Electromagnet design 1

[jpysam]

Hi,

I'm too new to magnetic things but I need to make an electromagnet to measure efficiency of my gearbox (electromagnetic brakes). I've done the calculations so far but there is a question on my mind.

So, I use 1018 mild steel for core material (best available option for me)
24 awg(0.51mm) copper wire for coil
1600 turns
0.5 Amp (max)
0.025m core length

In that case H is about 32000 [A-T/m] and the permeability for the 1018 is about 36.5 [H/m] and the magnetic flux density is 1.46 T. (I've got the values from a graph)

1- If I use 36 mm of magnetization surface( there is no any hole on the surface, full section, area of the surface is 0,00101 m2), the pulling force is about 88kg to my calculations.

2- If I use a hollow surface with the same out diameter, mean if there is a hole of 12 mm across the magnet length on the center (di=12mm, do=36mm, area of this one is 0,000904 m2), pulling force is 78 kg.

Obviously the area decreased, so was the force.

What I'm curious is that, because I drilled a hole in the magnet will there any increase at the flux density? And if will, might the core saturate earlier from 1.6 Tesla? (saturation point is about 1.6 T for 1018)

I heard the saturation point does not depend on the surface area from somewhere, but for example, two materials with the same outer diameter of 50 mm, one with a full cross-section and the other with an empty cross-section, (the empty one is like the same ring, for example with a wall thickness of 2 mm) when the same H is applied to these magnets, will the same amount of magnetic flux be induced in both? or will it have more magnetic flux induced because there is less material in the ring? I'm missing something and this really bothers me.

Thank you.


EDIT**

So, I just went deeper and saw that the magnetic circiut law and that says MMF= flux * Reluctance (flux= B* Area, Reluctance=Length/(mu*area))

MMF=B*Area*L/(mu*Area) Appearently, mmf doesnt depend on area. And if we think physically, mean with the consant mmf, when we remove A of area from the magnetization part, normally the magnetic flux density should increse but at the same time the reluctance of the core material increases at the same ratio and so in the end, B is always constant.

I will be glad if someone confirm me.

 

[MagMike]

I modeled the first design in FEA and calculated 1.7 kg pulling force.

For pulling forces, a single steel cylinder and coil is not the best geometry. One suggestion is to use a steel cylinder in a steel cup with a coil in between. These designs are much more efficient and there are a lot of standard designs available.

Here is an image of a typical design:

 

[jpysam]

Hi Mike, thank your for reply.

Of course, I made a design like the picture you posted, with a magnetized part in the middle and a steel wall on the outside.

I think there is a mistake in your calculation or I said something wrong. Because I lifted 10-15 kg even with only 220 windings and 0.5 amps. I literally machined it on the lathe and tried it. (I used 430 Steel at that time, but in the end it is the B value that matters)

By the way, so you think that am I right in what I wrote above in my 1st text or am I making a mistake?

 

[MagMike]

Jpysam,

Sorry, it sounded like you were designing a simple cylinder electromagnet. I didn't realize it was a round cup design. That said, the numbers for Case 1, with 88 kg appear reasonable, maybe a bit optimistic.

I don't understand what you are describing in Case 2, can you provide a sketch/picture?

 

[jpysam]

2.case is that there is a hole in the magnetized area. What I wonder in this case is written in the first post. Could you please take a look there. Thank you

 

[MagMike]

Jpysam,

In Case 2, if everything else stasys the same, the inner core will saturate earlier since it has less steel involved in the magnetic circuit. The total mmf will likely be less because of this.

 

[jpysam]

my first thought was in this direction, but I think the mmf should remain constant.

mmf=flux*reluctance (flux=B*area, reluctance=L/(mu*area))
mmf=B*area*(L/(mu*area))

so, yes I removed some A of area in the material and the total area decreased but also because of this, the reluctance of the magnetic circuit increased and these are at the same ratio, right?

And so in the end if the mmf is still the same, B has to be same, should stay the same.

and if B depended on the area, besides the material I think those B-H graphs would have to change according to each diameter.

 

[hacksaw]

There are four diameters involved, and don't forget it is not a linear problem.

 

[jpysam]

Hey hacksaw,

Could you pls explain your said, I didn't get it exactly, am I thinking wrong?