Faraday's law on circular wire

[cabraham]

You are wrong.
I have no more to add...going round in circles

If I am wrong, please inform us where I went wrong. I know it's been 7 weeks, but it takes more than just telling someone they are wrong to prove them wrong. Please examine the following and state where you disagree.

A wire loop immersed in a time-varying magnetic flux, ∅, will incur induction. The loop has a resistance R, and an inductance L, and if open, the gap will determine a capacitance C. Each quantity, R, L, and C, will contribute to the loop total impedance value Z.

Z = R + jX_L + 1/jX_C.

When open circuited, the voltage is measured and denoted as "V_oc". When short circuited, current is measured and denoted as "I_sc". Unless the gap is very small, the area of the wires is generally too small to result in substantial capacitance. At low and medium frequencies, "X_C" can usually be neglected.

So to simplify things, we can approximate loop impedance Z as the following:

Z = R + jX_L. Of course, X_L = ωL. So then:

Z = R + jωL, which is expressed in rectangular (Cartesian) form. To express Z in polar form:

|Z| = √R² + (Lω)², angle Z = arctan (Lω/R).

When loop is open, Z is infinite due to gap in wire so that V = V_oc, I = 0. When a loading resistance R_load, is placed across the gap in the loop, what is the measured V & I at the load?

There is a voltage divider here. If R_load >> R_wire, we still have to deal with X_L = Lω.

Ignoring R_wire, V_load is as follows:

|V_load| = |V_oc|*R_load/√((R_load)² + (Lω)²)

The above gives the magnitude of voltage at load in terms of open circuit voltage, R, ω, and L. Since Lω is X_L, if R_load >> X_L, then the radical in the denominator is approximately equal to R_load. So we have V_load = V_oc*R_load/R_load, or V_oc = V_load.

For R_load >> Lω, we can assume that V_load will remain nearly equal to V_oc for any R_load value >> Lω. Also, I_load = V_oc/R_load, as long as R_load >> Lω.

Any motor/generator text will affirm this. Induction motors with squirrel cage rotors discuss this analysis. Look up "deep bar rotors" or "double squirrel cage rotors". These utilize the relation between R and Lω to optimize starting torque and run torque. At standstill, ω = ω_line (ac line frequency i.e. 2∏*50/60 Hz), so X_L = maximum, so the impedance is computed taking both R and X_L into account.

When running ω is a small fraction of ω_line, the "slip" frequency, typically 2% to 5% of line frequency. So here X_L is very low, much less than R. Hence R dominates the impedance of the rotor.

The following texts affirm what I've posted:

Electromagnetic and electromechanical machinery, Leander Matsch

Electric Machinery, Fitzgerald, Kingsley, Umans.

I will elaborate if desired. Best regards to all.

Claude

 

[technician]

If I am wrong, please inform us where I went wrong. I know it's been 7 weeks, but it takes more than just telling someone they are wrong to prove them wrong. Please examine the following and state where you disagree.

A wire loop immersed in a time-varying magnetic flux, ∅, will incur induction. The loop has a resistance R, and an inductance L, and if open, the gap will determine a capacitance C. Each quantity, R, L, and C, will contribute to the loop total impedance value Z.

Z = R + jX_L + 1/jX_C.

When open circuited, the voltage is measured and denoted as "V_oc". When short circuited, current is measured and denoted as "I_sc". Unless the gap is very small, the area of the wires is generally too small to result in substantial capacitance. At low and medium frequencies, "X_C" can usually be neglected.

So to simplify things, we can approximate loop impedance Z as the following:

Z = R + jX_L. Of course, X_L = ωL. So then:

Z = R + jωL, which is expressed in rectangular (Cartesian) form. To express Z in polar form:

|Z| = √R² + (Lω)², angle Z = arctan (Lω/R).

When loop is open, Z is infinite due to gap in wire so that V = V_oc, I = 0. When a loading resistance R_load, is placed across the gap in the loop, what is the measured V & I at the load?

There is a voltage divider here. If R_load >> R_wire, we still have to deal with X_L = Lω.

Ignoring R_wire, V_load is as follows:

|V_load| = |V_oc|*R_load/√((R_load)² + (Lω)²)

The above gives the magnitude of voltage at load in terms of open circuit voltage, R, ω, and L. Since Lω is X_L, if R_load >> X_L, then the radical in the denominator is approximately equal to R_load. So we have V_load = V_oc*R_load/R_load, or V_oc = V_load.

For R_load >> Lω, we can assume that V_load will remain nearly equal to V_oc for any R_load value >> Lω. Also, I_load = V_oc/R_load, as long as R_load >> Lω.

Any motor/generator text will affirm this. Induction motors with squirrel cage rotors discuss this analysis. Look up "deep bar rotors" or "double squirrel cage rotors". These utilize the relation between R and Lω to optimize starting torque and run torque. At standstill, ω = ω_line (ac line frequency i.e. 2∏*50/60 Hz), so X_L = maximum, so the impedance is computed taking both R and X_L into account.

When running ω is a small fraction of ω_line, the "slip" frequency, typically 2% to 5% of line frequency. So here X_L is very low, much less than R. Hence R dominates the impedance of the rotor.

The following texts affirm what I've posted:

Electromagnetic and electromechanical machinery, Leander Matsch

Electric Machinery, Fitzgerald, Kingsley, Umans.

I will elaborate if desired. Best regards to all.

Claude

Did you read my post26 ?
Do you agree with this basic calculation seen in many standard textbooks?

 

[cabraham]

Did you read my post26 ?
Do you agree with this basic calculation seen in many standard textbooks?

Yes I read #26. What standard textbooks are you referring to? In your example, you simply assume that the loop impedance is defined entirely by R, w/o regard to what L or ω are. This cannot withstand scrutiny. As I've already stated repeatedly, your method will work but only with the condition that R is higher in value than X_L by a factor of 5 or more. Let's look at your example.

If dB/dt is 1.0 tesla/sec, A_loop = 1.0 sq meter, and R = 1.0 Ω, how do we compute I and V? Let's assume square wire, because it is easier to compute, but round wire could be used w/ an adjustment factor, but for now square will do. Also we use a circular loop. Let the wire thickness be 1.0 mm. Since area = 1.0 meter², computing L is easy.

L = μ₀N²A_loop/h, where "h" is the height of the loop, or 1.0 mm here (wire thickness), and "N" is one, the number of turns.

L = (4∏10^-7henry/meter)(1 turn)²(1.0 m²)/(10^-3 m),

L = 1.2566 millihenry. So that at a frequency of 1.0 radian/second, X_L = 0.0012566 ohm. Because this value is much less than 1.0 ohm R value, for all practical purposes, ignoring X_L and using R only is close enough. Hence I is very close to 1.00 amp.

But if ω = 1,000 rad/sec, and flux amplitude is correspondingly reduced by a factor of 1,000, the X_L value is now 1.2566 ohm, which is greater than the R value of 1.00 ohm. Thus the total loop impedance magnitude |Z| is √(1.00)² + (1.2566)². |Z| = 1.6060 ohm. Hence |I| = 1.00/1.6060 = 0.6227 amp.

The load voltage is |V| = 0.6227 volt. As you can see, there's more to it than just assuming that the entire loop impedance consists only of R. When R attains a low value, the inductance of the loop comes into play. The example you gave just happens to work out w/o considering X_L because the radian frequency ω is extremely small, 1.0 radian/second.

But in cases of induction, that is a very low frequency, around 0.159 Hz. As I showed, if ω is 1,000 rad/sec, only 159 Hz, then X_L cannot be ignored. In fact at 300 rad/sec (47.75 Hz), we must consider X_L. Power line frequencies of 50/60 Hz are high enough so that you cannot ignore loop inductance.

I hope this example serves to illustrate that there is usually more than 1 thing going on in a problem like this. Faraday's law simply conveys a math relation between NET flux and emf. But the "∅" in the equation is not merely the external flux cutting the loop, it is the composite of external plus internal flux due to loop current and its own self inductance.

The problem with your approach is that if the loop is very low R value, say 0.001 ohm, the answer you get for I is 1,000 amp! Clearly this is not the case. With 0.001 ohm R value, and 0.0012566 X_L value, the I value magnitude is 622.7 amp. So the 1,000 amp value is high by 61%.

Something to ponder. I will elaborate if desired. Best regards to all.

Claude

 

[technician]

I am a teacher...the number of textbooks that I could refer to is countless! The ones I use that you may find helpful are Nelkon & Parker, Duncan, Breithaupt, Young & Friedman and AQA exam textbooks.
Any answer/explanation I have provided is to be found in any of these. I also use the Hyperphysics web site. I am ONLY interested in the original post... a permant magnet and a circular coil. I think that has been covered.
I am not interested in sqirrel cage motors (this post has more to do with generators than motors)
I do not see where your dB/dt = 1 tesla/sec and a 'frequency' (?) of 1 rad/sec and then an ω = 1000rad/sec come into the explanation.
Sorry... I stick with my textbooks. If you want more discussion on this topic I suggest you start a fresh post (I will probably not contribute) because this is now nothing to do with the original post, in my opinion.

 

[cabraham]

Circuit theory is a well researched science, with little new in the last century. Impedance being the phasor sum of resistance and reactance cannot be disputed. You have not addressed my R + jX explanation. I will research the books you mentioned, but frankly, I am wondering if they actually say what you claim they say. I will find out and comment. BR.

Claude

 

[cabraham]

I would ask technician to seek a credible authority on e/m field theory and ask them to affirm or refute my statements as follows.

A loop with a gap having negligible capacitance is placed in a time varying mag field. The open circuit voltage Voc is measured. The gap is then connected across a resistance R. What is the current? Claude says that I is the ratio of Voc to Zloop. And that Zloop = √R² +X_L².

Why is this wrong? Technician, your answer is that as R decreases, I and power P increase unbounded. This flies in the face of energy conservation. Decreasing R does result in increased I only until R is much less than X_L. The maximum loop current, I, is Voc/X_L. Decreasing R to 0 results in this maximum current value.

The power in R is I²R, or Voc²/R. The mag field has limited power, so dissipation in R is limited. My result affirms this well known fact.

Later I will construct a paper detailing this. BR.

Claude

 

[technician]

here is a worked example from Young and Freedman (it is a very common example!)
If anything is wrong with this example then there is something seriously wrong with standard textbooks !
I, personally, am not going to reproduce standard textbooks here...they are available for all to refer to.
This answers the original post.

 

[milesyoung]

I would ask technician to seek a credible authority on e/m field theory and ask them to affirm or refute my statements as follows.

A loop with a gap having negligible capacitance is placed in a time varying mag field. The open circuit voltage Voc is measured. The gap is then connected across a resistance R. What is the current? Claude says that I is the ratio of Voc to Zloop. And that Zloop = √R² +X_L².

I think you should consider that technician might be talking about the steady-state solution for the induced current when the rate of change of flux through the loop is constant in time. The current is then indeed 1 A for a rate of change of flux of 1 Wb/s through a closed loop with a resistance of 1 ohm, regardless of its reactance.

Why is this wrong? Technician, your answer is that as R decreases, I and power P increase unbounded. This flies in the face of energy conservation.

It does not. As the induced current increases, so does the magnitude of the induced magnetic field. This in turn increases the force you'd have to apply to the magnet to produce whatever displacement profile that's needed to induce the constant rate of change of flux through the loop. Decreasing the resistance of the loop increases the mechanical power supplied, as it must, to make up for the increase in Joule heating in the conductor.

 

[cabraham]

here is a worked example from Young and Freedman (it is a very common example!)
If anything is wrong with this example then there is something seriously wrong with standard textbooks !
I, personally, am not going to reproduce standard textbooks here...they are available for all to refer to.
This answers the original post.

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

 

[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. BR.

Claude

I look forward to your new thread. I am happy that this one is finished.
'THE AUTHOR' = all authors
Ps the induced emf is sometimes called a 'back emf' and the induced current ( if there is one) is sometimes called an eddy current.
It is worth reading about these, their relative directions (Lenz's law) and energy conservation

 

[cabraham]

I think you should consider that technician might be talking about the steady-state solution for the induced current when the rate of change of flux through the loop is constant in time. The current is then indeed 1 A for a rate of change of flux of 1 Wb/s through a closed loop with a resistance of 1 ohm, regardless of its reactance.


It does not. As the induced current increases, so does the magnitude of the induced magnetic field. This in turn increases the force you'd have to apply to the magnet to produce whatever displacement profile that's needed to induce the constant rate of change of flux through the loop. Decreasing the resistance of the loop increases the mechanical power supplied, as it must, to make up for the increase in Joule heating in the conductor.

That is where we disagree. Reactance matters. Regarding increasing the force, you have now changed the op problem. Two identical loops are examined, one has 10 ohms, one has 1.0 ohm. If both are open circuited, the induced voltage, Voc, is the same for both.

If both are loaded w/ 10/1.0 ohm load resistances, depending on reactance, the currents are not always simply Voc/Rload. Regarding your statement about increasing force, that would also increase Voc. In order to make up for increased induced mag field you are increasing force. But that in itself increases the open circuit emf Voc. The problem has changed.

I will begin a new thread tonight, and we can take this further. I will conclude with this. To increase the force will inevitably result in a new value of Voc, hence the problem is not as simple as I = Voc/Rload. That is my point. With original force value, we get emf (open circuit) = Voc1. If R << X_L, so that reactance cannot be ignored, and we increase force to assure that we restore original net flux value, we now must compute current using a new value of Voc, namely Voc2.

With this increased force, if load is suddenly removed, Voc2 > Voc1. That is inevitable. Thanks.

Claude

 

[milesyoung]

Let's just first agree that we're considering an example where you have a conducting loop, fixed in space, within a time-varying magnetic field, such that the rate of change of flux through the loop is constant in time. Assume a rate of change of flux of 1 Wb/s.

If both are open circuited, the induced voltage, Voc, is the same for both.

That's fine. The induced emf would be 1 V.

If both are loaded w/ 10/1.0 ohm load resistances, depending on reactance, the currents are not always simply Voc/Rload.

In the case of a closed loop with a resistance of 1 ohm, the steady-state current will be 1 A.

In the case of a closed loop with a resistance of 10 ohm, the steady-state current will be 0.1 A.

You could just as well consider the step reponse of a RL-circuit to a constant voltage. Its steady-state current is certainly not a function of the reactance of the circuit.

Regarding your statement about increasing force, that would also increase Voc.

It would not. You're displacing the magnet such that the rate of change of flux through the loop is 1 Wb/s. That determines the open-circuit voltage. The motion of the magnet, regardless of the resistance of the loop, is the same. What changes is the mechanical power you must supply to produce this motion. It's an effect of reducing the resistance of the loop.

To increase the force will inevitably result in a new value of Voc, hence the problem is not as simple as I = Voc/Rload. That is my point. With original force value, we get emf (open circuit) = Voc1. If R << XL, so that reactance cannot be ignored, and we increase force to assure that we restore original net flux value, we now must compute current using a new value of Voc, namely Voc2.

I really can't make sense of any of this. The reactance has nothing to do with the steady-state current in the circuit.

 

[technician]

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.

It is a great book, the bible of A-level physics...did you find that it is more or less the same as Young and Freedman in its explanation?
I prefer Duncan. Have a look at that one.
PS... I am a grammar school 'instructor'

 

[cabraham]

Let's just first agree that we're considering an example where you have a conducting loop, fixed in space, within a time-varying magnetic field, such that the rate of change of flux through the loop is constant in time. Assume a rate of change of flux of 1 Wb/s.


That's fine. The induced emf would be 1 V.


In the case of a closed loop with a resistance of 1 ohm, the steady-state current will be 1 A.

In the case of a closed loop with a resistance of 10 ohm, the steady-state current will be 0.1 A.

You could just as well consider the step reponse of a RL-circuit to a constant voltage. Its steady-state current is certainly not a function of the reactance of the circuit.


It would not. You're displacing the magnet such that the rate of change of flux through the loop is 1 Wb/s. That determines the open-circuit voltage. The motion of the magnet, regardless of the resistance of the loop, is the same. What changes is the mechanical power you must supply to produce this motion. It's an effect of reducing the resistance of the loop.


I really can't make sense of any of this. The reactance has nothing to do with the steady-state current in the circuit.

Well of course the "steady state" current is zero regardless of R as well as L. If we're talking a magnet passing through a coil, then "steady state" is zerp period. The transient is all that happens. I will start a new thread based on a continuous time varying field, such as a generator, xfmr, antenna, etc.

FWIW, I don't think there is any dispute that Voc is determined by the change rate of flux of 1.0 web/sec. Here is my beef. A rate of "1.0 weber/second", can be d∅ = 0.1 web, dt = 0.1 sec; or d∅ = 0.01 web, dt = 0.01 sec; or d∅ = 0.001 web, dt = 0.0001 sec, etc., so that the rate is always 1.0 web/sec.

Here is my difference in thinking. If d∅ and dt are very large, the effective time constant is very large, and the L value has less influence. But if d∅ and dt are both very small, like 1.0 microweber per microsecond, then the time constant is very short, thus L has a significant influence.

If the L/R time constant of the loop is 1.0 millisecond, do you agree that the results are different for a 1.0 second time constant for the magnet crossing the loop as opposed to a 1.0 microsecond time constant. If the L/R time constant is smaller than the magnet motion time constant, you and I likely agree that L has little influence, and only R need be considered. We likely agree there.

Where we likely DISagree is the following. If the magnet moving through the coil takes place in 0.1 seconds, but the L/R time constant is 1.0 seconds, 10 times longer, we cannot ignore the influence of L. Do you follow? Thanks.

Claude

 

[milesyoung]

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.

Well of course the "steady state" current is zero regardless of R as well as L. If we're talking a magnet passing through a coil, then "steady state" is zerp period. The transient is all that happens.

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

If the L/R time constant of the loop is 1.0 millisecond, do you agree that the results are different for a 1.0 second time constant for the magnet crossing the loop as opposed to a 1.0 microsecond time constant?

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.

 

[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