As I was trying to find an answer to another problem, I was informed that the equation I was using would not work.

The equation I was using was $$F = (NI)^2\mu_0\frac{\text{Area}}{2g^2}$$

If this is indeed wrong, could someone please explain how it is wrong? I have spent some time trying to figure out what's wrong, and I can't seem to find the issue.

It seems to me, like my only problem is the one I asked about in my original question.

TIA, to anyone who can clarify this problem.

Edit: I may have just solved my own problem. I think I may have written the equation in my other question wrong. Looking at it, I don't know how I missed this before, it looks like I wrote it as $$F = (NI)\mu_0\frac{\text{Area}}{2g^2}$$ and not the correct equation of $$F = (NI)^2\mu_0\frac{\text{Area}}{2g^2}$$ While I believe this is correct, could someone please verify? I am going to be rearranging it to find the number of coils I require in my solenoid, and I really don't want to get this wrong.

Edit: So, in order to help this along, I am going to do an example. As an aside, I was asking my other question because I was getting some funny numbers out of this when I used a value of Area taken from plane b.


So, if I am presented with a cylinder 25mm long, and 5mm around, and I am asked to find the number of turns a solenoid that exerts a total force of 1 Newton/second on the cylinder at a distance of 5mm, and that uses 3 Amperes current must have, I would start by rearranging my solenoid force equation.

$$ \begin{align} F &= (NI)^2\mu_0\frac{\text{Area}}{2g^2} \\[10px] % \frac{F}{\mu_0} &= (NI)^2\frac{\text{Area}}{2g^2} \\[10px] % \frac{F}{\mu_0\frac{\text{Area}}{2g^2}} &= (NI)^2 \\[10px] % \sqrt{\frac{F}{\mu_0\frac{\text{Area}}{2g^2}}} &= NI \\[10px] % \frac{1}{I}\sqrt{\frac{F}{\mu_0\frac{\text{Area}}{2g^2}}} &= N \\[10px] \end{align} $$

Where F is the force exerted on the cylinder in Newtons, N is the number of turns, I is the current passing through the coil, μ₀ is 4π x 10^-7, Area is is the area of the cylinder in plane b(See photo), and g is the distance separating. the cylinder from the coil.

Example of planes passing through a cylinder:

$$ \begin{align} \sqrt{\frac{(1)}{(4π · 10^{-7})\frac{\text{Area}}{2(15)^2}}}/(3) &= N \\[10px] \sqrt{\frac{1}{4π · 10^{-7}·\frac{\text{(19.634375)}}{2(15)^2}}}/3 &= N \\[10px] \end{align} $$

And we know area is about $19.634375$ because area is $\pi r^2$ and the radius was $2.5 \, \mathrm{mm}$.

$$ \begin{align} \sqrt{\frac{1}{4π · 10^{-7}· 19.634375}}/3 &= N \\[10px] \sqrt{\frac{1}{0.000024673283}}/3 &= N \\[10px] \sqrt{40529.66927830399}/3 &= N \\[10px] 201.3198183942753/3 &= N \\[10px] N &= 67.1066061314251 \\[10px] N &= 67.1~\text{Turns} \end{align} $$

I don't believe I missed anything but if I did something wrong, please tell me. I hope this helps determine if this is the correct equation or not.

Edit: I don't suppose anyone can check my math?

  • $\begingroup$ The source of your formula should explain its derivation and limitations. $\endgroup$
    – R.W. Bird
    May 11, 2021 at 14:00
  • $\begingroup$ @R.W.Bird IIRC, "should" is very different than "does." $\endgroup$
    – CoilKid
    Oct 21, 2021 at 23:30
  • 1
    $\begingroup$ Dimensionally your formula is correct (once one realises that $g$ is a distance). But how can the force not depend on the relative permeability of the ferromagnetic material (or its thickness)? It's also fishy that the the length of the solenoid isn't involved. I think that to derive a useful formula (even an approximate one) would be difficult. I'd do some experiments! $\endgroup$ Jan 9, 2022 at 0:25
  • $\begingroup$ Or... you could use magnetostatics software and get the correct answer... $\endgroup$ May 19, 2023 at 3:38

1 Answer 1


The force on a magnetic moment will depend on the field gradient. For a magnetic dipole on axis with a thin coil, the problem is fairly easy to solve. (Assume axis is in the z-direction.)

  1. Calculate the $B$ field from a thin coil as a function of $z$.

  2. Differentiate with respect to axis direction ($z$) to find the field gradient.

  3. Calculate force, $F = u\frac{dBz}{dz}$.

The force will depend on the distance between the coil and magnetic moment.
(A magnetic moment in a uniform $B$ field experiences no force - only a torque.)

  • $\begingroup$ I don't know if you are or not. I was looking for the force in Newtons a solenoid(electromagnet) would exert on a piece of steel(ferromagnetic material). I had assumed the equation to be at least close to what I was looking for, as it operates using the current, number of turns, area of something(my other question was about it was the area of), and distance from coil. It seems logical that anything involving the force exerted by a solenoid would require inputs for those values. While it sounds reasonable, I don't know if it is correct. If it helps, I could put an example edit in. $\endgroup$
    – CoilKid
    Aug 21, 2014 at 20:40
  • $\begingroup$ Yes, I think I'll work out an example of my equation in an edit. Then perhaps someone could tell me if I missed anything? $\endgroup$
    – CoilKid
    Aug 21, 2014 at 20:46
  • $\begingroup$ Hmm, well when I write down an equation, the first thing I like to do is check the limits. If g is the distance, then I would expect the force to go to zero when the two dipoles are centered. For the case of an induced dipole, I think the theory says that the force should go as the sixth power of the distance...(when the centers are well separated.) but that assumes all sorts of linear behavior, real life may be different. $\endgroup$ Aug 22, 2014 at 1:29
  • $\begingroup$ I honestly don't know the theory really well. If it helps, I can cite the sources of my equation. (I would have to spend some time looking over my browser history, but hey. That's why its there) $\endgroup$
    – CoilKid
    Aug 22, 2014 at 1:33
  • $\begingroup$ I don't know if it helps, but I can cite my sources for my equation. Source 1, Source 2, Source 3 - I don't know if this one is accurate. It says Area is the cross sectional area of the coil. That doesn't really make sense, because you are using the current and number of turns already, Source 4 $\endgroup$
    – CoilKid
    Aug 22, 2014 at 1:47

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