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Re: Faraday induction



At 09:51 AM 5/30/01 -0400, Wolfgang Rueckner wrote:
Suppose we have a
single loop of wire lying in the plane of the page (or monitor screen) and
the loop is split -- that is to say, it's not a complete circuit.

OK.

Also imagine an increasing magnetic field into the page (monitor).

OK.

Faraday's/Lenz's law tells us that an emf will be induced such as to
produce a CCW current which generates a magnetic field out of the page.

It does?

I know of no such law.

I consider Lenz's law to be a rule of thumb that tells what "_tends_" to
happen in consequence of Maxwell's equations, Ohm's law, et cetera.

1) In this case, the relevant Maxwell equation is
Voltage = flux dot
i.e.
Voltage = (d/dt) flux

... referring to the non-potential voltage around a loop, as a function of
the flux through the loop.

2) Ohm's law, when applied to a gap of any reasonable size, says that no
current flows around the split ring.

So there will be a Faraday-induced voltage, but no Faraday-induced current.


Now here's the question. How large can this "split" become?

That's not a question about magnetism. It's a question about
capacitance. Once the gap is big enough to create a significant capacitive
impedance, no current flows around the ring. End of game. The real
question is how *small* the gap has to be, so that the capacitance thereof
can carry the induced current. Work it out. I'll bet you find it's a
reeeeeally small gap.

For example, suppose we open up the split so that it's as large as the
diameter of the loop -- what area does one use to calculate the magnetic
flux?

You can calculate the flux through any abstract loop. You can have a gap
that's so big that there's no wire at all, just a pure arbitrary loop, and
still there's flux through the loop.

1) This flux induces a voltage.
2) Whether or not this voltage drives a current is a separate question.


Does one just imagine a "short" between the two open ends of the loop so
that one has a quasi enclosed area?

That's the capacitance question again.