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Topic: testing out the Malus law replacement in Ohm's law of the Faraday law
Chapt15.34 explaining Superconductivity from Maxwell Equations #1174 New
Physics #1294 ATOM TOTALITY 5th ed

Replies: 2   Last Post: Jan 23, 2013 10:00 PM

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 plutonium.archimedes@gmail.com Posts: 18,572 Registered: 3/31/08
testing out the Malus law replacement in Ohm's law of the Faraday law
Chapt15.34 explaining Superconductivity from Maxwell Equations #1174 New
Physics #1294 ATOM TOTALITY 5th ed

Posted: Jan 23, 2013 1:05 PM

Testing out the Malus law replacement in Ohm's law of the Faraday law.

Now then, in my prior post I showed where Faraday's law is Ohm's law:
In Faraday's law as stated by:
emf  = -N dB/dt
where 1/N is Resistance, and dB/dt is Voltage and
emf is current i, and where we delete the negative sign.

i = V/R

Now for superconductivity, the R becomes another function of the Malus
law I' = I" cos^2 A.

So we replace R in i = V/R with the Malus law

i = V/(I"cos^2A)

Now let us check out in an experiment whether that is what happens in
physics.

On page 740 of Halliday & Resnick, Fundamentals of Physics, 3rd
edition, 1988, shows experiment one
of what is likely a remake of what Faraday did in the 1830s of his
famous Faraday law. On page 740 shows a closed loop wire connected to
a Galvanometer, G, and a hand holding a bar magnet aimed at moving
into the center of the closed loop of wire and by the motion, the G
should register a small electric current.

Now, to test out whether the Malus law is applicable, we re-do that
experiment and hold the bar magnet so that it does not move into the
center at a perpendicular to the wire loop cross section area. We move
the bar magnet at a oblique angle, just as in Malus law, we have a
oblique polarized filter.

So, the question is, how does the current in G register with the bar
magnet motion at oblique angles?

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