Faraday's law on circular wire

[gralla55]

In my examples on Faraday's law in my book, they use a drawing of a magnet approaching a circular wire. The changing magnetic flux then induces an emf on the circle of wire, which in turn causes a current to flow.

I'm wondering if a current will flow in that wire without it having an element to provide resistance, or if it's just drawn that way for simplicity. Perhaps the internal resistance of the wire itself is good enough.

 

[cabraham]

Current will exist with or w/o R. A changing B results in a Lorentz force on the wire's free charge carriers, namely electrons. As the electrons move around the loop, if no R is present no lattice collisions take place and no photons are emitted. So the wire stays cool. If R is present, collisions result in electrons droppint from conduction band into valence band, a lower energy state. To conserve energy photons are emitted in the infrared region which we feel as heat. In a superconductor there is an induced current but no heat since no R is present.

The voltage around the loop will be related to the current per the inductive reactance value. In circuit theory lumped parameters are employed. In field theory distributed parameters must be considered because that is the real world.

 

[technician]

what would happen if there was a gap in the wire loop?

 

[cabraham]

what would happen if there was a gap in the wire loop?

You would have a small capacitance between the ends of the wire loop, resulting in a small displacement current. The induced voltage would be calculated per law of Faraday, and the current can be computed as the open circuit voltage divided by the loop total impedance. This impedance Zloop, is the wire resistance plus the inductance plus the capacitive reactance. Did I help?

Claude

 

[technician]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

 

[cabraham]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

R = ρl/A

C = ε₀A/d

L = μ₀N²A/l

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jX_L - jX_C + R

Claude

 

[technician]

R = ρl/A

C = ε₀A/d

L = μ₀N²A/l

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jX_L - jX_C + R

Claude

The formula you have given for inductance relates to a toroidal coil !
This question is about a wire loop.

 

[Andrew Mason]

How does what you say relate to a magnet approaching a wire loop?
I think impedance and reactance are associated with AC circuits and cannot see where they help here.
How would you calculate the impedance of a wire loop with a gap and use it to analyse the effect of an approaching magnet?

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

 

[cabraham]

The formula you have given for inductance relates to a toroidal coil !
This question is about a wire loop.

N=1! The wire cross section can be assumed rectangular, if round a correction factor is needed. The equation is valid for a solenoid, 1 to N turns.

Claude

 

[cabraham]

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

The current will exist longer than a "very brief time", but rather continuously. If open circuit voltage Voc is 10 volts, with R of 10 kohm, X_C of -j1.0 Mohm, and X_L of +j10 kohm, then the current I is given by

10 / (10k - j1.0M + j10k) = 10.100 microamp with angle 89.42 degree.

Claude

 

[technician]

Inductive and capacitive reactance are associated with any time dependent emf. If there is some small capacitance in the loop some small current will flow for a very brief time due to the fact that the induced emf around the wire loop will not be constant. I think that is all cabraham is saying.

AM

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

 

[cabraham]

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

R = ρl/A_wire

C = ε0A_wire/d, where "d" is the space between the ends of the wire

L = μ0N²A_loop/l, where "l' is the length of the solenoid, or height of a single turn loop.

V(open circuit) = -Nd(phi)/dt
I = Voc/Z
Z = jXL - jXC + R

I hope this helps.

Claude

 

[technician]

height of a single turn loop
do you mean diameter of the loop?

 

[cabraham]

height of a single turn loop
do you mean diameter of the loop?

Diameter of wire if round, times adjustment factor. Or, for a rectangular wire, height of wire, which is thickness in direction normal to loop plane.

Claude

 

[tech99]

I assume the magnet is on the polar axis of the loop, as the effect will vary depending on geometry. Classically, we would say that as the magnet approaches the loop, field lines cut the loop and induce an EMF. If there is a "conducting" path around the loop the electrons can move and a current will flow. Post Maxwell, we say that the field of the moving magnet creates an electric field, and this encompasses the conductor of the loop, creating an EMF round the loop. Then, if a conducting path exists around the loop, a current can flow.
When the electrons start to move they are accelerating, so they will radiate an EM wave.
The radiated wave will have E and B fields. In addition, the magnetic field associated with the movement of electrons will constitute the magnetic part of the antenna induction near field. The electric field driving the electrons will constitute the electric part of the antenna induction near field. The energy lost to radiation will be permanently taken from the kinetic energy of the magnet. The reaction on the magnet of the induction fields will constitute stored energy which can be given back to the magnet.

 

[technician]

Diameter of wire if round, times adjustment factor. Or, for a rectangular wire, height of wire, which is thickness in direction normal to loop plane.

Claude

Ok... I assumed a wire of cross sectional area 1mm² with a gap of 1mm.
This wire forms a loop of rea 1m²
I used C =ε₀A/d and got C = 8.86x10^-12F

I took the area of the loop formed by the wire to be 1m²and the wire to be square 1mm x 1mm (l = 1mm in the equation for L)
This gives me L = 4∏x10^-7x10^-3 H = 4πx10^-10H

How do you use these values of C and L to calculate X_c and X_L?

 

[tech99]

The problem you have is that the EMF induced in the wire in your case is a complex function containing many frequencies, rather than, say, a sine wave which it would be for a loop rotating in a uniform field. In the rotating case, you find the reactance at the rotation frequncy and calculate the current from Ohm's Law. The total reactance of the loop will be Xl - Xc. You find reactances as follows: Xl = 2 pi F L and Xc = 1/2 pi F C.

 

[cabraham]

Of course we should not forget to include loop resistance R as well. The capacitive & inductive reactances are per tech99's post above. R + jX_L - jX_C = Z_loop.

Claude

 

[tech99]

Hi Cabraham
I was trying to keep it simple, as resistance is small compared with reactance. But if you want to use total impedance you have to add reactance and resistance vectorially,
so Z = sqrt (R^2 + X^2)

 

[technician]

Of course we should not forget to include loop resistance R as well. The capacitive & inductive reactances are per tech99's post above. R + X_L - X_C = Z_loop.

Claude

You see the values I got for C and L...do these look reasonable?...how would you calculate X_land X_c so that I can take this further... I am OK with R
Edit. I do not see any calculation of XL or Xc !

 

[Andrew Mason]

These calculations look interesting ! is A the area enclosed by the loop (I assume this !) or the cross sectional area of the wire making up the loop?
For the toroidal coil A is the cross sectional area of the coil...not the area enclosed by the coil !

A is the area enclosed by the coil and any line between the ends of the coil.

AM

 

[technician]

Ok... I assumed a wire of cross sectional area 1mm² with a gap of 1mm.
This wire forms a loop of rea 1m²
I used C =ε₀A/d and got C = 8.86x10^-12F

I took the area of the loop formed by the wire to be 1m²and the wire to be square 1mm x 1mm (l = 1mm in the equation for L)
This gives me L = 4∏x10^-7/10^-3 H = 4πx10^-4H

How do you use these values of C and L to calculate X_c and X_L?

These are the values I got for C and L...do they look reasonable?
How are these used to find X_cand X_l

 

[cabraham]

X_C = -1/Cω

X_L = Lω

ω = 2∏f

Claude

 

[technician]

X_C = -1/Cω

X_L = Lω

ω = 2∏f

Claude

What is ω in the context of this post?
I recognise ω in the context of AC theory.

 

[cabraham]

The "ω" is radian frequency in radian/second. If the induced emf is a special function like rectangular, sawtooth, etc., then it can be decomposed via Fourier series. If it is a transient event, a Fourier transform will give a continuous spectrum of frequencies.

For such a case, it may be easier to just write a 2nd order differential equation and solve in the time domain. An interesting exercise it is.

Claude

 

[technician]

The original post relates to a magnet approaching a circular coil.
If the coil is a single loop having a resistance of 1ohm and the area of the loop is 1 square metre and it is experiencing a changing flux of 1tesla/sec is the induced emf simply 1volt and the induced current simply 1Amp?
If there is a gap in the wire the emf will still be 1 volt but the current will be zero?

 

[cabraham]

The original post relates to a magnet approaching a circular coil.
If the coil is a single loop having a resistance of 1ohm and the area of the loop is 1 square metre and it is experiencing a changing flux of 1tesla/sec is the induced emf simply 1volt and the induced current simply 1Amp?
If there is a gap in the wire the emf will still be 1 volt but the current will be zero?

Not necessarily. The loop has inductance as well as resistance. We have been discussing this. Please review the whole thread. If the resistance is 1.0 ohm, but the inductive reactance is close to or greater than 1.0 ohm, the current will be less than 1.0 amp. The inductive reactance is the "X_L" mentioned above. Please review.


Claude

 

[technician]

My answer in post 26 is correct! This is an elementary Faraday's law calculation as seen in textbooks. (e = -d∅/dt)
I have followed the previous posts closely and got as far as calculating L and C for a loop of wire with a gap (quoted in post22) but then reached a dead end because of some uncertainty about ω.
Are you sure that you are not confusing the original post with AC anaysis?

 

[cabraham]

My answer in post 26 is correct! This is an elementary Faraday's law calculation as seen in textbooks. (e = -d∅/dt)
I have followed the previous posts closely and got as far as calculating L and C for a loop of wire with a gap (quoted in post22) but then reached a dead end because of some uncertainty about ω.
Are you sure that you are not confusing the original post with AC anaysis?

Even when ω is not clear, L & C enter the picture. The differential equation is:

Voc = L*di/dt + R*I + (1/C)*integral (i dt)

Solving the diff eqtn for I(t) produces the result needed. The inductive & capacitive reactances cannot be neglected simply because ω is not obvious. To simply divide Voc by "R" to get the current is wrong. Using that equation, as R went down, I would increase w/o limit.

A Voc of 1.0 V, with R = 1.0 ohm, does not always give I = 1.0 A. That would be 1.0 watt of power. If the power moving the magnet through the coil is less than 1.0 W, then the electric power cannot be 1.0 W. Energy conservation is immutable. Internal loop value of L means that as soon as current is set up, a counter-emf is present. So the loaded or closed loop voltage must always be less than open circuit value.

Claude

 

[technician]

My answer is correct. This is a standard textbook elementary example and exam question !
I think you are confused with AC circuit theory and the application of Faraday's law.
The induced emf comes about as a result of a changing magnetic flux linkage and I think you have taken this induced emf as a voltage applied to an L, C, R circuit.
I have not seen anything like your analysis in any of my textbooks and there is plenty of experimental evidence that the values in my calculation are correct.
Can you give me a reference where your explanation, especially the equation, is published so that I can check it?
I can't believe that all of my textbooks are mistaken !

 

[cabraham]

I'm confused? Dude who are you? How far did you get in EE, BS, MS, Ph.D.? Honestly I don't believe what you're saying. If the emf open circuit is 1.0 V, are you claiming that to compute current we ignore L & C and just divide by R? Every text in the world uses the RLC diff eq for the solution. That is my "reference". Are you full EE student/grad, or are you EET? What texts are you relying on?

 

[technician]

I'm confused? Dude who are you? How far did you get in EE, BS, MS, Ph.D.? Honestly I don't believe what you're saying. If the emf open circuit is 1.0 V, are you claiming that to compute current we ignore L & C and just divide by R? Every text in the world uses the RLC diff eq for the solution. That is my "reference". Are you full EE student/grad, or are you EET? What texts are you relying on?

It should not matter who I am. This is an elementary A-level question so nothing more than A-level is needed to get the answer.
If you cannot agree that, using the values quoted above, emf = 1 volt and current = 1Amp then you are in serious disagreement with A-level teaching and A-level textbooks.
Whatever you do...do not criticize teachers or the textbooks they use! If the answer is correct at A-level...it is correct.
If you do not agree with the calculation then I urge you to give an answer to this question. There are countless examples in a whole range of physics textbooks.

 

[technician]

I'm confused? Dude who are you? How far did you get in EE, BS, MS, Ph.D.? Honestly I don't believe what you're saying. If the emf open circuit is 1.0 V, are you claiming that to compute current we ignore L & C and just divide by R? Every text in the world uses the RLC diff eq for the solution. That is my "reference". Are you full EE student/grad, or are you EET? What texts are you relying on?

Do you accept that explanations and answers to questions should conform to textbooks used in teaching?
The question I posed, and I believe the original post refers to, is a standard A-level type question with a very straightforward answer given in all A level textbooks.

 

[cabraham]

Do you accept that explanations and answers to questions should conform to textbooks used in teaching?
The question I posed, and I believe the original post refers to, is a standard A-level type question with a very straightforward answer given in all A level textbooks.

The equation v = -N*d∅/dt is a broad statement. What is "∅"? It is not only the external flus linking the loop, but the loop's own internal flux as well. Look up "inductance" and see for yourself.

As long as the loop has L, which it must have, a portion of the total flux inside the loop area will be due to its own self inductance L. This affects "v". Otherwise conservation of energy is violated. If I = Voc/R, with no regard to L & C, you're saying that a copper loop with heavy wire, 0.001 ohm R value, with Voc - 1.0V, will result in 1000 amp??!? I don't think so. If a superconductor is used, what is the current technician?

Your references do not go into the depth of detail needed for these questions. Any EE w/ a motor theory background knows what I'm saying.

Claude

 

[technician]

The equation v = -N*d∅/dt is a broad statement. What is "∅"? It is not only the external flus linking the loop, but the loop's own internal flux as well. Look up "inductance" and see for yourself.

As long as the loop has L, which it must have, a portion of the total flux inside the loop area will be due to its own self inductance L. This affects "v". Otherwise conservation of energy is violated. If I = Voc/R, with no regard to L & C, you're saying that a copper loop with heavy wire, 0.001 ohm R value, with Voc - 1.0V, will result in 1000 amp??!? I don't think so. If a superconductor is used, what is the current technician?

Your references do not go into the depth of detail needed for these questions. Any EE w/ a motor theory background knows what I'm saying.

Claude

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

 

[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.