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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[electricpete]

Given:
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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[electricpete]

In my last post, I forgot to state all my symbols. I will repost with it cleaned up a little to define the symbols.

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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[hacksaw]

pete, thanks for the explanation, I was worried that my mu-metal shilds weren't going to work

 

[electricpete]

Hacksaw – I think you are saying that it’s obvious that the force in a motor doesn’t act on the conductor because the conductor is shielded from flux by the slot.

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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[prex]

electricpete,
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 B₀. 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 B₀. The body has a slot in it: it is a hole with rectangular section, one side parallel to B₀, 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 B₀/[μ]_r and in the slot it is proportional to B (for paths of equal lengths).
Hence in the slot B=B₀/[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=I₀B₀/[μ]_r, and the force on the shield F_cs=I₀B₀(1-1/[μ]_r).

prex

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

 

[electricpete]

yes, I agree 100%.

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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[electricpete]

On second reading, your discussion 3 Mar 07 16:39 is a little different than what I said 3 Mar 07 17:13.

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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