How to read the pull force of a magnet

[Gillhk]

I have a chart and a graph for a ferrite magnet grade 5 / Y30, but I do not understand them and want to know what is the pull force of the magnet and how to read it on these 2 documents?

[Sensei]

mT = mili Tesla unit

KGs = Kilo Gauss unit

KA/m = Kilo Ampere per meter unit

KJ/m³ = Kilo Joules per m^3 unit

KOe = Kilo Oersted

 

[Gillhk]

Thank you for the explanation. Do any of these columns show the pull force of the 20mm x 3mm  grade 5 ferrite disk magnet?

[Bender]

It depends in rather nonlinear ways on the distance to the object pulled and the material and geometry of that objects and the objects around it.

Even if all those are known, it is only possible to calculate the force with a numeric finite element model, except for some specific cases.

[Enthalpy]

The company that produces the magnets sometimes gives a pull force in the data sheets. If not, it gets complicated: no algebraic solution.

If the magnets are long and narrow, I've derived an algebraic solution that fits manufacturer's data rather well, there
https://www.scienceforums.net/topic/59338-flywheels-store-electricity-cheap-enough/?do=findComment&comment=876426

Do not trust formulas like B²S/µ, they are wrong.

[Bender]

On 23/7/2018 at 5:02 PM, Enthalpy said:

Do not trust formulas like B²S/µ, they are wrong.

No it's not.

It might not accurate, usually because B isn't constant over S and because B is not easy to calculate, it can get really complicated to integrate B over S. That does not make it wrong.

[Enthalpy]

21 hours ago, Bender said:

No it's not.

It might not accurate, usually because B isn't constant over S and because B is not easy to calculate, it can get really complicated to integrate B over S. That does not make it wrong.

You believe what you want. I don't need any more to experiment and learn on this topic. Hint: B²S/µ  or 0.5 B²S/µ might have been correct if B didn't depend on the gap between the magnets. This isn't the case with permanent magnets. My suggestion is that you measure magnet forces or search for manufacturer's data and check how wrong such formulas are.

[Bender]

57 minutes ago, Enthalpy said:

You believe what you want.

I don't need to. I designed and built a solenoid actuator for my PhD using this formula, and the experimental data matched my model nicely.

58 minutes ago, Enthalpy said:

Hint: B²S/µ  or 0.5 B²S/µ might have been correct if B didn't depend on the gap between the magnets.

The formula does not require B to be constant. Integrating B(x)dS/mu works great, if you know B(x). S or mu can even depend on the air gap, and it would still work.

1 hour ago, Enthalpy said:

This isn't the case with permanent magnets.

This isn't the case with electromagnets either. It just gets worse with permanent because they behave like an air gap themselves.

1 hour ago, Enthalpy said:

My suggestion is that you measure magnet forces

Exactly

1 hour ago, Enthalpy said:

search for manufacturer's data and check how wrong such formulas are.

Such data is pretty worthless to check the formula.

[Enthalpy]

If you orient the magnets to repel an other, you get a force, and it's nearly the same as for the attractive orientation despite the induction is zero.

So formulas based on B²S/µ are just b*llocks.

More subtle computations based on volume integration as a function of the magnet separation are not manageable by hand and are heavy to program. Integrating over the current sheaths is faster to program and to run, and in some cases it even gives an algebraic solution. Clean comparison.