Is MIT Prof. Lewin wrong about Kirchhoff's law?


by sarumonkee
Tags: kirchhoff faraday
hikaru1221
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#325
Apr5-11, 12:14 PM
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Quote Quote by yungman View Post
The static E is small
Please prove this. I never know that the static E field is really that small.
Many texts ignore this fact I believe.
yungman
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Apr5-11, 01:27 PM
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Quote Quote by hikaru1221 View Post
Please prove this. I never know that the static E field is really that small.
Many texts ignore this fact I believe.
[tex]\vec E=\frac{\vec J}{\sigma} \approx \frac {\vec J}{5\times 10^7}[/tex]

It is approximation only. I did not border to look up conductivity of copper.

I am refering to induced E when current change.
hikaru1221
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Apr5-11, 01:50 PM
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The E in that equation is the total E field. You are not proving anything about the static E field in particular.

There is a model to prove that equation (E = J/sigma), and it starts with the force of the total E field (aside from the damping force) that exerts on electron. Google for Drude model, you will see it.

I also started with that equation (E = J / 10^7 or E = J / something very large) to prove that static E field cancels out induced E field. Static E field is never that small to be neglected in any case. It is the total E field that is small.
yungman
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Apr5-11, 02:01 PM
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Quote Quote by hikaru1221 View Post
The E in that equation is the total E field. You are not proving anything about the static E field in particular.

There is a model to prove that equation (E = J/sigma), and it starts with the force of the total E field (aside from the damping force) that exerts on electron. Google for Drude model, you will see it.

I also started with that equation (E = J / 10^7 or E = J / something very large) to prove that static E field cancels out induced E field. Static E field is never that small to be neglected in any case. It is the total E field that is small.
I don't know exactly what static E you refer to, In this thread, static E is the longitudinal field of a wire developed by the voltage drop when current pass through. Other than that, I am really not interested in the detail definition.

My only interest here on this thread is how does electrodynamics accomodate the induced voltage. THis is the 700lb gorilla here. Maybe I should start a new thread as most people avoid this thread!!
hikaru1221
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Apr5-11, 02:19 PM
P: 802
static E is the longitudinal field of a wire developed by the voltage drop when current pass through.
No, it isn't. It's not about the definition either. It is due to charge cummulation. It happens all the time, even in the simplest circuit as a resistor connected to a DC power supply.

The way I see it, this static E field is the way to solve your gorilla issue. This is how the circuit and everything reacts to that induced voltage/ induced E field: it builds up static E field and, if inductance of the circuit is significant, its own B field and induced E field. The static E field is built in accordance with the intrinsic characteristics of everything involved (resistors, wire). The current is then built up in accordance with the total E field, by J = sigma*E.
cabraham
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Apr5-11, 05:45 PM
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Quote Quote by yungman View Post
I don't know exactly what static E you refer to, In this thread, static E is the longitudinal field of a wire developed by the voltage drop when current pass through. Other than that, I am really not interested in the detail definition.

My only interest here on this thread is how does electrodynamics accomodate the induced voltage. THis is the 700lb gorilla here. Maybe I should start a new thread as most people avoid this thread!!
Faraday: v = -N*d(phi)/dt.
Ohm: J = sigma*E.
Lorentz: F = q*(E + uXB).

The induced voltage is described in Faraday's Law, FL. But we must be careful. A time changing mag flux is related to the emf (voltage) induced in the loop per FL. But the flux "phi" is a net flux, not the external flux exciting the loop.

When the time varying, herein "ac", mag flux phi, excited the loop, there is charge motion per Lorentz force law, LFL. Free electrons in the wire are moved in a direction determined by E & B. E acts tangentially, B acts normally. Hence a rotational field condition is present & the electrons circulate around the loop. But hold on. The moving e- constitute a current, which produces another mag field. If the external mag flux density is called "Be", & internal is "Bi", then Bnet = B = Be + Bi. Of course, the law of Lenz, LL, tells us that the induced or internal B field opposes the external B in polarity.

In addition, we have another thing going on. As e- transit through the wire, they crash into lattice particles & lose some energy. This energy is radiated in the form of photon emission. It is around 5 to 7 micron in wavelength, & is felt as heat in the infrared region. Charges accumulate due to said collisions, & these charges have their own associated E field, since e- carry E fields due to their own charge.

This accumulation of e- charge gives rise to an ir-rotational (conservative) E field. This E field has no curl & is not an emf, but a drop. It is a polarizing type of force, its curl is zero, as it can not drive electrons around the closed loop.

So what is the voltage? How is it determined? The voltage V, is simply the line integral along a particular path of E*dl. But E has 3 components, the induced E field due to external B, the E field due to the current in the wire & its ac B field, Bi, & the static ir-rotational E field due to charge accumulation incurred via electron collisions w/ the lattice structure.

The voltage from a to b, is the line integral of the composite E fields along a chosen path. Inside the Cu wire, what is the voltage value, Vcu? We are measuring voltage from terminals a & b, w/ the path as the inside of the Cu winding.

The external ac mag flux density Be, gives rise to an induced E field, Ee, such that electrons in the Cu wire are moved along the wire, w/ said Be acting normal. This Ee is rotational.

But the induced current, Iloop, has its own B field, ac, we will call Bi. Bi is an ac field, & it has the opposite polarity of Be. It induces an E field & voltage or emf in the loop. This is the self-inductance of the loop. This is Ei, also rotational. For a very low loop resistance, the Bi cancels the Be almost entirely. What remains is the E field due to charges accumulating due to collisions between electrons & lattice structure. Call this one Ec. as it is due to charged particles.

Hence Enet = E = Ee + Ei + Ec. But Ee & Ei nearly cancel perfectly inside a low resistance conductor. If the loop were open, Ei tends toward zero, & the loop voltage is maximum due to no cancelling between Ee & Ei. Closing the loop via a resistive load results in current & a counter-balancing E & B fields, Ei & Bi. So inside the copper we still have Ec. If the entire loop was very low resistance so that Ee & Ei nearly cancel entirely, we still have Ec.

The Cu sec winding in my example is 0.1 ohm. The 10A current times the 0.1 ohm results in the voltage drop of 1.0V. The line integral of the composite E field along the Cu path results in Ee & Ei almost cancelling, & Ec*dl giving us around 1.0V. In other words, as soon as the ext mag field, Be, enters the Cu wire, current is induced. The induced current has a strong B field, Bi, that cancels the external, & equilibrium is reached.

You seem to be looking for the 120V induced emf inside the copper in distributed form. But don't forget that there is a counter-emf generated as well. The distributed emf sources are nearly perfectly cancelled by the distributed counter-emf sources. But the Ec component does not get cancelled. It accounts for the 1.0V drop inside the Cu.

To better visualize this, consider a low resistance loop of wire, 0.010 ohm, closed & immersed in an ac mag field, Be. The Lorentz force moves the free electrons in the wire. This is current. But the current generates its own mag field, Bi. The law of Lenz tells us that they oppose each other. If the current is 1.0A, w/ a 0.010 ohm loop resistance, the voltage around the loop is 0.010 volt.

But we now open the loop, keeping the flux & area the same. The current plummets to near zero, but the voltage increases to 10V. Here, the ac mag flux produces an induced emf of 10V in the open state. When the loop closes, the net loop voltage is a mere 0.010V.

Why the difference? Of course, it is the cancellation. The external & internal parts of B & E account for the drastic difference in voltage between high & low impedance conditions. With a high-Z loop, the external B field is unopposed. Without induced current, the E field is due to the external B field, & the full voltage is realized since there is no loop current to cancel it.

When the loop resistance is low, the current generates cancellation of the external fields. The voltage is the line integral of all 3 phenomena. How can the Cu wire have just 1.0V, when the heater load has 119V, when their paths start & end at the same 2 points, a & b? They are in parallel, yet differing voltages are found.

Why? Answer is cancellation, & charge accumulation due to differing resistance values. I believe I've made my case, but someone other than me should affirm.

Claude
yungman
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#331
Apr5-11, 07:30 PM
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Thanks both of you, there is a lot of info here. I have to print them out, read first and think about all that. I have my own thing to study at the moment and can't put undivided attention to this right at the moment.
yungman
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Apr6-11, 02:07 AM
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Quote Quote by cabraham View Post
Faraday: v = -N*d(phi)/dt.
Ohm: J = sigma*E.
Lorentz: F = q*(E + uXB).

The induced voltage is described in Faraday's Law, FL. But we must be careful. A time changing mag flux is related to the emf (voltage) induced in the loop per FL. But the flux "phi" is a net flux, not the external flux exciting the loop.

When the time varying, herein "ac", mag flux phi, excited the loop, there is charge motion per Lorentz force law, LFL. Free electrons in the wire are moved in a direction determined by E & B. E acts tangentially, B acts normally. Hence a rotational field condition is present & the electrons circulate around the loop. But hold on. The moving e- constitute a current, which produces another mag field. If the external mag flux density is called "Be", & internal is "Bi", then Bnet = B = Be + Bi. Of course, the law of Lenz, LL, tells us that the induced or internal B field opposes the external B in polarity.

In addition, we have another thing going on. As e- transit through the wire, they crash into lattice particles & lose some energy. This energy is radiated in the form of photon emission. It is around 5 to 7 micron in wavelength, & is felt as heat in the infrared region. Charges accumulate due to said collisions, & these charges have their own associated E field, since e- carry E fields due to their own charge.

This accumulation of e- charge gives rise to an ir-rotational (conservative) E field. This E field has no curl & is not an emf, but a drop. It is a polarizing type of force, its curl is zero, as it can not drive electrons around the closed loop
.
Don't see how charge accumulate, this is conductor, no charge can accumulate inside. Is all this conductance loss? Why is it irrotational? How do you determine which component of E is rotational and irrotational? Please give formula.
So what is the voltage? How is it determined? The voltage V, is simply the line integral along a particular path of E*dl. But E has 3 components, the induced E field due to external B, the E field due to the current in the wire & its ac B field, Bi, & the static ir-rotational E field due to charge accumulation incurred via electron collisions w/ the lattice structure.
Please give formulas of Ec, Ei and Ee respect to their source.
The voltage from a to b, is the line integral of the composite E fields along a chosen path. Inside the Cu wire, what is the voltage value, Vcu? We are measuring voltage from terminals a & b, w/ the path as the inside of the Cu winding.
Can you draw a diagram showing the path?
The external ac mag flux density Be, gives rise to an induced E field, Ee, such that electrons in the Cu wire are moved along the wire, w/ said Be acting normal. This Ee is rotational.

But the induced current, Iloop, has its own B field, ac, we will call Bi. Bi is an ac field, & it has the opposite polarity of Be. It induces an E field & voltage or emf in the loop. This is the self-inductance of the loop. This is Ei, also rotational. For a very low loop resistance, the Bi cancels the Be almost entirely. What remains is the E field due to charges accumulating due to collisions between electrons & lattice structure. Call this one Ec. as it is due to charged particles.

Hence Enet = E = Ee + Ei + Ec. But Ee & Ei nearly cancel perfectly inside a low resistance conductor. If the loop were open, Ei tends toward zero, & the loop voltage is maximum due to no cancelling between Ee & Ei. Closing the loop via a resistive load results in current & a counter-balancing E & B fields, Ei & Bi. So inside the copper we still have Ec. If the entire loop was very low resistance so that Ee & Ei nearly cancel entirely, we still have Ec.
Again please show in drawing for open and closed loop.
The Cu sec winding in my example is 0.1 ohm. The 10A current times the 0.1 ohm results in the voltage drop of 1.0V. The line integral of the composite E field along the Cu path results in Ee & Ei almost cancelling, & Ec*dl giving us around 1.0V. In other words, as soon as the ext mag field, Be, enters the Cu wire, current is induced. The induced current has a strong B field, Bi, that cancels the external, & equilibrium is reached.
I don't see this, Like all transformers, the secondary always has a range of voltage because Lens Law is a voltage equation. Output voltage varies with load but mainly due to internal resistance and also the the internal resistance of the primary. That if you draw more and more current from the secondary, you draw more and more current from primary and the effective primary voltage dropped due to loss in the resistance of the primary.
You seem to be looking for the 120V induced emf inside the copper in distributed form. But don't forget that there is a counter-emf generated as well. The distributed emf sources are nearly perfectly cancelled by the distributed counter-emf sources. But the Ec component does not get cancelled. It accounts for the 1.0V drop inside the Cu.
I don't get this, why then you physically can measure 119V between the two terminal.
To better visualize this, consider a low resistance loop of wire, 0.010 ohm, closed & immersed in an ac mag field, Be. The Lorentz force moves the free electrons in the wire. This is current. But the current generates its own mag field, Bi. The law of Lenz tells us that they oppose each other. If the current is 1.0A, w/ a 0.010 ohm loop resistance, the voltage around the loop is 0.010 volt.
Again, in the drawing, show how you measure the 0.01V.
But we now open the loop, keeping the flux & area the same. The current plummets to near zero, but the voltage increases to 10V. Here, the ac mag flux produces an induced emf of 10V in the open state. When the loop closes, the net loop voltage is a mere 0.010V.
Where is the 10V coming from?
Why the difference? Of course, it is the cancellation. The external & internal parts of B & E account for the drastic difference in voltage between high & low impedance conditions. With a high-Z loop, the external B field is unopposed. Without induced current, the E field is due to the external B field, & the full voltage is realized since there is no loop current to cancel it.

When the loop resistance is low, the current generates cancellation of the external fields. The voltage is the line integral of all 3 phenomena. How can the Cu wire have just 1.0V, when the heater load has 119V, when their paths start & end at the same 2 points, a & b? They are in parallel, yet differing voltages are found.
Please show a drawing how you set up the probe to measure the 1V and 119V with two meters.
Why? Answer is cancellation, & charge accumulation due to differing resistance values. I believe I've made my case, but someone other than me should affirm.

Claude
I am loss after half way. I'll wait for more equation and drawing before I can continue. I still would like to know how you measure the 1V drop due to internal resistance in your transformer. I need to see the path how you measure the different voltages in the non conservative case you present.

I am making a first attempt to make sense of the E components:

[tex]\nabla \times \vec E_e =-\frac{\partial \vec B_e}{\partial t} \;\hbox { to get Ee from Be.} [/tex]

[tex] \nabla \times \vec B_i = \mu_0\sigma_{Cu} \vec E_e + \mu_0\epsilon\frac{\partial \vec E_e}{\partial t} \;\hbox { to get Bi from Ee.}[/tex]

[tex]\nabla \times \vec E_i =-\frac {\partial \vec B_i}{\partial t} \;\hbox { to get Ei from Bi} [/tex]
cabraham
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Apr6-11, 09:50 AM
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Quote Quote by yungman View Post
I am loss after half way. I'll wait for more equation and drawing before I can continue. I still would like to know how you measure the 1V drop due to internal resistance in your transformer. I need to see the path how you measure the different voltages in the non conservative case you present.

I am making a first attempt to make sense of the E components:

[tex]\nabla \times \vec E_e =-\frac{\partial \vec B_e}{\partial t} \;\hbox { to get Ee from Be.} [/tex]

[tex] \nabla \times \vec B_i = \mu_0\sigma_{Cu} \vec E_e + \mu_0\epsilon\frac{\partial \vec E_e}{\partial t} \;\hbox { to get Bi from Ee.}[/tex]

[tex]\nabla \times \vec E_i =-\frac {\partial \vec B_i}{\partial t} \;\hbox { to get Ei from Bi} [/tex]
You're on the right track. I type slowly & long posts take a lot of time. I gave a xfmr example, but my explanation was eventually focused on a loop immersed in an ac B field, like that of an antenna receiving rf. A xfmr has one more thing going on.

A loop immersed in an rf B field in free space is subjected to induction. But said loop has an area which receives a specific amount of radiated power. This is induction w/ constant power. In the open circuit state, v = -N*d(phi)/dt. Also, phi = Ac*B, where Ac is the cross-sectional area of the loop, & B is mag flux density.

When the loop is open, Be, the external mag field, is related to the loop voltage Vloop, as follows. Vrms = Bpk/(4.443*f*Ac*N), where f is frequency, N is turns, per Faraday's law, FL. But if we close the loop in a high value of resistance R, we get current.

This current generates a field which opposes Be, so we call it Bi. I covered the rest previously. An equilibrium is reached when the loop resistance R is low enough so that the Bi cancels Be. Lowering R further does not increase the current. The voltage reduces as R is lowered, i.e. loop voltage decreases w/ decreasing R & increasing current.

It has to be this way per conservation of energy law, CEL. This is a constant power condition. The loop receives a limited amount of rf power, & the loop power cannot exceed the incident power per CEL. Hence Bi cancels Bi when R is low enough.

Now the xfmr is examined. When open secondary is measured, 120V appears at the terminals. Let's use these parameters for the xfmr including core. Vrms = 120V, Ac = (5cm X 5cm) 25 cm^2, f = 60 Hz, N = 120 turns both pri & sec, lc = core path length = 50 cm, Rsec = 0.1 ohm, & mu_r = relative permeability of core including incidental gap = 1000.

The B field in the core computes to 1.501 tesla per FL (15,010 gauss, a typical value for a grain oriented silicon steel material). To get H, we divide by mu, where mu = mu0*mu_r. Since NI = integral H*dlc, we get a magnetizing current of 0.498 amp, or 0.5A rounding off.

So we have a xfmr w/ sec open, 120V rms, & 0.5A magnetizing current, Imag. What happens when we connect the 11.9 ohm heater load across the secondary. The 0.1 ohm sec winding resistance is in series w/ the 11.9 ohm heater, for a total of 12.0 ohm, & the sec current, Isec = 10A. The terminal voltage drops by 1.0V to 119V rms.

The 1.5 tesla is the core when open is Be in this case. It requires an H to sustain it, w/ Imag of 0.5A. If the sec is loaded, that load current, induced by Be/Ee, tends to produce a mag flux, or "mmf", opposite in polarity to Be. This is Bi. Only 0.5A of counter-mmf will cancel the 1.5 tesla of Be. So where does the 120V come from, as you just asked?

A xfmr is not operating under a constant power condition like a loop in free space. A xfmr operates w/ constant voltage. The primary is connected across a good strong well-regulated constant voltage source, CVS. The power company goes through great effort to insure the voltage at our outlet is 120V rms.

As soon as load current is drawn at the sec, the counter-mmf produces Bi cancelling Be, resulting in a drop in terminal voltage. But the xfme primary is connected to a CVS. Said CVS then outputs an increase in current which counters the counter-mmf. The additional primary current provides "counter-counter-mmf". Just as the counter-mmf (or "Bi" if you prefer) resulted in counter-emf & a drop in voltage, the counter-counter-mmf produces a counter-counter-emf & an increase in voltage.

As long as the primary is excited by a good solid CVS which has the power capability to meet the load demands, said CVS will offer any current necessary to keep Vpri at 120V rms. Thus the cancellation of Be by Bi, is countered by increased Ipri.

But the mag flux still cannot enter the Cu sec winding to any large degree. Since the sec Cu resistance & the heater load are in series, their current is identical. Hence the 120V is divided between the 0.1 ohm & 11.9 ohm. When current exists in 2 different resistances in series, the higher resistance material incurs more electron to latiice collisions, & more accumulated charges. The charges provide their E field, Ec. When all 4 components of E are evaluated, we get 1.0V in the Cu sec, & 119V in the heater load.

In a nutshell, the CVS sets up a core flux, which sets up a sec E field & voltage. When loaded, the sec current produces Bi/Ei which cancels the Be/Ee. The CVS then forces an equilibrium condition by providing just enough primary current so thet Vloop = 120V. It's a CVS, that is what it does. Lattice collisions take place more frequently in the higher resistance medium. But being in series the current in each of the 2 media must be equal. Hence charges build up & the E field due to charges, Ec, adds or subtracts to/from Ee, Ei, & counter-Ee.

Dr. Lewin has a good paper on Ec which I'll dig up & post. Did this help?

Claude
yungman
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#334
Apr6-11, 10:54 AM
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Thanks for your time. Again, I'm going to have to take time to read before I reply. Can you give some drawings if you have time at the mean time.

One thing concern me so far in this analysis is: If I follow this theory, then Ei and will cause another Bi2 and then Bi2 will cause Ei2 and Ei2 will cause Bi3 and to Ei3 and so on. So it is going to be an approximation of an infinite series.

I am not saying I agree with you, You are knowledgable, I want to try to understand your point of view before going any further. Actually I am going to stop my own study and review the Lenz law today first because the EM books don't really go too deep into Lenz. I am going to read through the ED book today.

Thanks
cabraham
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#335
Apr6-11, 05:21 PM
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Quote Quote by yungman View Post
Thanks for your time. Again, I'm going to have to take time to read before I reply. Can you give some drawings if you have time at the mean time.

One thing concern me so far in this analysis is: If I follow this theory, then Ei and will cause another Bi2 and then Bi2 will cause Ei2 and Ei2 will cause Bi3 and to Ei3 and so on. So it is going to be an approximation of an infinite series.

I am not saying I agree with you, You are knowledgable, I want to try to understand your point of view before going any further. Actually I am going to stop my own study and review the Lenz law today first because the EM books don't really go too deep into Lenz. I am going to read through the ED book today.

Thanks
I see your point, & I feel you are thinking in good terms, but the underlined "cause" words in the highlighted text give me concern. I believe, & most of the science community believes, that E & B are mutually inclusive, cannot exist independently, & that neither can be the cause nor the effect of the other. Special relativity describes E & B forces both as Coulomb interaction forces between charges taking special relativity into account. E & B are 2 views of the same action from different frames of reference.

But you are correct that the net E field is the sum of Ee from the distant source, Ei from the induced current, counter-Ee from the CVS exciting the primary, & Ec due to charge build-up in boundary regions. The Ec concept is explained by none other than Prof. Lewin in the attached lecture paper. It should help immensely. BR.

Claude
Attached Files
File Type: pdf induction_mit_lecsup41.pdf (467.6 KB, 10 views)
yungman
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#336
Apr6-11, 06:13 PM
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I understand about the chicken and egg thing. I was just following your notion of Be cause Ee and then Bi and Ei step by step. Which make sense.

Too much reading materials!!! They really don't go deep into Lenz in EM books, they more dive into EM waves and Tx lines. I am reading the Lenz in Griffiths and talking about 3 E also. I might not be back tonight, too much reading!!!

One thing I am thinking.

[tex] Emf = -\frac {d\Phi}{d t} [/tex]

This show flux induce a voltage, not current like what you use. You seems to start with say the secondary providing 10A with internal resistance of 0.1 ohm. then you show the short circuit around the secondary is 1V where it is 10A X 0.1 ohm!!


Thanks
cabraham
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#337
Apr6-11, 09:29 PM
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Quote Quote by yungman View Post
I understand about the chicken and egg thing. I was just following your notion of Be cause Ee and then Bi and Ei step by step. Which make sense.

Too much reading materials!!! They really don't go deep into Lenz in EM books, they more dive into EM waves and Tx lines. I am reading the Lenz in Griffiths and talking about 3 E also. I might not be back tonight, too much reading!!!

One thing I am thinking.

[tex] Emf = -\frac {d\Phi}{d t} [/tex]

This show flux induce a voltage, not current like what you use. You seems to start with say the secondary providing 10A with internal resistance of 0.1 ohm. then you show the short circuit around the secondary is 1V where it is 10A X 0.1 ohm!!


Thanks
Well. if the loop resistance is non-infinite & non-zero, then current & voltage are both induced. For a shorted loop we have I = integral H*dl. For an open loop emf = -d(phi/dt).

If the loop resistance is relatively high then phi_e is all that accounts for the emf. But w/ a low value of loop resistance, phi_i cancels phi_e. Phi = Ac*B. The Lorentz force states that there is a force acting on electrons, moving them. In the process of moving they incur collisions & a voltage drop occurs. Also, while moving they generate an internal B field, Bi which generates another emf.

So the emf equation you cited has multiple parts. Of course emf = -d(phi)/dt, but remember that phi is the external, plus internal, plus another external if the problem involves a CVS as a source, plus the E field due to charges Ec.

The emf equation is correct but there is a lot going on here.

Claude
yungman
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#338
Apr6-11, 09:44 PM
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In the shorted circuit, the resistance of the primary come into play, you draw so much current the primary voltage drop due to the drop in the primary resistance. This together with the secondary resistance will limit the max current. But that does not mean that it become a current equation. If you put very heavy primary winding and secondary, I don't think you can use the idea of set current of say 10A and get only 0.01 V in the secondary.

Can you give the diagram how you measure the different voltage drop with different path. that would be the ultimate proof. In order to see that you get 119V in one path and 1V in the second, we are going to have to be able to read that physically.
yungman
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Apr7-11, 01:27 PM
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Hey Claude

Thanks for your time. I understand a lot more about Faradays law in the last two days from your posts and from the book I read. I started another post concentrate on what you wrote because I think it warrant more discussion that is not related to this professors case.

I understand the constant flux [itex]\Phi[/itex] in antenna vs the transformer plugging into the wall plug. Please join in the other post to continue that point. And I think what you said has no bearing to Levin's case.



Now back to the MIT professor.

I think the transformer is a lot more easier to talk about than the Levin's single loop case. The reason is because both example reflect the same phenomena. In Levin case, he adjusted the input flux to get 1V total output so it's like your transformer that the power company maintain constant primary voltage and increase current to give constant voltage at the secondary and maintain 1V.

The difference is in your case, your secondary is say 120 turns, so the induced voltage onto the measuring probe's ground lead only is 1/120 of the overall voltage. So the measurement has only about 1% error vs the single loop secondary of Levin's experiment. The ground lead of the probe would produce 100% error like what I shown in #224 and #227.


Now this is my observation: I don't see where in your transform example give any voltage source to give 120V. In order to draw out the circuit diagram, you are going to have to give the source of the voltage. All the explaination of the magnetic field and electric field are like creating voltage source and counter voltage sources and so on. But how do you draw it in the circuit diagram? Just like Antiphon said, Levin mix circuit drawing and real EM circuit and it does not work that way. If you have a voltage, you have to represent it in the circuit diagram. Levin is wrong to take the circuit diagram literaly and continue on with the physical observation.

In order for Levin or anyone else to justify the path dependent, they have to justify where the voltage come from. It is ABSOLUTELY not enough to just say we "STIPULATE" the voltage exist and then turn around and use the circuit diagram without it and justify the path dependent thing. You want to justify the non conservative thing, you have to justify the voltage source ( or the lack of it). If you want to draw the circuit diagram to represent the real physical circuit, you are going to have to draw where the voltage come from.

Also, the good thing about using the transformer, we have absolutely no problem doing any measurement now. Now lets redraw the circuit, redefine the point A and D and then lets do the measurement of the two path, they WILL be the same using the transformer in the middle. There is no if and buts about this, you cannot treat the transformer as a single note D anymore. In Levin's case, he need to draw in the transformer before he can use the circuit diagram to justify the path dependent statement. This is what I have been arguing about for a long long time, maybe not in these exact words........Because never in my wildest imagination that ED people here don't consider induced voltage source and we end up going around and around, I kept drawing circuit based of equivalent voltage sources and a lot of you guys just ignor my drawing!!!




Well? Anyone else, I know from looking and the number of people reading this thread that there are plenty of people interested in this. What is your opinion?
yungman
yungman is offline
#340
Apr7-11, 02:45 PM
P: 3,844
This is the drawing of Levin's circuit and Cabraham transformer. It is basically the same circuit.



I did not draw the primary of the transformer, it is understood the primary is just to provide the [itex]\Phi[/itex] and like the power company, Levin adjust the driver of the solenoid to get 1V on the two resistors. This constant output voltage really simplify all the arguments of Faraday's law, secondary internal B and E because both case adjust the primary to get the desired voltage. The theory become very very simple..........It become a VOLTAGE SOURCE!!!!

As you can see in part A, the circuit that Levin drew should really be the right hand side drawing that include a transformer loop that have flux going through it and voltage induced on the loop.


In B, this is Cabraham's transformer circuit that he described. The internal resistance of the secondary is 0.1 ohm and the load resistance is 11.9 ohm. The circuit draw 10A and the voltage across the load is 119V. You can see the left side of B is drawn without any voltage source and treat the secondary of the transformer as a note D just as Levin drawn in his circuit diagram. But if you draw the transformer in like on the right side, then the circuit is completed.


Bottom line, Levin is WRONG to use this as an example to prove non conservative voltage that is path dependent because he left out the transformer part in the circuit. And further he call the transformer as note D.

Feel free to comment on my finding.




To StevenB:

With the 120 turns transformer secondary. There should be no problem measuring the voltage using a probe no matter how it swing, what direction the probe come in. BECAUSE, the loop formed by the probe is only a single turn only, so it would introduce about 1% of errour however which way you swing. AND you know and I know that now, if you tape on anywhere of the secondary of the transformer, you will get different voltage in the measurement. It would not be like the experiment we did that we are trying to measure along the single turn transformer while the loop formed by the probe ground keep fighting us.

I did not work on finding a way to measure the experiment because I figure no matter how I do it, there is always a reason to argue against and it's not fruitful. I am thankful someone dug this up and ANTIPHON shine the light on the difference between circuit diagram and real components.
cabraham
cabraham is offline
#341
Apr7-11, 04:17 PM
P: 998
I'll draw a pic later. I've been tidying up the house today because I have guests arriving tonight. Been quite busy. Tonight or tomorrow, I'll have an illustrative pic. But here is a quick comment.

The pic you showed has the 0.1 ohm sec Cu resistance & the xfmr sec inductor depicted as separate lumped parameters. You earlier argued that the reality is distributed parameters including emf sources.

The emf in the xfmr case is forced to be roughly the value of CVS exciting the primary. Ee & Ei will oppose each other, but the CVS provides whatever primary current is needed to force a flux density of 1.5 tesla, resulting in 120V rms terminal voltage. But the voltage along the Cu path is 1.0V, & along the heater path is 119V. Why?

I'll have a pic later. For now please refer to the Dr. Lewin paper I recently posted. At the Cu-heater boundary, we have differing resistances. But the current in the Cu & the heater are the same, being connected in series.

So per Ohm's Law, OL. Jcu = sigma_cu*Ecu, & Jhtr + sigma_htr*Ehtr. But we know that Jcu = Jhtr = J. So that sigma_cu*Ecu = J = sigma_htr*Ehtr. We know that the sigma values differ, as Cu has much lower resistivity. But the CVS driving the xfmr primary forces an emf of 120V rms. We know this. This emf is distributed, i.e. 1.0 volt/turn. But why is there 1.0V across the Cu inside the wire, & 119V across the heater?

What if the sec winding was made with high resistance wire, so that both the Cu sec & the heater were 6.0 ohms each? There would be no difference in terminal voltage regardless of path. It would be 60V across the Cu & 60V across the heater. The 120V emf, forced by the CVS at the xfmr primary is distributed around the loop, & since the resistances are identical for each path, no Ec exists, i.e. the E field in each section is due to emf only, no static charges are present.

But if sec Cu is 0.10 ohm, heater is 11.9 ohm, we still have 120V emf for the sec loop regardless of resistance values. But due to the differing resistances, there is a charge build-up at each interface. Refer to Dr. Lewin's diagram. One boundary (between Cu & heater) has an accumulation of "+ve" charge, the other boundary "-ve" charge.

These discrete charges have their own E field, Ec. The Ec has a polarity as follows. In the Cu, Ec is oriented so as to subtract from Ee. In the heater Ec adds to Ee. So the loop emf stays at 120V because this Ec field has zero curl. Going around the loop integrating Ec only results in a rise & a drop of equal magnitude. Around the loop, Ee integrates to 120V, & Ec to zero.

In the Cu, Ee integrates to half (assuming symmetry) which is 60V, & Ec to -59V. The net voltage in the Cu is the difference of 1.0V. But in the heater, Ee integrates again to half, or 60V, but Ec integrates to +59V. The net voltage in the heater is the sum of 119V.

Meanwhile, the CVS is oblivious to Ec. Take the 2 cases I detailed, 6.0 ohm Cu sec with 6.0 ohm heater, & the other case being 0.1/11.9 ohm resp. The net load is 12 ohm, 10A, either way, likewise for the 120V emf. The CVS & the process of induction are oblivious to the exact resistance distribution across the secondary. In both cases, the emf is 120V, & the CVS, the core flux, etc. is unaware of how the 120V is distributed between the Cu resistance & load resistance.

Nature fixes this with the accumulation of static charges at the interface between the 2 media. This charge field has its own Ec field. The Ee type of field has curl. Around a loop it has a non-zero integral & non-zero emf. But the Ec field is non-curly, having zero net emf around the loop. In other words the CVS is oblivious to Ec. Ec has zero curl & hence it cannot "mess up" the 120V emf, mandated by laws of induction & the nature of a CVS.

But the Ec field exactly fixes the voltage distribution while preserving Ohm's Law, Faraday, Ampere, conservation of energy, Lenz, etc. By definition, the voltage from xfmr sec terminals a to b, is the integral of the composite E field, whose parts are Ee, Ei, counter-counter Ee provided by the CVS, as well as the all-important Ec.

Do NOT neglect Ec. It does not show up in the induction laws, & has no interaction w/ the CVS. It merely distributes the 120V emf in accordance w/ the resistances preserving Ohm's law.

J = sigma*E, so that if J is the same for 2 regions, which it is, & E were the same, then we would have a paradox. How can J & E both be the same while the sigma values differ for each region. The answer is that if we combine Ee & Ei & the CVS Ee all into a composite Ee', that value of Ee' is uniform. It would integrate to the same value in Cu & heater. Hence we get a paradox. The energy per unit charge to transport charge through copper is but 1.0V, vs. 119V for heater.

Ec fixes all that. Symmetry would result in 60V in each section. Ec provides -59V in the Cu & +59V in the heater. All laws are upheld.

Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!!!. Cheers.

Claude
stevenb
stevenb is offline
#342
Apr7-11, 05:49 PM
P: 697
Quote Quote by cabraham View Post
Not so intuitive is it? Great minds stumble with this concept. It takes a lot of thought to sort it out. Prof. Lewin is spot on dead right!!!.
I second that comment.

I'll add one of my own too. Thanks to the great mind of Faraday, we don't need a great mind, nor should we stumble, when we study this subject. We need only trust in his Law. And, even if we want to get into these nitty gritty details, how much easier even that will be if we start with an acceptance that Faraday's Law trumps our intuition.


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