Faraday's law on circular wire

[cabraham]

The premise was that the rate of change of flux through the loop was constant in time. I asked you to consider what the steady-state solution for the current would be in that case, since the solution would be consistent with the one given by technician in post #26. It's also a common example in introductory physics texts, so it seemed probable that you weren't arguing from the same premise.


For a magnet passing through a coil, the rate of change of flux through the coil isn't constant in time.


The premise for my example, and the one given in #26, is that the coil is within a time-varying magnetic field, such that the rate of change of flux through the coil is constant in time. 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. Only then are we discussing equivalent systems.

But we can simplify as follows. Let's say the magnet passes through the coil such that we get a trapezoidal open circuit voltage Voc = 1.00 volt at the plateau, with a duration of 0.10 seconds. What is I the loop current? You mentioned "steady state", but can we assume that the system reaches steady state?

If the L/R time constant of the circuit is 0.010 seconds, then we can discuss the steady state current as settling to Voc/Rload. If Rload is 10 ohm, steady state current is 0.10 amp. Likewise, for any larger pulse wave duration, 1.0 sec, 10 sec, etc., the L/R time constant determines the transient response, but the steady state is Voc/Rload.

But if the open circuit voltage wave duration is 0.10 sec as above, but L/R time constant is 1.0 sec, then the system never reaches steady state. Any good text, Millman/Taub/Schilling "Pulse Digital & Switching Circuits" & "Digital Integrated Electronics" will show the detailed math. With 0.10 sec Voc at 1.0 volt, L/R, I will ramp up at a rate (Voc/L)*t. When the current is about 1/10th its "steady state value", the Voc is gone and current decays never having attained its steady state value.

With ac continuous excitation, the criteria is that R >> X_L, so that Iload = Voc/R. With a single pulse, the criteria for Iload = Voc/R, is as follows:

duration of Voc pulse >> L/R time constant.

Best regards.

Claude

 

[milesyoung]

But we can simplify as follows. Let's say the magnet passes through the coil such that we get a trapezoidal open circuit voltage Voc = 1.00 volt at the plateau, with a duration of 0.10 seconds. What is I the loop current? You mentioned "steady state", but can we assume that the system reaches steady state?

That's not simplifying, it's an entirely different scenario.

Again, the premise for my example, and the one given in #26, is that the coil is within a time-varying magnetic field, such that the rate of change of flux through the coil is constant in time. Regardless of how you go about producing such a field, that's what you have to consider. In that case, the steady-state current is the ratio of the open-circuit voltage to the resistance of the conducting loop.

But if the open circuit voltage wave duration is 0.10 sec as above, but L/R time constant is 1.0 sec, then the system never reaches steady state. Any good text, Millman/Taub/Schilling "Pulse Digital & Switching Circuits" & "Digital Integrated Electronics" will show the detailed math. With 0.10 sec Voc at 1.0 volt, L/R, I will ramp up at a rate (Voc/L)*t. When the current is about 1/10th its "steady state value", the Voc is gone and current decays never having attained its steady state value.

I'm well aware of what influence the time constant has in your example, but if you want to discuss a scenario where the rate of change of flux through the coil isn't constant in time, then:

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.

 

[technician]

The author assumed that X_L << R, so that would be correct only under that specific condition. The motor/generator texts specifically include X_L in the equivalent circuit and use this for calculations. I will post an illustration tonight, but I will start a new thread so we can elaborate w/o bothering this thread.

By the way, the Nelson & Parker is the book I looked at. Some of the contributions were made by college instructors, some were high school, and one contributor was a grammar school instructor. BR.

Claude

This is a highly respected textbook ( all of my textbooks are!) ...are you seriously questioning these books with an assertion that the authors are making some unqualified 'assumption' ?
Do you believe the answer is only correct within certain assumptions ?
I believe that you are confusing AC circuit analysis with physics explanations!

 

[cabraham]

Yes, I'm suggesting that this text makes assumptions, which in some cases are not unreasonable. It's like Galilean relativity in HS physics. If 2 cars approach each other w/ speeds v1 and v2, the relative speed is v1+v2. But in uni physics, if 2 space ships at speeds within an order of magnitude of light speed, Galilean approach is no good, and relativity applies.
My position is that sometimes L is too small to matter. Your approach works in that case. But in cases where L is large enough, a different result is obtained, your approach needs modified.

 

[technician]

No no no no no... I wondered when the speed of light was going to come into it !
None of this is 'MY' approach that needs modifying.
It is standard physics textbook references. The complications caused by the speed of light are dealt with appropriately in these books at the appropriate time.
I still believe that you are confusing circuit analysis where changing currents cause changing magnetic fields with the physics of changing magnetic fields causing induced emfs.
Your introduction of ω into your explanation confirms this for me.
Have you considered the problem from a conservation of energy approach?
Some tips... E = -d∅/dt ...I = E/R ... E = Blv ...F = BIl...power = i²R

 

[cabraham]

Cons of energy supports me. E =-dphi/dt must consider flux from external source as well as internal loop flux due to self inductance.
If E is constant regardless of R, then P = I^2*R which equals E^2/R. If E stays fixed, R decreasing results in unlimited increase in power. Decreasing R will increase power until R equals X, where max power occurs. Decreasing R belowthe value of X will decrease power. Current I, is determined by Voc/Z, where Z=R+jX. Otherwise energy conservationis violated. I'll draw a pic later. BR.
Claude

 

[milesyoung]

If E is constant regardless of R, then P = I^2*R which equals E^2/R. If E stays fixed, R decreasing results in unlimited increase in power.

Say I apply a constant voltage across a resistor with resistance R. Are you saying that, since I'm able to increase the power delivered to the resistor by decreasing R, I'm somehow violating the law of conservation of energy?

 

[cabraham]

If the voltage is induced via magnetic field, then there is a limited amount of power in the field. Reducing R will demand more of that power. When R = X, power in R is half that of mag field. Decreasing R results in less heating power in R. More later.
Claude

 

[milesyoung]

If the voltage is induced via magnetic field, then there is a limited amount of power in the field. When R = X, power in R is half that of mag field. Decreasing R results in less heating power in R.

Let me try another example:

I'm moving a bar magnet towards a closed conducting loop such that the the rate of change of flux through it is constant in time. I'm able to maintain this for a very long time, far longer than, say, 4-5 time constants of the equivalent RL circuit for the loop, so I have a very long period of time where the current in the loop is constant in magnitude. Let's call this the steady-state current.

Note that time-varying does not mean, for instance, that the magnitude of the magnetic field has to vary sinusoidally. In my example, the magnitude of the magnetic field could be increasing monotonically at a constant rate with respect to time.

Assume a rate of change of flux through the loop of 1 Wb/s, so the open-circuit voltage is 1 V. Say the resistance of the loop is 1e-3 ohm, so the power delivered to the conductor is 1000 W.

There's potential energy stored in the magnetic field produced by the steady-state current in the loop, but this has nothing to do with the power delivered to the conductor through the interaction of the induced magnetic field around the conductor and that of the bar magnet. It takes significant work, 1000 W worth to be precise, to maintain the motion of the bar magnet that produces the constant rate of change of flux through the loop. If the resistance of the loop was 1e-6 ohm, you'd have to supply 1e6 W mechanical power to be able to maintain the motion of the bar magnet that produces the constant rate of change of flux through the loop, and so on. There's nothing about this that violates the law of conservation of energy.

Now, please don't start with the time constants again. I know what you mean, but it has nothing to do with my example. I also know what complex, real and reactive power means, but again there's no reactive power involved in my example.

 

[cabraham]

Now you're talking motor theory. With motors or generators, to increase power requires more fuel to be spent. CEL (conservation of energy law) is ok as long as the mechanical power to generator shaft equals output electrical power plus losses. But mu point is that the added energy comes from somewhere. 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. If the load R is 10 ohm, I=0.10 amp, Voc = 1.0 volt, let's call these Iload1, Voc1, and R1. Let R2 = 0.001 ohm, but force on magnet and work increase by 10^4 to make up for loading effect.

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.

In generator design we call this condition a "load dump".

Claude

 

[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