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Re: induced emf again



"John S. Denker" wrote:

The total voltage around the loop is unchanged. The local
voltage-drops are re-arranged: a little more voltage-drop
here and a little less there.

This is easily visualized by considering a C-shaped conductor
in a changing magnetic field. Choose a loop that follows the
shape of the C, inside the material except where it has to cross
the gap. There will be no field in the bulk of the conductor.
There will be a concentrated field in the gap.

This was a very useful example. The induced emf is just the
total voltage around a loop; how it is distributed depends on
the R of different segments. The gap has a large R and the
maximum drop of voltage is there (the induced current is
never zero).

It is clear that the derivation (b) is not desirable. It makes
one think that the sliding rod is like a battery; free carriers
travel uphill (electrically speaking) in the rod and downhill
(electrically speaking) in the rest of the wire frame. But this
is wrong, the electrical uphill does not exist in our closed
loop. Many students probably think that the electric field
lines inside the rod are from the positive end to the negative
end. They view the rod as a generator in which mechanical
work is used to separate positive charges from negative
charges. A new physics education research topic?

FROM THE INITIAL MESSAGE:

I am puzzled by motional emf (v*B*L) produced when a
metal rod of length L slides with the speed v along the right
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.
Ludwik Kowalski