I'm trying to solve a practical problem, as follows:

Three magnetic coils placed in a row, pole to pole. The left and right ones are separated by 5 centimeters and face opposite each other. The middle magnet has an arrow with the letter F attached to it, pointing to the right.

I want to make a system that consists of three electromagnets arranged as in the picture. The two side ones can’t move, but the middle one can. Now I want to know how many winds and how much current must flow trough each magnet in order to have a minimum force of about $10\,\mathrm N$ exerted on the middle magnet. There are however many variables and limitations that make it difficult for me to solve. I also can't find much information online on how these magnets interact.

The two magnets on the sides are oriented in opposite directions, one of them repelling and one attracting the middle one. The middle magnet is probably going to be smaller and weaker. I also assume that the minimum force is in the middle of the two magnets, at about $2.5\,\mathrm{cm}$ from each one.

What I do have is an approximate formula for the force of an electromagnet on a piece of iron in function of the distance, windings, current and area. However, does this reflect the force of three magnets placed together well? I tried to use this and got 1000 windings, $1.5\,\mathrm A$ and a circular area $5\,\mathrm{cm}$ in diameter. However, this would require $170\,\mathrm m$ of copper wire. So the costs would be a bit much. It also seems a bit off.

In case you are wondering, the field will oscillate to make the middle magnet go back and forth. Therefore I also assume that a permanent magnet will quite quickly lose its strength.

What am I missing and are there any pointers you can give me so I can make a better approximation to get materials?


closed as off-topic by David Z Jan 4 at 4:47

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  • $\begingroup$ See kjmagnetics.com/calculator.asp?calcType=block $\endgroup$ – Alex Trounev Jan 3 at 22:58
  • $\begingroup$ Welcome to Physics.SE! Your question is precise and well-written, but I think it might not be on-topic for this particular SE site, but rather for one of the sister sites, like EE.SE. I’ve gone ahead and asked the moderators to look at it; if they agree, it will be migrated there, so you won’t need to ask it anew. $\endgroup$ – Alex Shpilkin Jan 3 at 23:06
  • 1
    $\begingroup$ @AlexShpilkin Don't migrate this. The current forum is the right one for this question. $\endgroup$ – user5713492 Jan 4 at 1:21
  • $\begingroup$ I don't see this as a stable situation. What prevents the central weaker magnet from joining the larger one which attracts it? $\endgroup$ – Dr S T Lakshmikumar Jan 4 at 1:47
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    $\begingroup$ I don't think this is on topic at Electrical Engineering, so I'm going ahead and putting it on hold rather than migrating it. But I'll check with people on that site. $\endgroup$ – David Z Jan 4 at 4:48

I am going to assume that the central magnet is attached to some kind of rails so that instead of having $6$ degrees of freedom (roll, yaw, pitch, $x$, $y$, and $z$) it only has one, $z$, assumed horizontal in your diagram. Also I am going to assume symmetry about the $z$-axis.
I think neodymium-iron-boron magnets are supposed to be quite insensitive to demagnetization unless the strength of the demagnetizing field is well over $1$ Tesla, which condition should be easily satisfied for your application. Thus we may consider using such a magnet.

The analysis would be simpler for a spherical central magnet, but these things are more awkward to mount. The simplest geometry of cylindrical magnet is one where the ratio of height to diameter is $\sqrt3:2$ but since that would be a custom magnet I select a $1"\times7/8" \text{N}50$ magnet. If we compare eqs. $(7)$, $(8)$, and $(9)$ of this paper we find that, for a magnetic dipole $m_z$ aligned in the $z$-direction on the $z$-axis, the magnetic force is $$\vec F=m_z\frac{\partial{B_z}}{\partial z}\hat k$$ Note that adding the forces due to $2$ magnets on a piece of iron would not work because the magnetic fields of the $2$ outer magnets would cancel out so the piece of iron would not be magnetized $m_z=0$ so there would be no force even though there is a magnetic field gradient.

Assuming constant magnetization vector $\vec M=M\hat k=\frac{B_r}{\mu_0}\hat k$ and using $B_r=1.43\,T$ for an $\text{N}50$ magnet from this table we get $$m_z=\frac{1.43\,\text{T}}{4\pi\times10^{-7}\,\text{T}\cdot\text{m}/\text{A}}\times\frac{\pi}4(1\,\text{in.})^2\left(\frac78\,\text{in.}\right)\left(\frac{0.0254\,\text{cm}}{\text{in.}}\right)^3=12.8\,\text{A}\cdot\text{m}^2$$ Then we want $$F=2\times12.8\,\text{A}\cdot\text{m}^2\frac{\partial{B_z}}{\partial z}=10\,\text{N}$$ Where the factor of $2$ is there because there are $2$ outer magnets. Thus we need $$\frac{\partial{B_z}}{\partial z}=0.390\,\text{T}/\text{m}$$ for each electromagnet.

Now, the problem would be solved if you knew how to find this gradient at a distance of about $1$ in. from the surface of your electromagnet, but this could be a bit of a complicated calculation because the magnetization of the core varies in a more or less unpleasant way with the current in the coil.

Let's start with this electromagnet which is supposed to have a pull force of about $55$ lb and a diameter of about $1.5$ in.. For comparison, this model magnet has a pull force of $42.10$ lb and a diameter of $1.5$ in. so, although the pull force of this magnet might not be measured the same way as for the electromagnet I am going to assume it is.

Then if we go to this force calculator and enter the parameters for the magnet in question and a distance of $1$ in. we get a force of $1.25$ lb which would be great if we can assume the magnetic dipole approximation is OK for this geometry considering that the actual surface-to-center distance is more like $1.125$ in. and the magnetic dipole moment of my candidate magnet is about half again that of this model magnet.

So I think that my candidate magnet with $2$ of my cited (or perhaps even smaller) electromagnets may be sufficient. For about $\$60.00$ you could buy them and lets us know whether this is so.

Oh yeah, you should put stops in front of the outer electromagnets because that central magnet will surely break if it slams into one of them.


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