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The voltage across an inductor isn't -Ldi/dt

You won't be able to maintain a steady-state DC voltage drop
across a conductor with zero DC resistance, because that would
imply infinite current. However, the transient or AC-voltage
case the story is entirely different. Consider, for example,
my 10kV 300kHz ferrite-core transformer, which has 22 turns on
its secondary. The single-turn primary has 450 volts across
its ends. Since it's a thick band of 1"-wide copper, with for
all practical purposes, zero DC resistance, by your obviously-
incorrect understanding this is not possible. BTW, the current
in my transformer's single thick strip of copper is modest and
due almost entirely to activities outside the copper strip.

Let's assume the secondary is open-circuit so that it has no effect on
the the primary. When you apply 450V across the primary, that doesn't
mean the primary winding has a voltage of magnitude 450V through it
because total EMF = 450V - primary = rate of change of magnetic flux =
Lp*d(Ip)/dt. With static I, total EMF = 450 - primary = 0, and so
primary = 450V which, as you pointed out, would give an infinite
current for a 0R primary.

Where there is a changing magnetic field, the electric field isn't
conservative and so you can't assume that components in parallel will
have the same volage through them. Suppose you have a ring of 1R in a
uniform magnetic field normal to the plane of the ring. Suppose it
changes at a uniform rate so that it induces an electric field in the
ring with the total EMF around the the ring being 1V. The voltage
across a 0.1R segment = 0.1V is in parallel with the rest of the 0.9R
= 0.9V segment, yet the voltage drops are not the same. In a
conservative field where total EMF around ring = 0V, you can say
V_0.1R = V_0.9R.

Some people recommend this paper. If you want I'll email it to you.

R.H. Romer: "What do 'voltmeters' measure?
Faraday's law in a multiply connected region",
Am. J. Phys., 50, 12 (1982), pp. 1089-93
 
W

Winfield Hill

Jan 1, 1970
0
jmc8 said:
Let's assume the secondary is open-circuit so that it has no effect
on the the primary. When you apply 450V across the primary, that
doesn't mean the primary winding has a voltage of magnitude 450V
through it because total EMF = 450V - primary = rate of change of
magnetic flux = Lp*d(Ip)/dt. With static I, total EMF =
450 - primary = 0, and so primary = 450V which, as you pointed out,
would give an infinite current for a 0R primary.

What are you, a TROLL? You wrote "I always thought the voltage
across an inductor was -Ldi/dt. Yet it cannot be ..." Silly boy.
Some people recommend this paper. If you want I'll email it to you.

R.H. Romer: "What do 'voltmeters' measure?
Faraday's law in a multiply connected region",
Am. J. Phys., 50, 12 (1982), pp. 1089-93

I usually enjoy Am. J. Phys. papers, especially when they cover
an interesting topic and demonstrate a clear pedagogic approach.
I subscribed for many years, starting as a physics undergraduate
but stopped before 1982. Certainly send me a copy. Thanks.
 
That lecture note is poorly written, confused, and useless. A voltage
drop, or more accurately potential drop, is understood to be a measure
of energy expenditure per unit of charge injected through the component.
There is no such energy expenditure in a pure inductance, it is all
storage, and therefore there is no potential drop across it. Faraday's
law introduces the concept of electromotive force due to induction, or
e.m.f., and quantifies its interaction with the electric field in closed
loops.
It is a statement about the net potential drops around the loop
external to the inductor and not the inductor itself.

integral_loop_E.dl = integral_area_B.dA. Why doesn't it work to
include the inductor?
You cannot have a
di/dt in the inductor without an external circuit loop, the loop will
always be present, but because it can take many forms, it is simpler to
consider that the algebraic sum of the external loop drops as appearing
across the inductor in the sense that this is where one would place the
measurement probes. Get real.

In a non-conservative field, the voltages across components in
parallel doesn't equal one another in general. Have a look at my reply
to Win.
 
What are you, a TROLL? You wrote "I always thought the voltage
across an inductor was -Ldi/dt. Yet it cannot be ..." Silly boy.

With hindsight, I think I have to agree with you. I should have said
"I always thought that an EMF = Ldi/dt was generated INSIDE the
conductor of an inductor. In fact the EMF is infinitesimally small
i.e. 0 if it has a resistance of 0R and carries a finite current."
I usually enjoy Am. J. Phys. papers, especially when they cover
an interesting topic and demonstrate a clear pedagogic approach.
I subscribed for many years, starting as a physics undergraduate
but stopped before 1982. Certainly send me a copy. Thanks.- Hide quoted text -

I sent it and I hope you enjoy reading it, if you have the time.
 
J

jasen

Jan 1, 1970
0
There is no electric field inside a conductor of zero resistance
carrying a finite current. If there was, the current would be
infinite. Therefore there is no electric field inside an inductance of
zero resistance carrying a finite current. The line integral of E
across the terminals of the voltmeter - its measured voltage - equals
the rate of change of magnetic flux through the circuit consisting of
the inductor, the leads to the voltmeter and the terminals of the
voltmeter.

the problem with that line of reasoning is that the voltmeter is outside
the conductor.


Bye.
Jasen
 
the problem with that line of reasoning is that the voltmeter is outside
the conductor.

How does that create a problem in my reasoning? I'm just using
Faraday's law for the loop consisting of the inductor, leads and
voltmeter. E.dl through inductance and leads is zero , with E.dl
through the voltmeter making up the rest = Ldi/dt.
 
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