# Magnetic pull on iron object

• B
• liquidFuzz
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.

#### liquidFuzz

I have a scenario where I would like to calculate the electronics needed to create a force (pull on) an iron object. The set up is a coil (solenoid) with a steel (iron) object. I use Maxwell's equation (Ampere's law) to estimate the magnetic field B of the solenoid.

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

liquidFuzz said:
I have a scenario where I would like to calculate the electronics needed to create a force (pull on) an iron object. The set up is a coil (solenoid) with a steel (iron) object. I use Maxwell's equation (Ampere's law) to estimate the magnetic field B of the solenoid.

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

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), H, throughout the iron object and the magnetic permeability of the iron object (##\mu##):

##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

vanhees71
Andrew Mason said:
If they are close so that the magnetic fields are uniform between them over a surface area A,
If the field is uniform, there is no force on a piece of iron. The force is proportional to the gradient of the field.

rude man and vanhees71

?? 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

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

vanhees71
I am not sure what you are saying. A magnetic field can be locally uniform and still produce a magnetic force. For example, a current-carrying conductor placed in a uniform magnetic field will experience a magnetic force according the Lorentz force: F = I L×B

AM

vanhees71
Andrew Mason said:
I am not sure what you are saying. A magnetic field can be locally uniform and still produce a magnetic force. For example, a current-carrying conductor placed in a uniform magnetic field will experience a magnetic force according the Lorentz force: F = I L×B

AM
Technically no. The "currents" in a ferromagnet all form closed loops. In a uniform field any "pull" is always counterbalanced by a "push"because the current loop is closed. Torque perhaps but no force net force.
In practice the field is never globally uniform and there is always a gradient that produces a net force. But magnet design is tricky

If the field is uniform, there is no force on a piece of iron. The force is proportional to the gradient of the field.
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.

Just to clarify:

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.

vanhees71
Yeah, a freeomagnet is just lkke an ordinary magnet, only cheaper.

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russ_watters, jtbell, vanhees71 and 2 others
Andrew Mason said:
So, first of all, you have to determine the magnetic flux induced in the iron object.
Oh no, no, no! By the pendulous yarbles of Zeus, no!
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 $\vec{B}$ level answer - best I can do. The energy density in a magnetic field in vacuum is $B^2$. 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:

$$U=-\int_V B^2 dV$$

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

$$F=\nabla \int_V B^2 dV$$

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.

rude man and berkeman
If you integrate a field over a spatial volume, the ##\vec{x}##-dependence is gone. From this calculation you thus get ##F=0## always.

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

malawi_glenn
If I wasn't clear, V is the volume of the object in the field. Basically, you treat the object as a hole in the magnetic flux density. Unfortunately, while the ideas are B-level, or at least lower I-level, the actually calculations are horrible if you can even do them at all.

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.

DaveE and vanhees71
Well, I still don't understand your notation then. ##\int_V \mathrm{d}^3 x \vec{B}^2(t,\vec{x})## is independent of ##\vec{x}## and thus its gradient is 0.

V is the volume of the object in the magnetic field, so U is a measure of the magnetic flux that would be there is the iron were not. If ir's a bolt, V is a bolt-shaped volume. If it's a nail. V is a nail-shaped volume. And so on.

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: $\vec{B} \approx (0,0,B_0/z^3)$. 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,z0). The energy removed (in appropriate units) is

$$B_0^2 \pi r^2 \int_{z_0 - L/2}^{z_0 + L/2} dz/z^6 = -5B_0 ^2\pi ^2 \left[ \frac{1}{(z_0+L/2)^5} - \frac{1}{(z_0-L/2)^5}\right]$$

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

Turning to the force (and replacing z0 with z for clarity):

$$F = \frac{\partial}{\partial z} \frac{5VB_0^2}{2z^6} = \frac{5VB_0^2}{12z^7}$$

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

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In iron it is effectively zero (let's assume a perfect ferromagnet and not worry about1% or less corrections,)
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 energy is less 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

hutchphd said:
The B field inside the material is not zero. In fact it is made larger by the iron.
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:
But we agree it ain't pretty
That's for sure!

hutchphd
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: $\vec{B} \approx (0,0,B_0/z^3)$. 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,z0). The energy removed (in appropriate units) is

$$B_0^2 \pi r^2 \int_{z_0 - L/2}^{z_0 + L/2} dz/z^6 = -5B_0 ^2\pi ^2 \left[ \frac{1}{(z_0+L/2)^5} - \frac{1}{(z_0-L/2)^5}\right]$$

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

Turning to the force (and replacing z0 with z for clarity):

$$F = \frac{\partial}{\partial z} \frac{5VB_0^2}{2z^6} = \frac{5VB_0^2}{12z^7}$$

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:
1. Think energetics and not forces; if you want forces, derive them from energies.
2. An actual calculation is in no way B-level.
Shouldn't there be a factor ##1/2## (for theoretical physicists) or ##1/(2 \mu_0)## (for experimental physicists)?

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 energy is less 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
Shouldn't the ##\chi_m## be inside the integral (under the gradient operator)?

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

vanhees71