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

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

(OP)

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)

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

_{c}=I_{0}B_{0}, F_{c}is the force per unit length acting on the conductor, and that for this formula to be valid, B_{0}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}=-I_{0}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

disturbancein 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 dΦ(B)=lB_{0}ds hence F_{c}l=-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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## RE: prove: adding mag shield doesnÆt change total force current in field

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"

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

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.

ht

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.

ht

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

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

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

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

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

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}=-I_{0}Φ(B), the full variation would be:dU

_{p}=-Φ(B)dI_{0}-I_{0}dΦ(B)=-Φ(B)dI_{0}-I_{0}lB_{0}ds-I_{0}AdB_{0}(the last term only approximate, supposing A is large)This shows that we should worry not only about any change in B

_{0}, but also about a change in I_{0}. 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 I_{0}). So we assume that during the displacement ds someone will change the setting of the generators of B_{0}and I_{0}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 withdU

_{p}=-I_{0}lB_{0}ds.Note also that this way of reasoning gives an independent proof of the second elementary law of Laplace (dF=I

_{0}B_{0}dl in scalar form): this should give a sufficient assurance on its exactness.prex

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

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

The correct equation is dU=-IdΦ=Fdx.

The change in stored energy because of dx is -IdΦ; 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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## RE: prove: adding mag shield doesnÆt change total force current in field

iscorrect, except that the final result is dU=0 as the work done to displace the wire goes into (or comes from) stored energy.http://www.xcalcs.com : Online tools for structural design

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

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

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:

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

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?

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

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

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

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

independentlyknow 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(IΦ)/dt, so dU_{elec}=IdΦ+ΦdI. 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}=-IΦ. With your sign conventions, the energy stored in the circuitdecreases: 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!

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

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

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.

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

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

phi1 = all flux linked by I1 (mutual and self)

phi2 = all flux linked by I2 (mutual and self)

phi12 = mutal flux linked by both I1 and I2

L1

L2

L12

x

x0

L = length

i1

i2

I1

I2

Polarities are such that phi1 and phi2 are positive in the same direction in their area of overlap and increasing either i1 or i2 increases Phi12

phi1 = L1*i1 + L12 * i2

phi2 = L2*i2 + L12 * i1

Phi12 = L12

dphi1 = L1 di1 + L12 di2

dphi2 = L2 di2 + L12 di1

L12 is a function of x, L1 and L2 are not.

We have originally established I2 and L12 in order to create associated flux B0. So

L12 * I2 = B0 * x0 * L

L12 = B0 * x0 * L / I2

Start with position x0, i1=i2=phi1=phi2=0. We track the energy stored as we accomplish three changes in order:

A – Increase i2 from 0 to I2

B – Increase i1 from 0 to I1

C – Move x from x0 to x0+dx

We will track the magnetic energy added in each of these steps as dWmagA, dWmagB, dWmagC

A - Find the magnetic energy stored in the system as we increase i2 from 0 to I2

dWmagA = Int i2 dphi2

since i1 is 0, dphi2=L2 di2

dWmagA = Int i2 L2 di2

(where the limits of integration for i2 are 0..I2)

dWmagA = 0.5 * L2 * I2^2

B - Find the magnetic energy stored in the system as we now increase i1 from 0 to I1

dWmagB = Int i1 dphi1 + Int I2 dphi2

dphi1 = L1 di1 + L12 di2 = L1 di1 (since I2 is not changing in this step)

dphi2 = L2 di2 + L12 di1 = L12di1 (since I2 is not changing in this step)

dWmagB = Int i1 L1 di1 + Int I2 L12di1

(where the limits of integration for i1 are 0..I1)

dWmagB = 0.5 * L1*I1^2 + L12*I1*I2

Adding together the results of steps A and B we now have stored magnetic energy

dWmagAB = dWmagA+dWmagB = 0.5 * L2 * I2^2+ 0.5 * L1*I1^2 + L12*I1*I2

as expected

C – increase x from x0 = x0+dx

In the previous steps we had only change in electrical input and change in stored magnetic energy, so those two changes had to be equal. In this step we have these two changes plus a mechanical change. We will compute the electrical energy input at each terminal along with the change in stored energy. We will use these results together with energy balance to deduce the mechanical energy.

L12 = Ph

========================

L12 = B0 * x * L / I2

dL12/dx = B0 * L / I2 = (L12/x)

dL12 = B0 * L / I2*dx = (L12/x)dx

phi1 = L1*I1 + L12 * I2 (I1 and I2 now treated as constant)

dphi1/dx = d(L12)/dx * I2

dphi1/dx = I2 * B0 * L / I2=B0*L

dphi1=B0*L*dx

dWelec1C = int I1 dphi1 = int I1 B0*L*dx

dWelec1C = I1 B0*L*dx

phi2 = L2*I2 + L12 * I1 (I1 and I2 now treated as constant)

dphi2/dx = d(L12)/dx * I1

dphi2/dx = (B0 * L / I2)*I1=B0*L*I1/I2

dphi2 = (B0 * L / I2)*I1=B0*L*I1/I2 dx

dWelec2C = int I2 dphi2 = int I2 * B0*L*I1/I2 dx

dWelec2C = I1 I2 * B0*L*dx

Change in stored energy in the magnetic field dWmagC

examinine the expression

Wmag = 0.5 * L2 * I2^2+ 0.5 * L1*I1^2 + L12*I1*I2 (derived above as WmagAB)

dWmag/dx = I1*I2* dL12/dx (since L1 and L2 do not vary with x).

dWmagC = I1*I2* dL12

substitute dL12 = B0 * L / I2*dx

dWmagC = I1*I2* B0 * L / I2*dx = I1*B0*L*dx

To summarize part C so far:

Now an energy balance for part C

energy in = increase in stored energy – energy out

energy out = energy in – increase in stored energy

dWmechC = dWelec1c + dWelec2c - dWmagC

dWmechC = I1*B0*L*dx + I1*B0*L*dx - I1*B0*L*dx

dWmechC = I1*B0*L*dx

The result is correct. The energy balance is a little tricky.

Interesting that the external field had to supply energy. I wonder how this problem would work with permanent magnets? Perhaps the permanent magnet field would have to decrease since it has no energy input?

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

I also forgot to give the overview of what I was doing. What I did was to replace the external field B0 with a second current loop I2.

The interesting result is that the current loop I2 must have energy input to maintain it's strength when the conductor loop is moved relative to it in a direction that increases the flux linkage. That was the flaw of my previous attempt... I assumed the external field did not participate in the energy balance.

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

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

I agree 100%. But the misconception is very widespread and sometimes it is difficult to convince people otherwise. You will find many books that show the picture of a current loop in a magnetic field and tell you that’s how a motor works... with forces acting ON the conductor.

I had quite an interesting experience on another forum where I raised this discussion. A PhD in engineering who claims to be an expert on motors (has written a few books, gives training, writes articles) jumped into the thread and quickly proclaimed I was wrong (the force acted on the conductors), that I didn’t know what I was talking about, and that all my proofs were wrong. He also posted his credentials in the thread as if that were supposed to make him right. The discussion went on for 3 months and 95 pages, not including many detailed attachments that I posted. The thread was eventually deleted by the forum administrators due to unprofessionalism and I was also asked to remove the copy of the thread from my website.

So, you can say it, explain the physics, and talk to you’re blue in the face, but some people won’t listen. I did do the video which I think convinced a lot of people that didn’t exactly understand the theory.

There are a number of reasons people grab onto the idea of force on the condcutors. It is very widely taught in textbooks. The F=qVxB is very commonly shown along with the right-hand rule showing the direction of force. (the force is sometimes called the motor effect and rule often called right hand rule for motors... to distinguish from gnenerators). Some authors are careful to mention that this gives the magnitude of the force but the force doesn’t actually act on the conductors. Others clearly don’t even understand the reality and tell us the the force acts on the conductors.

Here is a sampling of what you will see in the literature on the subject:

http://en.wikipedia.org/wiki/Electric_motor

“The fundamental principle upon which electromagnetic motors are based is that there is a mechanical force ON any current-carrying wire contained within a magnetic field”

Modeling and High Performance Control Of Electric Machines”

“1.1 – Motors work on the basic principle that magnetic fields produce forces on wires carrying currents”

(this is the third sentence of the book.... he does clarify much later in chapter 4 that this force on conductors applies only under the assumption that the conductors are assumed to be located in the airgap)

“Modern Electric, Hybrid Electric, and Fuel Cell Vehicles”

6.1.1 Principle of Operation and Performance

The operation principle of a DC motor is straightforward. When a wire carrying electric current is placed in a magnetic field, a magnetic force acting ON THE WIRE is produced. The force is perpendicular to the wire and the magnetic field as shown in Figure 6.3. The magnetic force is proportional to the wire length, magnitude of the electric current, and the density of the magnetic field; that is, F = BIL.

Shows the picture of the wires sitting between two magnets

Electric Circuit Theory and Technology by Byrd

In an electric motor, conductors rotate in a uniform magnetic field. A single-loop conductor mounted between permanent magnets is shown in Figure 21.1. A voltage is applied at points A and B in Figure 21.1(a). A force, F, acts ON THE LOOP due to the interaction of the magnetic field of the permanent magnets and the magnetic field created by the current flowing in the loop. This force is proportional to the flux density, B, the current flowing, I, and the effective length of the conductor, l, i.e. F D BIl. The force is made up of two parts, one acting vertically downwards due to the current flowing from C to D and the other acting vertically upwards due to the current flowing from E to F (from Fleming’s left hand rule). If the loop is free to rotate, then when it has rotated through 180°, the conductors are as shown in Figure 21.1(b). For rotation to continue in the same direction, it is necessary for the current flow to be as shown in Figure 21.1(b), i.e. from D to C and from F to E.

ac – “Similar forces are applied to all the conductors on the rotor, so that a” torque is produced causing the rotor to rotate.”

Practical Troubleshooting of Electrical Equipment

“The rotating magnetic field induces emf in the rotor by the transformer action. Since the rotor is a closed set of conductors, current flows in the rotor. The rotating fields due to stator currents react with the rotor currents, to produce forces ON the rotor conductors and torques.”

First Course in Power Electronics and Drives

11-3 – Introduction to Electrical Machines and the Basic Principles of Operations: 11-3-1: Electromagnetic Force: Consider a condutor of length L in figure 11-6a. The conductor is carrying a current i and subjected to an externally-established magnetic field of a uniform flux-density B perpendicular to the conductor length. A force Fem is exterted on the conductor due to the electromagnetic interaction between the external magnetic field and the conductor current. The magnitude of this force Fem is given by F= B I L”. (There is no more clarifying discussion of force in all of section 11-3, so a reader would naturally assume that this principle is explaining the basic principle of operation of electrical machines.

EE Ref Book 16e Newnes Laughton (

20.3 The elements discussed [above] indicate that there are two methods of developing a mechanical force in an electromagnetic machine

Interaction - The force fe ON A CONDUCTOR carrying a current

i and lying in a magnetic field of density B is fe = Bxi

per unit length, provided that the directions of B and i are

at right angles; the direction of fe is then at right angles to

both B and i. This is the most common arrangement.

(2) Alignment. Use is made of the force of alignment between

two ferromagnetic parts, either or both of which may be

magnetically excited. The principle is LESS OFTEN APPLIED,

but appears in certain cases, e.g. in salient-pole synchronous

machines and in reluctance motors.

"Standard Handbook for Electrical Engineers"

(S.H.E.E.), 13th ed, edited by Fink and Beaty. Page 20-10 section 21 gives theory of operation of a synchronous motor and states "The electromagnetic torque acting between the rotor and the stator is produced by the interaction of the main field Bd and the stator current density Ja, as a J x B force on each unit volume of stator conductor".

Schaums Electromagnetics – has a whole chapter on torque and force. Talks only about F=qVxB!

Operation and Maintenance of Large Turbo Generators

This basic law is attributed to the French physicists Andre Marie Ampere (1775– 1836), Jean Baptiste Biot (1774–1862), and Victor Savart (1803–1862). In its simplest form this law can be seen as the “reverse” of Faraday’s law. While Faraday predicts a voltage induced in a conductor moving across a magnetic field, the Ampere-Biot-Savart law establishes that a force is generated on a currentcarrying conductor located in a magnetic field. Figure 1.14 presents the basic elements of the Ampere-Biot-Savart’s law as applicable to electric machines. The figure also shows the existing numerical relationships, and a simple hand-rule to determine the direction of the resultant force.

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

excuse me if I prefer to leave to you such theoretical wanderings. After all I think that we both know the conclusion, the one that is in my first post.

I'll make now a further step, determining the separate forces that act on the conductor and on the shield. I didn't check in depth the Yan's proof, so I don't know whether the method is equivalent, but the result is of course the same.

Let's figure out a body with a relative permeability μ

_{r}immersed in an uniform field of intensity B_{0}. For what matters here both the body and the field may be considered as infinitely extended in all directions: what matters is that for a body of sufficient extension the field intensity in regions far from edges will also be B_{0}. The body has a slot in it: it is a hole with rectangular section, one side parallel to B_{0}, the extension in the third direction being unimportant. We want to determine the field intensity B in the central portion of the slot, where edge effects are of minor importance.This is readily done by observing that the magnetomotive forces over paths in parallel must be the same. In the body the magnetomotive force is proportional to B

_{0}/μ_{r}and in the slot it is proportional to B (for paths of equal lengths).Hence in the slot B=B

_{0}/μ_{r}. To be noted that this conclusion is not tightly connected to the shape of the slot, and remains valid for the central portion of a round one or for any other shape.The result is already obtained: for a conductor placed in the center the force per unit length will be F

_{cc}=I_{0}B_{0}/μ_{r}, and the force on the shield F_{cs}=I_{0}B_{0}(1-1/μ_{r}).http://www.xcalcs.com : Online tools for structural design

http://www.megamag.it : Magnetic brakes for fun rides

http://www.levitans.com : Air bearing pads

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

Compare a current-carrying conductor sitting on the inside surface of the airgap to a conductor sitting in a slot.

Integral H dot dl = Ienclosed.

The portion within the iron has H~0 because H = B / mu and mu is very large. So the portion of any contour within the airgap will be the same in both cases. The airgap flux pattern will be the same for the conductors on the inside surface as for the conductors in the slots.

If airgap flux pattern is the same for both cases, we conclude torque is the same for both cases in a number of ways:

1 - intuitively we know the power is transferred through the airgap by the flux pattern. If flux pattern is the same, power transferred through the airgap must be the same.

2 - we can apply maxwell's stress tensor. It will tell us mathematically that if the flux pattern is the same, the torque is the same.

3 - We can apply T = mmf_s * mmf_r * sin(delta_sr). where mmf_s = stator mmf, mmf_r = rotor mmf, and delta_sr is the angle between. If the currents are the same in both cases, the mmf's will be the same as shown above, and the torque will be the same. This also has the intuitive appeal of resembling a permanent magnet on the stator leading a permanent magnet on the rotor... a picture most people can relate to even if they are not familiar with electromagnetism.

The above is probably the simplest, most straightforward explanation I know of (and I have had it in my back pocket for awhile). But I am also trying to attack the problem from a number of directions to look at different proofs. By now means would I include the detailed derivation of how F=dW/dx applies to this case, I just want to satisfy myself that it is correct before I use it.

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

What you said 3 Mar 07 16:39 is based on the setup shown in Jan's proof part 2, we can deduce the flux in the conductor is mu_r below B0, and accordingly the force on the conductor lower by the same amount.

What I said 3 Mar 07 17:13 is a different angle that I intend to present in my article. For a very high permeability, the flux pattern in the airgap will be the same, regardless of whether the conductor is sitting in the gap (on the surface of the core) or embedded in a slot. So the torque in the two cases are the same.

Different parts of the same picture.

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