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How to calculate the force applied by a magnetorquer on a piece of a perfect ferromagnetic material at a particular distance in vacuum given that I know the actuatuon (in Am2)

https://www.isispace.nl/product/isis-magnetorquer-board/

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  • $\begingroup$ Calculation wont be easy, this question would fit better into physics group. $\endgroup$
    – Uwe
    Nov 30, 2021 at 15:47
  • $\begingroup$ @Uwe can you check the answer provided by me? $\endgroup$ Nov 30, 2021 at 16:08

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That's a neat piece of equipment - almost makes me want to build a cubesat!

I'm not sure what's meant by a 'perfect' ferromagnetic material. Is is hard or soft? Are we talking about something in the vicinity of the coils, or far away?

If you're far away (compared to the dimensions of the coils), you can just assume the field from the equivalent dipole. The term 'actuation' apparently refers to the maximum magnetic dipole moment of ~$0.2$ $A.m^2$ of the coil at full current.

If you're in the near field, it's more difficult. I see there are two long coils with ferrite cores, so the field from those can be approximated by an isolated magnetic charge at each end of the rod. It sounds like the third coil is a large flat air-cored rectangular coil and I don't think there's an easy analytic expression for the field unless you happen to be near the coil axis.

If the small piece of ferromagnetic material if hard (a permanent magnet), the force goes as the dot product of its magnetic moment times the gradient of the field from the coil.

If the small piece of ferromagnetic material is very soft (high permeability, no hysteresis, no saturation), then there will be an induced magnetic moment proportional to the field from the coil and the volume of the ferromagnetic piece. In this case the resulting force is proportional to both the field magnitude and the field gradient.

It's useful to remember that force is field gradient in $A/m^2$ times magnetic moment in $A.m^2$ times the permeability of free space (which has units of newtons per square ampere).

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I found the answer. Force on a ferro-magenetic piece of material placed at distance s from pole of the electromagnet with actuation M carrying current I is given by

F = (M*I*u)/(2*(s^2))

Here, u= magnetic permeability of the medium, e.g. in vaccum u = magnetic constant

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  • $\begingroup$ I dont believe this formula is applicable to your question. The force does not depend on the size of the ferro-magenetic piece of material, the force would be the same for a tiny and a large piece of material. $\endgroup$
    – Uwe
    Nov 30, 2021 at 16:23
  • $\begingroup$ You are right and as per the formula force is not function of size or mass of the ferromagnetic material. Why do you think the formula states force is function of the size of the ferro-magenetic material? The s is distance not size. Distance between pole of the magnet and the ferro-magenetic material. $\endgroup$ Nov 30, 2021 at 16:30
  • $\begingroup$ @Uwe please check the comment $\endgroup$ Nov 30, 2021 at 16:59
  • $\begingroup$ If the force would not depend on size or mass, an infinitesimal small piece would get the same force as a large piece. Where did you find that formula? Do you have a web source? $\endgroup$
    – Uwe
    Nov 30, 2021 at 17:53
  • $\begingroup$ @Uwe yes, small and big mass of ferro-magenetic material both will get exerted by same amount of force but the acceleration will be inversely proportional to their mass. $\endgroup$ Nov 30, 2021 at 17:55

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