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

That's nice. Now could you take a few minutes out of your busy schedule to learn that "it's" means IT IS?

That *was* a dead giveaway, wasn't it?

Next, he's going to explain how to use sheep's bladders to prevent earthquakes.

:-}

Bob

You can do that?

-- Boris

--
Posted via a free Usenet account from http://www.teranews.com

I always thought the voltage across an inductor was -Ldi/dt. Yet it

** The date on the top of that text is a worry ....

...... Phil

Nope, it's E = + L di/dt using conventional notation.

The electric field is nonzero because there's a changing magnetic field, which creates potential. The magnetic field is changing because of... dI/dT !

Same thing, if you have a voltmeter. Same thing if you don't.

You might do that.

John

Yeah, 2002 was a bad year for me too. Lost my job and stubbed my big toe. Nail fell right off.

** Google Groper half wit alert !!

"Phil Allison"

** Yawn .....

What a tedious moron.

..... Phil

OK

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.

This is not obvious to me. With the voltmeter and leads connected to the inductor, only V across the voltmeter contributes to the total E.dl around the loop, since E is zero inside the leads and the inductor. So the voltage across the voltmeter equals the total E.dl. Take the leads away and E.dl across the voltmeter no longer equals the total E.dl around the loop because E is no longer zero where the leads were.

What is your impression of the paper if you have read it?

At least you have made an attempt to answer/correct my points. Thanks.

You need to read up on it. You are thoroughly confused. I hope you are not a student ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

In a volume of space that is subject to a changing magnetic field, there *is* a corresponding electric field. If there's a conductor in that space, the electric field is present inside the conductor, too. That's why you can have voltage across the ends of a transformer winding.

No. Place a coil or loop in a time-varying magnetic field. Connect its ends to a twisted pair or a coaxial cable and run that cable a mile away, far from the magnetic field zone, and use the voltmeter there. Same voltage.

Of course, if the voltmeter and its leads are in the field, you'll have to take the additional potentials into account. In the case of a voltmeter connected to the secondary of a transformer, the voltmeter isn't exposed to much field, but it still measures the winding voltage.

As noted, you can move the voltmeter far away, out of the field, and accurately measure the non-zero coil or loop voltage. And a 100-turn coil pumps move voltage into the meter than a 5-turn coil, or a bigger loop more than a small one, so there must be voltage within the coil.

There *is* an electric field inside a conductor which is immersed inside a changing magnetic field. "Ohm's law" is trashed by electrodynamics. The electric field is like a gravitational field... it permeates everything. If that were not true, every generator and transformer and motor on the planet would instantly stop working.

In Hertz's seminal experiments, he detected em fields by using a single metal wire that *almost* made a full ring, leaving just a tiny gap. When exposed to a sufficiently strong em field, he observed sparks jumping across the small gap. No voltmeter needed.

Classic electromagnetics.

John

Read it more carefully, you still are confused.

(Physics 8.02 is MIT FRESHMAN physics, 2nd semester ;-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

If the current is finite and resistance is 0, then the charge inside the conductor rearranges itself to create an electric field that cancels the applied electric field. The net electric field is therefore zero. Since the electric field is the measured force on a unit positive charge placed there and an infinitesimally small i.e. 0 force is required to move it because the reistance is zero, then the electric field has to be zero.

I think the answer is that Faraday's law is true for any passive region that modifies the electric field, and the electric field in Faraday's law is the total electric field in the passive region and not just the induced electric field. So that the electric field through the terminals of the voltmeter is that due to an infinitesimally small current that generates a voltage across it's internal resistance.

For a finite I , E.dl through 0R leads is still zero, from definition of what an electric field is.

Suppose a conductor requires 0.01V applied across it to push 1A through it. If it is now wound into an inductor and pushes 1A through

1R, the voltage through the 1R will be 1V, and through the inductance still 0.01V. An electron in the inductance will still require the same force to move it for the same current, whether that force is from a static electric field or a changing magnetic field. If the force is the same, the electric field there is the same.

The electric field inside a conductor depends on the resistance of the conductor. But for a conductor of 0 resistance carrying a finte current it is zero. Concentrate on the definition of an electric field.

Yes. And the electric field in the gap was several orders of magnitude greater than that in the metal wire.

Well, he makes the point that E.dl through an inductor is zero, yet you have difficulty understanding why. The paper was written by a distinguished professor at MIT.

Once again, thanks for attempting to discuss my points.

[snip]

You're mixing up DC and AC components with complete abandon.

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC\'s and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

It's not. Just quit being silly. If that were so, nothing would work.

Since the electric field is the measured force on a

OK, OK, I give up. Believe whatever you insist on believing.

And don't send me a resume.

John

I hate to tell you, but ohm's law is not trashed. Please consider the resistance of the coil and the current thru the coil, along with the (remote?) load.

"John Larkin"

** You did notice the date on the paper the OP quoted - didn't you ?

....... Phil

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.

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. 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.

"Fred Bloggs"

** You did notice the date on the paper the OP quoted - didn't you ?

Wake up.

......... Phil