prove: adding mag shield doesnÆt change total force current in field 1

[electricpete]

Let’s say that you have a current-carrying conductor carrying current I0 in air in an a uniform external field B0. The force as we know is Fconductor = I0 x B0

Now enclose that current carrying conductor in a cylindrical “shield ring” of permeability MuR and expose it to the same external uniform field B0.

As we increase MuR, the force on the conductor Fconductor decreases, and the force on the shieldring Fshieldring increases. We can see this qualitatively by imagining the flux lines which pass less and less near the conductor as we increase MuR.

Now an interesting fact:. the total force among both conductor and shield ring doesn’t change as we change MuR.
Ie. Fconductor + Fshieldring = I0*B0 (regardless of MuR)

I have read that in an article. I know it to be true based on various indirect observations.

Can anyone explain to me a proof why it should be that
Fconductor + Fshieldring = I0*B0
(Regardless of MuR)


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[prex]

Recalling that, in your formula above F_c=I₀B₀, F_c is the force per unit length acting on the conductor, and that for this formula to be valid, B₀ must be orthogonal to conductor axis, I'll give my try.
A first observation is that considering an isolated current carrying conductor is a mathematical abstraction, because only a closed circuit may carry an electrical current. So what we have to consider is a loop of which at least a portion is immersed in a substantially uniform magnetic field (before applying the electromotive force).
To simplify things, but without loss of generality (because the same conclusions hold for a circuit of any shape), let's assume that the portion of conductor of interest is straight with length l, that this portion is orthogonal to the uniform field, and that the entire loop is planar with enclosed area A. Though also this is not necessary, let's also assume that at least the area A is covered by the uniform flux (that will be extended also in the surrounding region).
Now we know from the basic laws of electromagnetism that the potential energy stored in the circuit is U_p=-I₀[Φ](B) where [Φ](B) is the flux of B through the surface of area A.
If we imagine that the portion of conductor of interest moves by a distance ds in a direction orthogonal to both its axis and B, we can observe that the disturbance in the preexisting uniform field caused by the flowing current will rigidly move of the same amount. And, as this disturbance dies out rapidly with the square of the distance, we can conclude that the change in the flux of B through the circuit will be d[Φ](B)=lB₀ds hence F_cl=-dU_p/ds and you get the formula for F_c.
Now you start to see where I want to get: if you add a ferromagnetic shield around the conductor, the same conclusion holds, that is the field disturbance (that is now different) moves with both and dies out rapidly. So the force stays the same, if you consider the conductor and the shield as bound to move together.
However I consider the interesting problem you raised as somewhat misleading. In fact another phenomenon comes into effect when you add the shield: the magnetic attraction between the conductor and the shield. This attraction is theoretically zero in cylindrical simmetry, but this is an unstable condition, the conductor will be attracted by the shield as soon as it is moved from the center position. This means that it would be difficult to experimentally prove your statement by a separate measurement of the forces acting on the shield and on the conductor.

prex

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[hacksaw]

pete,

what bothers me about this question is that the current carrying conductor (sans shield) experiences a force transverse to that of the shield.

with the shield in place, the field seen by the conductor is considerably reduced but remains transverse to that on the shield.

i'd be interested in your "indirect observations"

 

[electricpete]

Thanks prex. I have been working a lot lately with field approach (energy density = 0.5*mu0*H^2). I had to go back and review the circuit approach to understand your proof. It makes good sense now and is probably a lot simpler than what I was working towards..

However I consider the interesting problem you raised as somewhat misleading. In fact another phenomenon comes into effect when you add the shield: the magnetic attraction between the conductor and the shield. This attraction is theoretically zero in cylindrical simmetry, but this is an unstable condition, the conductor will be attracted by the shield as soon as it is moved from the center position. This means that it would be difficult to experimentally prove your statement by a separate measurement of the forces acting on the shield and on the conductor

I don’t consider that a problem at all. Consider the shield/conductor as one system. There are two separate sets of forces. One is equal/opposite forces between conductor/shield which is internal to the system and produces no net force on the system and does not change magnitude as external field magnitude varies. The other is interaction between the system and the external field which produces a net force on the system F = L B0 I0.

To do an experiment:
Support === LoadCell1 === Shield === LoadCell2===Conductor.

Assume the current is into the paper, external field acts up, resulting force acts to the right. Assume conductor is displaced slightly to the left of center of shield and so attractive force acts to the right on the shield and to the left on conductor.

Do two experiments, a and b with external field Ba and Bb

In trial a
LoadCell1 reads F=L Ba I0
LoadCell2 = LC2a

In trial B
LoadCell1 reads F=L Bb I0
LoadCell2 = LC2b

LC2a = Fatt - k * L I* Ba where Fatt is the attractive force between conductor and shield and k is constant describing fraction of the total external force that acts on the condutor

LC2b = Fatt - k * L I* Bb

LC2a-Fatt = -k*L*I*Ba
LC2b-Fatt = -k*L*I*Bb
Take the ratio of the previous two equations:

(LC2A-Fatt)/(LC2b-Fatt) = Ba/Bb
We can solve Fatt since it is the only unknown, but I’m not going to bother with the algebra.

The force on the shield is L I* Bb - Fatt

Or else we could just center it perfectly so Fatt=0 as you say.

Hacksaw – one indirect observation is that if you look at a motor with conductors in the slots and PRETEND that the conductors are located in the airgap and calculate the sum of all the F=L * I * Bgap (where L = length, I = current, Bgap =airgap flux), then you get the correct total force corresponding to the torque produced by that motor (even though the force now acts on the iron because the conductor is shielded from the flux). Apparently moving the conductor between the external field (the gap) and the shield (the slot) does not change the total force appreciably.

If you look at item D4 here, you will see an example calc of this type where the correct motor torque is calcualted by “pretending” the conductors are in the airgap.

http://electricpete1.tripod.com/torque_web/main1.htm

Also if you look at item D5-1, you will see a colleague’s proof of the question I posed in this thread.

Even better, item D5-2 provides the means to find out what fraction of the force acts on the iron (shield) and what fraction acts on the conductor. Specifically, equation 15C gives the fraction of torque on the iron is 1- [1/muRelative]. Ad muRelative approaches infinite, the fraction on the iron approaches 1. As muRelative approaches 1, the fraction on the iron approaches 0

I have been looking for a more intuitive way to explain it. I like prex’s way. I have another approach that I’m working on, but it’ll take awhile to make it presentable.

The parent page for the above is here.

http://electricpete1.tripod.com/torque_web/index.htm

It contains some other related links including a video lab demo that I did which showed the force acts on the iron.

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[hacksaw]

the original problem statement makes no mention of a closed loop, so how does this affect the argument outlined by prex

 

[electricpete]

I don’t have any objection to adding a return path for the current. As long as that path is outside the field it shouldn’t change anything (or as prex said far enough away that movement of the conductor by distance dx makes no appreciable change in flux in the neighborhood of the return conductor).

Once again I like the simplicity of prex’s solution.

I want to back up and make sure I understand the fundamental energy conservation at work.

dW/dt = Pelec_in – Pmech_out

dW/dt = I dPhi/dt - F *dx/dt

dW = I dPhi – F dx
where W is stored energy associated with a singly-excited circuit.

So two questions:
1 – Is it necessary to account for extra energy that may be transferred to the the external field? Perhaps this needs to be treated as a multiply-excited system with another circuit to generate B0?

2 – When we take derivative of the energy
Fc*L=-dUp/ds
isn’t it necessary to hold Phi constant?

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[hacksaw]

seems that these arguments are contrary to the lorentz force acting on the current flow, am I missing ssomething?

 

[prex]

electricpete,
first a correction to a mistake in my post above, that you probably already noticed: the field generated by a current flowing in a conductor dies out with the distance (or is proportional to the inverse of it), not the square. Of course the reasoning doesn't change though.
Concerning your questions I'll try again:
Given the formula U_p=-I₀[Φ](B), the full variation would be:
dU_p=-[Φ](B)dI₀-I₀d[Φ](B)=-[Φ](B)dI₀-I₀lB₀ds-I₀AdB₀ (the last term only approximate, supposing A is large)
This shows that we should worry not only about any change in B₀, but also about a change in I₀. However we are not interested in the effects connected with these changes, that in turn depend on other here unrelevant parameters (like the internal resistance of the generator that feeds I₀). So we assume that during the displacement ds someone will change the setting of the generators of B₀ and I₀ so that both are kept constant. Of course this will require to inject or to drain some energy, but this will exactly compensate the first and the third term in the equation above. So we are left with
dU_p=-I₀lB₀ds.
Note also that this way of reasoning gives an independent proof of the second elementary law of Laplace (dF=I₀B₀dl in scalar form): this should give a sufficient assurance on its exactness.

prex

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[electricpete]

hacksaw - Lorentz force requires a flux at the location of the conductor. Let's say the iron shield has very high permeablity. Then the shield prevents the flux from reaching the conductor, so there is no significant Lorentz force on the conductor (the same applies for conductor in a slot in a rotating machine). And there is a force on the sheild resulting fomr interaction of the fields associated with the current and the external field.

The final result is that there is no force on the conductor from interaction with the external field and the amount of force on the shield is the same force that would have existed on the conductor if it were exposed directly to the external flux.

Many textbooks explain motor operation by showing a copper wire between two permenant magnets, stating F=qVxB, and stating or implying that the torque producing force in a motor acts on the conductors. That last statement or implication is incorrect if the conductors are in slots.

I have provided a lot of supporting discussion at the link above, as well as a video of a lab demonstration where you can see these conclusions with your own eyes.

prex - thanks for your response... I'm still digesting.

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[electricpete]

Returning to my question #2:
dU = I d? – F dx
U(?,x) is a path-independent function of variables ? and x

So to calculate F, we need
F = -dU/dx evaluated holding ? constant
But the dx that we posed when determining dU/dx introduced a change in ? which seems to violate the requirement to hold ? constant.

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[prex]

Your equation dU=Id[Φ]-Fdx appears incorrect to me.
The correct equation is dU=-Id[Φ]=Fdx.
The change in stored energy because of dx is -Id[Φ]; to obtain this change you must add (or subtract) mechanical work to the system, and this will be Fdx that must equal the change in stored energy (with the sign changed).

prex

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[prex]

Correcting myself: your equation is correct, except that the final result is dU=0 as the work done to displace the wire goes into (or comes from) stored energy.

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[electricpete]

I am by no means criticizing, but instead playing the devil's advocate. I am hopeful to reach a good simple proof.

For the system containing stored energy U=F(Phi, x), I am pretty sure that we need to perform integrations and derivatives along a path where only one of those variables (Phi or x) is changing at a time.

Starting again with the energy balance
U = energy stored in the system
dU/dt = Pelec_in – Pmech_out
dU/dt = I * e(t) - F*v(t)
dU/dt = I dPhi/dt - F *dx/dt
dU = I dPhi – F dx

I = dU/dPhi. dU/dPhi can only be evaluated with x held constant (not a particularly useful formula)
F = -dU/dx. dU/dx can only be evaluated with Phi held constant

The geometric analogy is a function f(x,y) =3*x(t)+4*y(t) where x and y are functions of time.
df/dt = df/dx *dx/dt + df/dy * dy/dt (partial derivatives)
df/dt = 3 * dx/dt + 4 * dy/dt
df = 3*dx + 4*dy.

If I wanted to find df/dx (x0,y0) by comparing the value of at two different points in the vicninity of (x0,y0), I would need to choose two points where y=y0.
For example I could use df/dx ~ [f(x0+dx, y0) - f(x0,y0)] / dx
where dx is small
The important thing is that I need to hold y constant while computeing df/dx.

Hopefully we can see if U plays the role of f, Phi plays the role of y, and x plays the role of x, then we need to hold Phi constant while computing dU/dx

Going back to our system. If we allow Phi to change, then , it appears that we're allowing energy into the system from an external electrical source (as can be seen by tracing backwards in the derivation to where I*dPhi is associated with electrical power into the system). Unless we can quantify that energy input, we can't learn anything from that equation.

Let's say the mass (inertia) is negligible. Movement was accomplished by applying force F in to achieve movement ds=s1 during a period t1 and a velocity v1 =s1/t1. As before the movement increases the area enclosed by dA. = s1*L

A voltage was induced in the loop during that time period: e1 = dPhi/dt = v1*L*B0

The voltage is in a direction given by lenz' law.

To maintain the current, we would need to supply from an external voltage source an equal opposite voltage e_externalsource = dPhi/dt in an opposite direction.

The energy transferred into or out of the circuit by that external voltage source would be
W = Int (Pelec)dt = int (I e) dt = int I dPhi = I * B0 * dA = I Phi L dx

I lost track of which energy was flowing in and out (mechanical or electrical). On the surface it looks like energy is passing through between the electrical side and the mechanical side without changing the stored energy, but I'm not sure.

The bottom line - I don't think we can compute F = dU/dx under a circumstance where Phi is allowed to change without considering the electrical power interchanged with the circuit.

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[electricpete]

I will mention that there is a proof of this scenario for the specific case of infinite permeability in "Electromagnetic Field Theory" by Zahn pages 370-373.

It is a very ugly solution where he uses the exact analytical solution for the field based on solution of a poisson's equation boundary value problem in three different regions including outside the shield.

It would be interesting if we could find a simple proof for something that was so tedious for that author to show. (or maybe I am deluding myself into thinking there should be a simple explanation?).

The conclusion of that section gives the qualitative insight:

The force on the [iron] cylinder [in the case of infinite permeability] is the same as that of an unshielded current carrying wire, given by F=I*B0*L. If the iron core has a finite permeability, the total force on the wire (Lorentz force) and on the cylinder (magnetization force) is again equal to F=I*B0*L. This fact is used in rotating machinery where current carrying wires are placed in slots surrounded by highly permeable iron material., Most of the force on the whole assembly is on the iron and not on the wire, so that very little restraining force is necessary to hold the wire in place. The force on a current-carrying wire surrounded by iron is often calculated using only the Lorentz force, neglective the presence of the iron. The correct answer is obtained but for the wrong reasons. Actually there is very little B field near the wire as it is almost surrounded by the high-permeability iron so that the Lorentz force on the wire is very small. The force is actually on the iron core.

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[prex]

Where I don't follow you is when you say that [Φ] must remain constant: in fact [Φ] changes just because the area through which it is computed has changed.
You want perhaps to keep B and I constant, and indeed they wouldn't be constant unless someone goes to the generators and adjusts the supply: this requires an energy exchange, but this also allows for not accounting in the energy balance the terms related to the change of B and I.
Hence: the mechanical energy spent to move the wire goes into potential energy stored in the circuit keeping B and I constant and having a change only in the area enclosed by the circuit.

What do you think of the argument that this way of reasoning allows to correctly derive the second law of Laplace?

prex

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[electricpete]

I went back and reviewed Cheng's "Field and Wave Electromagnetics" 2nd ed Chapter 6 pp252-255.

Based on that review, the short answer is that Fmag=dUmag/dx will give the correct magnitude in cases when Phi varies and I stays the same, where I have assumed Fmag*dx is positive in the direction that delivers mechanical energy outside of the system (movement occurs in the same direction of the magnetic force and opposite direction of opposing external mechanical force). Note this is opposite of the sign we would expect if the transaction were simple transfer of energy from stored magnetic energy to external mechanical energy (we would expect Fmag=-dUmag/dx in that case). The reason is that it is not simple transfer between the stored magnetic energy and the mechanical energy. There is also energy IdPhi which must delivered into the system by the external circuit in the case where we execute a virtual displacement which changes flux under assumed constant current. (remember input Pelec =I dPhi/dt so input dUelec = I dPhi which is non-zero if we change Phi will holding I constant)

The reason that we get the same magnitude as the simple magnetic-to-mechanical energy transaction even though the physical energy transfer is much different is that the electrical circuit will put in TWICE as much energy as the mechanical system removes, resulting in an increase in stored magnetic energy exactly equal to them amount of the mechanical energy removed from the system.

i.e. for dUmech=Fdx taken out of the system by mechanical force exterted with changing flux and constant current, we have
dUelec = I dPhi = electrical energy into the system
dUmech = (1/2) I dPhi = mechanical energy out of the system
dUmag = (1/2) I dPhi = increase in stored magnetic energy of the system.

If you doubt this, try to make the energy balance between the field and the external force work without considering input power. Assume we have rectangular loop with current flowing CW. On the top leg of the loop (current flowing to the right), we have a length L of the conductor/shield which is immersed in external field B0 flowing into the page. The resulting magnetic force is F in the upward direction. If we allow that magnetic force to do work by moving in the direction it wants (transferring energy to some external force), then the energy stored in the system must also increase. On the surface it looks like a contradiction to conservation of energy (the system energy increased at the same time that the system did work against an external force). But the conservation of energy is restored when you include the energy added by the electrical circuit.

I have assumed in the terminology above that we allow the magnetic force to do work against an external force through distance dx. If we reverse the situation and push the external force to do work pushing through a distance opposite the magnetic force, the energy flows reverse. (We have mechanical energy input (1/2) I dPhi into the system, stored magnetic energy decrease of (1/2) I dPhi, and a total of the external electrical voltage source absorbs and energy I dPhi.)

Perhaps this was already taken into account in Prex' original proof at some level. It was not obvious to me until I looked it up. Sorry if I took something simple and made it complicated.

So, I think with some minor clarification regarding the energy transfer, the proof is starting to look pretty good. Any comments?

p.s. I have thought about my previous question #1 (Is it necessary to account for extra energy that may be transferred to the the external field?) and I think there is no energy change associated with that external field as long as we keep the geometry described by prex. Model the external field a separate set of large-area coils or permanent magnets. As long as they don't move, there is no mechanical work done by them. And as long as the total Phi linking that coil doesn't change, there is no electrical energy change associated with that circuit. This is satisfied as long as the conductor/shield is moving around somewhere in the center of the field, not the periphery, also remembering the 1/r drop in field as discussed above. No electrical or mechanical input/output means no change in stored energy of that circuit and it cannot affect the energy balance of the other circuit.

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[electricpete]

I now have a factor of 2 to track down.

F dx = 1/2 I dPhi = (1/2) I B L dx

F = (1/2) I B L

Something is missing somewhere


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[prex]

electricpete,
you didn't answer my last question. My point is: if you, by some deduction, can prove that a statement is true, and you already and independently know that such statement is true, then in fact you prove the correctness of your deduction.
Coming back to your reasoning, I agree that you are complicating something simple.
1) I can't see how you are deriving the coefficient 1/2
2) More exactly P_elec=d(I[Φ])/dt, so dU_elec=Id[Φ]+[Φ]dI. In fact the electrical energy to be input in the system is exactly what is required to keep I constant, or, in other words, if there is no electrical energy input, the the current would inevitably change in association with a change in [Φ]. As we assumed I to be indeed constant, we don't need to consider U_elec in the balance.
3) Your energy balance above has an incorrect sign, as U_p=-I[Φ]. With your sign conventions, the energy stored in the circuit decreases: this explains why work may be done. Remember the simple experiment explained in many textbooks, where one side of your rectangular loop is free to slide over long conductors: with the current flowing CW as in your example, the slide would move outwards, as in this case the flux of B is positive, and its increase, by decreasing the stored energy, would do the work; if the current flows CCW, the slide would move inwards, but in this case the flux of B is negative, so again it would increase, and the energy decrease, to do the work.
If your eyes see flying stars now, so do mine!

prex

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[electricpete]

1) I can't see how you are deriving the coefficient 1/2

Where did the ½ come from? Cheng's "Field and Wave Electromagnetics" 2nd ed Chapter 6 pp252-255.

I will also try to provide my own proof from first principles here.

Let
dWelec = electrical energy into the system
dWmag = change in energy stored in the ssytem
dWmech = mechanical energy out of the system
(the above follows the convention that electrical energy is positive going in and mechanical energy is postive going out – these positive directions are the directions of energy flow associated with an electric motor)

Start with
v = dPhi/dt
Pelec(t) = dWelec/dt = i * v = i * dPhi/dt
dWelec/dt = i * dPhi/dt
dWelec = i dPhi [EQUATION 1]

Equation 1 represents the electrical energy into the circuit. We can use it in two different situations: when i is a function of Phi and when i is assumec constant w.r.t. Phi. We will use both.

First, assume linear magnetic system with FIXED mechanical position x.
Then i = Phi / L0 where L0 is inductance.

Find the energy which will be put into the magnetic field bringing it from zero energy i=0 to a final state i=I

Welec = int i(Phi) dPhi = int (Phi/L) * dPhi = (1/2) Phi^2 / L
substitute in I = Phi/L0
Welec = (1/2) * Phi * I = (1/2) * L0 * I^2

Since x was fixed during the above transaction, there was no mechanical energy Fdx exerted, and the change in magnetic energy must be the same as the energy input
Wmag = (1/2) * Phi * I

This expression for energy stored in the magnetic field Wmag is valid independent of how we got there.

Now use this result Wmag = (1/2) * Phi * I and impose the requirement that current I must be constant as in Prex’ assumption (a valid assumption)
dWmag = (1/2) * I * dPhi [Equation 2]

We will now vary x. Equation 1 and Equation 2 remain valid where it is understood Phi is a function of x and so dPhi = L * B * dx where L is length.

What is the mechanical energy removed when we move by distance dx? By conservation of energy, it is the electrical energy input minus the increase in stored magnetic energy
dWmech = dWelec – dWmag
from equation 1 (I constant this time) and equation 2, we know the two terms on the rhs
dWmech = I dPhi – (1/2) I dPhi
dWmech = (1/2) I dPhi

2) More exactly Pelec=d(I?)/dt, so dUelec=Id?+?dI. In fact the electrical energy to be input in the system is exactly what is required to keep I constant, or, in other words, if there is no electrical energy input, the the current would inevitably change in association with a change in ?. As we assumed I to be indeed constant, we don't need to consider Uelec in the balance.

The expression that you use for dWelec/dt = Pelec=d(I?)/dt seems incorrect to me. I’m not sure where you got it from. I think maybe you assumed the magnetic energy stored in the circuit is I?. I disagree... it is (1/2)I? as can be verified with the familiar expression for energy in an inductor ½ * L *I^2. I have developed my own expression for dWelec/dt above. Let me know if you don’t agree with it.

3) Your energy balance above has an incorrect sign, as Up=-I?. With your sign conventions, the energy stored in the circuit decreases: this explains why work may be done. Remember the simple experiment explained in many textbooks, where one side of your rectangular loop is free to slide over long conductors: with the current flowing CW as in your example, the slide would move outwards, as in this case the flux of B is positive, and its increase, by decreasing the stored energy, would do the work; if the current flows CCW, the slide would move inwards, but in this case the flux of B is negative, so again it would increase, and the energy decrease, to do the work.

I believe my signs are exactly the way I intended them. I believe you and I disagree over the nature of the underlying energy transaction. I don’t agree that in holding I constant, we have a simple transfer of energy from the magnetic field to the mechanical system doing/taking the work without any involvement of the elctrical circuit. When we include the electrical circuit, we see an opposite sign appear. This is not a proof of the involvement of the electrical circuit, just a consequence of it.

Am I missing something within the big picture? Must be. If it gives a factor of two off for force, there must be something wrong with the principle or the application of it, or the assumptions we have built into the experiment.

Does your solution give the right answer? Yes. That is a necessary but not sufficient condition for a correct proof.

Please answer three questions:
1 – T/F: Force = dW/dx is based on conservation of energy (equivalently W = Int Fdx)
2 – T/F Changing Phi while holding I constant will require electrical input energy to the system IdPhi as I proved above from first principles.
3 – T/F we must include electrical energy input into the system if we are going to do any calculation involving conservation of energy within the system, such as item #1.


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[electricpete]

The equations I mentioned
dUelec = I dPhi = electrical energy into the system
dUmech = (1/2) I dPhi = mechanical energy out of the system
dUmag = (1/2) I dPhi = increase in stored magnetic energy of the system.

applied to a singly excited system. We can see where the half began was integrating dWmag= i(Phi) * dPhi = Phi/L dPhi and wehn we integrate we get Phi/2.

Prex' system is a doubly excited system. I have to re-examine how those equations apply there, but I think the half will disappear. I noted the following expression for energy in a doubly-excited system:
Wmag = 0.5*L1 *I1^2 + 0.5*L2 *I2^2 + L12 * I1*I2
the factor of 0.5 is gone from the cross term which is the one of interest. I'm pretty sure if I track back the derivation of this expression for energy in a doubly-excited system, it will help erase the factor of 0.5. I'm still working on it.


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