Faraday's law on circular wire

[milesyoung]

Now you're talking motor theory.

The only difference between my two examples is that I now specifically included:

You'd have to consider, for instance, a bar magnet approaching (not passing through) the coil in such a way that it produces a rate of change of flux through the coil that is constant for a period of time long enough for any transient to decay away.

to get the discussion away from the transient state of the current in the conducting loop, which was irrelevant to the example I gave in #47.

How is my example now otherwise different from the one given in #47?

But my point is that the added energy comes from somewhere.

Where in my examples am I adding energy from nowhere?

But remember that the open circuit voltage does change. If you load the loop with 0.001 ohm, and you increase the force on the magnet, you have really changed Voc. The value of Voc increases with increased force to magnet.

It does not. The open-circuit voltage is determined by the motion of the bar magnet. Since the motion of the bar magnet is the same regardless of the resistance of the loop (that was the premise for my example, remember?) the open-circuit voltage is always 1 V.

The new Voc2 cannot now still be 1.0V, w/ Iload = 1000A. If the load resistor has 1,000A at 0.001 ohnm R value, the voltage at the load resistor is indeed 1.0V. But if we suddenly open the load from the loop, Voc2, the open circuit voltage w/o load, but with increased force on magnet, is much greater than 1.0V, more like 1e4 volts.

If you suddenly change the resistance of the loop, the open-circuit voltage is still 1 V because the motion of the bar magnet does not change.

Also, if you completely disregard the premise for my example, this still doesn't make any sense. If t is the time at which you change the resistance of the loop, the velocity of the bar magnet is the same at t- and t+, so the open-circuit voltage is the same at t- and t+.

The torque exerted on the shaft of a generator is not what determines the open-circuit voltage present at its terminals. It's determined by the generators angular velocity and design parameters.

 

[cabraham]

The only difference between my two examples is that I now specifically included:
to get the discussion away from the transient state of the current in the conducting loop, which was irrelevant to the example I gave in #47.

How is my example now otherwise different from the one given in #47?Where in my examples am I adding energy from nowhere?It does not. The open-circuit voltage is determined by the motion of the bar magnet. Since the motion of the bar magnet is the same regardless of the resistance of the loop (that was the premise for my example, remember?) the open-circuit voltage is always 1 V.If you suddenly change the resistance of the loop, the open-circuit voltage is still 1 V because the motion of the bar magnet does not change.

Also, if you completely disregard the premise for my example, this still doesn't make any sense. If t is the time at which you change the resistance of the loop, the velocity of the bar magnet is the same at t- and t+, so the open-circuit voltage is the same at t- and t+.

The torque exerted on the shaft of a generator is not what determines the open-circuit voltage present at its terminals. It's determined by the generators angular velocity and design parameters.

Actually, they are inter-active. The torque is directly related to current, speed related to voltage. But when a generator is loaded, current in the stator winding generates a torque counter to the applied shaft torque. This new torque results in speed dropping, and that reduces Voc. Torque, angular speed, current, and voltage interact.

Let's say the generator is a simple dynamo/magneto made by spinning a bar magnet inside a coil. Of course the angular speed ω, and the flux ∅ determine Voc. But once Voc is established w/o a load, then we load the coil, what happens? The current due to loading generates counter-torque which reduces ω, as well as reducing ∅. These result in reduction of terminal voltage.

Of course if we increase torque at shaft we can overcome this counter-torque and restore ω to original value. But is Voc restored to original value? The total flux ∅ is that due to magnet minus that due to load current per law of Lenz. Although ω is restored to original no load value, ∅ is not its original value. How do we restore ∅?

If the generator is wound rotor instead of permanent magnet, we increase the field current. Since the stator winding load current magnetic flux cancels some of the rotor flux, increasing rotor current and flux restores ∅ to original value. But if the load is suddenly removed, the Voc is different from before due to increased field current. With a permag rotor, we cannot increase flux, so we increase speed to get needed flux to maintain constant voltage at terminals. Again, removing load suddenly results in a Voc larger than before due to increased speed.

Again, I am only pointing out that there is much interaction here and that Mother Nature does not provide free voltage regulation. Many on these forums have stated that with induction, Voc is determined by ∅ & ω, and that Rload and Iload do not affect Voc, but I assure you that is not the case. What my opponents have been stating is that Mother Nature provides automatic voltage regulation free of charge w/ no effort required on our part.

Anybody familiar with commercial power generation/distribution knows otherwise, as does anybody experienced with car & aircraft alternators. When load current changes, field current must change to maintain constant terminal voltage. Any good machines book covers this in detail with good math to illustrate numerically.

Believe me when I tell you that if what you state is true, we would never need voltage regulators at all. I will elaborate on any point to clarify. BR.

Claude

 

[milesyoung]

This is getting ridiculous. It's such a simple example.

This thread has run its course, the OP is long gone.

 

[technician]

This is getting ridiculous. It's such a simple example.

This thread has run its course, the OP is long gone.

I agree,and when standard texts are dismissed bad physics results.

 

[cabraham]

I agree,and when standard texts are dismissed bad physics results.

Are you saying my physics is bad? You cannot refute one sentence of mine, yet you insist you have it right. Every motor/generator text affirms me. You're a grammar school teacher, you refer to texts not even collegiate level, do you understand that? Where in my treatise did I err? Instead of pointing out my alleged error, you just say how ridiculous this discussion is.

Show me the error in this. Inductive reactance is as real as the resistance in any circuit. The total impedance of a loop is the phasor sum of both R & X_L. The open circuit induced voltage is Voc. This Voc is divided across both R as well as X_L. The impedance which Voc is across is Z, not just R. Because your "standard texts" do not mention X_L, you assume it's irrelevant. When sources conflict, the only way to resolve is to apply circuit theory and field theory to obtain answer. I did just that. I will elaborate but for you to just say I'm wrong doesn't count at all. Show your proof. You merely assert that R is relevant while X_L is not.

Claude

 

[technician]

[QUOTE=You're a grammar school teacher, you refer to texts not even collegiate level,

I perfectly understand what this means, by the way some of my texts are university level. The authors provide their email addresses for feedback... Especially if you find anything amiss in their analysis...do you want me to pass these onto you if you feel there is a need to enlighten these authors, I think they would appreciate this.
This is irrelevant...text books are the ultimate reference and the 'truth' is not suddenly revealed at some future date. More detail maybe but that is another issue (speed of light??)
What about this approach:
Force on a charge carrier in a magnetic field = Bqv...OK?
This leads to force on a current carrying wire in a magnetic field of F = BIl...OK?
This wire on parallel rails connected to an emf in a magnetic field will require a force F = BIl to hold it in place or move it with constant velocity 'v'...OK?
The mechanical power required to pull this wire along at velocity v = Fv. = BIlv...OK?
The electrical power supplied to the wire = EI...E = the emf connected to the parallel wires...OK?
So (assuming no other energy complications)
EI = BIlv...OK?
So the emf = Blv
This is exactly the emf induced in a moving wire of length l in a magnetic field B moving at velocity v.

Ps... In post 42 a worked example is given from a university textbook...do you agree with this solution without reservation?

 

[cabraham]

In open circuit yes I agree. The difference here is what happens when current exists. The time changing flux from the external source is unopposed in open circuit. Hence Voc is determined as the texts describe. But once you load the circuit with R & X_L, you must consider the following.

Just as an external magnetic flux varying in time has an associated induced voltage, so does a magnetic flux due to internal loop current. Inductance in the loop, i.e. self-inductance, is just as subject to law of Faraday as is external flux. We can lump the distributed loop inductance into a single value L.

The inductive reactance for the loop, is X_L = Lω. The equivalent circuit is a constant voltage source of value "Voc", an inductance "L", with a resistance "R". Let's say that using your computations above the Voc value is 1.00 volt (open circuit).

What happens when R closes the loop? We have a series network, Voc source, L, and R. You claim that I is simply Voc/R. But if R = 1.00 ohm, with L = 1.0 henry, and ω = 1.00 radian/second, what is I? If I was 1.0V/1.0 ohm = 1.0 amp, we have contradiction w/ laws of physics.

The source voltage Voc = 1.00 V must equal the sum of the voltage across L and that across R, 90 degrees out of phase. If I = 1.00 amp (Voc/R), then we get 1.00 volt across R per Ohm. So 1.00 volt at the source, 1.00 volt across R, leaves 0 volts across X_L? But jX_L is j1.00 ohm, which when multiplied by 1.0 amp gives j1.00 volt.

The only way I could be 1.00 amp is for induced emf to have an open circuit value of 1.00 + j1.00 = √(1.00)² + (1.00)² = √2 = 1.414 V. But if Voc is 1.00V, then I has to be 1/√2, or 0.7071 amp.

The method for computing Voc as you described is not what is being challenged. I am well aware of what you are saying. My point is the law of Lenz. As soon as current exists in the loop another flux, varying in time, is added to this problem. By definition, the flux linkage per unit current, weber-turns/amp, is inductance, in henries.

Although this problem is of a distributed field nature, we can lump inductance into a single element value L. Is this making sense? Maybe I am not as good at explaining things as I would like to believe? BR.

Claude