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My question is, how to estimate the force on an iron object that is affected by the magnetic field.

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In summary, to calculate the force on an iron object that is affected by the magnetic field, you first have to determine the magnetic flux induced in the object and then use Maxwell's equation to calculate the force.

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My question is, how to estimate the force on an iron object that is affected by the magnetic field.

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It is rather complicated. The pull between the solenoid and the iron object is the result of the interaction of the magnetic field of the solenoid and the induced magnetic field of the iron object. (The induced magnetic field in the iron object being the result of the alignment of the magnetic dipoles in the iron when placed in the solenoid field).liquidFuzz said:

My question is, how to estimate the force on an iron object that is affected by the magnetic field.

So, first of all, you have to determine the magnetic flux induced in the iron object. That field is a product of the strength of the magnetic field strength (of the solenoid),

##B_{induced}=\mu H##

But to find the force between the solenoid and iron actuator is not trivial. If they are close so that the magnetic fields are uniform between them over a surface area A, it should be just ##\mu H^2A^2##. But it would be much simpler just to measure it for a given solenoid, current and iron actuator.

AM

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If the field is uniform, there is no force on a piece of iron. The force is proportional to the gradient of the field.Andrew Mason said:If they are close so that the magnetic fields are uniform between them over a surface area A,

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?? I was referring to the field in the centre between the N and S poles. The left magnet represents the solenoid and the right magnet is the iron object with the induced magnetic field. It is somewhat analogous to electric charges in a parallel plate capacitor.

AM

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Is not uniform.Andrew Mason said:the field in the centre

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AM

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Technically no. The "currents" in a ferromagnet all form closed loops. In aAndrew Mason said:F =IL×B

AM

In practice the field is never globally uniform and there is always a gradient that produces a net force. But magnet design is tricky

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Yes. Think of an iron slug attracted into a solenoid. Where does it "want" to go? Right in the center where the field is strong but the gradient is balanced between the two directions. There's a symmetry argument that says it has to be this way, if your iron is equally attracted to both positive and negative magnetic poles.Vanadium 50 said:If the field is uniform, there is no force on a piece of iron. The force is proportional to the gradient of the field.

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Is the issue the effect of a uniform magnetic field on a freeomagnet?

Or is there a question about whether there is a force proportional top a gradient.

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TIL a new word!Vanadium 50 said:the effect of a uniform magnetic field on a freeomagnet?

https://www.google.com/search?clien...vDEkQ8BYoAXoECAEQNg&biw=1122&bih=512&dpr=1.09

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Yeah, a freeomagnet is just lkke an ordinary magnet, only cheaper.

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Oh no, no, no! By the pendulous yarbles of Zeus, no!Andrew Mason said:So, first of all, you have to determine the magnetic flux induced in the iron object.

While that will eventually get to the answer, it is the path of maximum horribleness.

First, there is no B-level answer. Sorry.

There is a [itex]\vec{B}[/itex] level answer - best I can do. The energy density in a magnetic field in vacuum is [itex]B^2[/itex]. In iron it is effectively zero (let's assume a perfect ferromagnet and not worry about1% or less corrections,) So the energy difference in the field is:

[tex]U=-\int_V B^2 dV [/tex]

The force on the object is the gradient of the potential energy:

[tex]F=\nabla \int_V B^2 dV [/tex]

That's the answer. Unfortunately, except in rare cases of exceptional symmetry, there is little more to be done here.

As an aside, note how if B is constant, F = 0.

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Perhaps we should first find out what the question really is?

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I don't think we're going to see the OP clarify the question. He hasn't been back since posting. But I think he's clear enough: he wants to know how much current he needs in a solenoid valve to open and close it, given all the other parameters. Calculating that is really, really hard.

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If a specific example will help: Suppose I have a thin cylinder-shaped piece of iron of length L and radius r constrained to move in z in a field that is dominantly in z: [itex]\vec{B} \approx (0,0,B_0/z^3)[/itex]. This is kind of like a dipole field far from the dipole. The field in the x and y directions is not exactly zero here - that doesn't follow Maxwell's Equations, but this doesn't matter because of the symmetry and the constraint:

Call the point the center of the cylinder (0,0,z_{0}). The energy removed (in appropriate units) is

[tex]B_0^2 \pi r^2 \int_{z_0 - L/2}^{z_0 + L/2} dz/z^6 =

-5B_0 ^2\pi ^2 \left[ \frac{1}{(z_0+L/2)^5} - \frac{1}{(z_0-L/2)^5}\right][/tex]

[tex]\approx 5B_0^2 \pi^2 \frac{L/2}{z_0^6} = \frac{5V B_0^2}{2z_0^6}[/tex]

Turning to the force (and replacing z_{0} with z for clarity):

[tex]F = \frac{\partial}{\partial z} \frac{5VB_0^2}{2z^6} = \frac{5VB_0^2}{12z^7} [/tex]

I'm sure I did some screwing up here, but the point is that this is calculable, but even in an oversimplified case, it's far from trivial. In reality the dependence is not this strong - it's only the 6th power of z. The reason is that if I had written a field that satisfied Maxwell's equations the amount of field eliminated is smaller. The field lines that were along z turn in x and y - they don't just disappear.

But the two points remain:

Call the point the center of the cylinder (0,0,z

[tex]B_0^2 \pi r^2 \int_{z_0 - L/2}^{z_0 + L/2} dz/z^6 =

-5B_0 ^2\pi ^2 \left[ \frac{1}{(z_0+L/2)^5} - \frac{1}{(z_0-L/2)^5}\right][/tex]

[tex]\approx 5B_0^2 \pi^2 \frac{L/2}{z_0^6} = \frac{5V B_0^2}{2z_0^6}[/tex]

Turning to the force (and replacing z

[tex]F = \frac{\partial}{\partial z} \frac{5VB_0^2}{2z^6} = \frac{5VB_0^2}{12z^7} [/tex]

I'm sure I did some screwing up here, but the point is that this is calculable, but even in an oversimplified case, it's far from trivial. In reality the dependence is not this strong - it's only the 6th power of z. The reason is that if I had written a field that satisfied Maxwell's equations the amount of field eliminated is smaller. The field lines that were along z turn in x and y - they don't just disappear.

But the two points remain:

- Think energetics and not forces; if you want forces, derive them from energies.
- An actual calculation is in no way B-level.

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I agree that energetics are a good way to do this but I don't think (as you note!) the analysis is correct.Vanadium 50 said:In iron it is effectively zero (let's assume a perfect ferromagnet and not worry about1% or less corrections,)

The B field inside the material is not zero. In fact it is made larger by the iron. The associated field energy

Assume for sanity that the the ferromagnet is simply linear. Then the Magnetic energy in the field everywhere can be written $$\frac 1 2 \mathbf B \cdot \mathbf H $$ and the force on the linear material is then $$\mathbf F =\frac {\mu \chi_m} 2 \int \nabla (H^2) $$ provided there are no free currents in the linear material.

But we agree it ain't pretty

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But the energy in the field is. The energy per field line is down by a factor μ/μ0 - maybe 1000. So what I actually calculated was the energy in the field in the space that would be there if the iron were not.hutchphd said:The B field inside the material is not zero. In fact it is made larger by the iron.

That's for sure!hutchphd said:But we agree it ain't pretty

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Shouldn't there be a factor ##1/2## (for theoretical physicists) or ##1/(2 \mu_0)## (for experimental physicists)?Vanadium 50 said:

Call the point the center of the cylinder (0,0,z_{0}). The energy removed (in appropriate units) is

[tex]B_0^2 \pi r^2 \int_{z_0 - L/2}^{z_0 + L/2} dz/z^6 =

-5B_0 ^2\pi ^2 \left[ \frac{1}{(z_0+L/2)^5} - \frac{1}{(z_0-L/2)^5}\right][/tex]

[tex]\approx 5B_0^2 \pi^2 \frac{L/2}{z_0^6} = \frac{5V B_0^2}{2z_0^6}[/tex]

Turning to the force (and replacing z_{0}with z for clarity):

[tex]F = \frac{\partial}{\partial z} \frac{5VB_0^2}{2z^6} = \frac{5VB_0^2}{12z^7} [/tex]

I'm sure I did some screwing up here, but the point is that this is calculable, but even in an oversimplified case, it's far from trivial. In reality the dependence is not this strong - it's only the 6th power of z. The reason is that if I had written a field that satisfied Maxwell's equations the amount of field eliminated is smaller. The field lines that were along z turn in x and y - they don't just disappear.

But the two points remain:

- Think energetics and not forces; if you want forces, derive them from energies.
- An actual calculation is in no way B-level.

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Shouldn't the ##\chi_m## be inside the integral (under the gradient operator)?hutchphd said:I agree that energetics are a good way to do this but I don't think (as you note!) the analysis is correct.

The B field inside the material is not zero. In fact it is made larger by the iron. The associated field energyisless for a given value of B because of the iron.

Assume for sanity that the the ferromagnet is simply linear. Then the Magnetic energy in the field everywhere can be written $$\frac 1 2 \mathbf B \cdot \mathbf H $$ and the force on the linear material is then $$\mathbf F =\frac {\mu \chi_m} 2 \int \nabla (H^2) $$ provided there are no free currents in the linear material.

But we agree it ain't pretty

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Oh, probably.vanhees71 said:Shouldn't there be a factor (for theoretical physicists) or (for experimental physicists)?

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