction and (b) from balancing two
forces. The second derivation bothers me it goes
like this:

The free carriers in the rod will be subjected to
F'=q*v*B force. Thus one end of the rod will
tend to become positive while another will tend
to become negative. The process will continue
till the electrostatic force F"=q*E (acting on free
carriers) becomes equal and opposite to F'. This
leads to E=v*B and emf=E*L=q*v*B. In other
words the rod is treated as a battery causing a
current in the conducting loop.

Is this derivation desirable? It seems to imply
that the electric current in the rigid frame is
due to static charges residing at the ends of
the sliding rod. But we know that this is not
true; we know that electric field lines in the
closed loop have no beginning and no end.

Suppose I remove the rigid frame over which
the rod was sliding. The same rod is moved in
a vacuum, still perpendiculary to the magnetic
field B. Are there going to be static charges of
opposite sign at the ends of the rod or not?
Ludwik Kowalski