Electromagnetism

[Trisorion]

I am trying to make the strongest electromagnet using two AA batteries. This means low voltage (1.2 volt in parallel, .03 ohm internal resistance, 40 amps maximum discharge when in parallel, .7 volts at full discharge). When I first started, I thought it was simple, just pile on as many turns as I can with as much current as I can. Then I heard about magnetic saturation. After reading the wikipedia article I am not sure what to do. **broken link removed**).

The magnetic saturation wikipedia article makes it seem like the physical size of the core does not matter. It makes it seem that it is just flat out impossible to get above 1.8 magnetic force regardless of what size or shape the core is. Doesn't the length of the core or it's cross sectional area matter?

I am limited to core materials of: (their permeability can be found in the wikipedia permeability article, but im not sure how that factors in)
Silicon Grain orient steel
Low Carbon Steel

I can use any size wire, although with such low voltage and high maximum discharge, the lower AWG range seems best because of it's low resistance. More facts like turns per inch and resistance of the wire can be found on the AWG wikipedia article.

If someone could help me understand the optimum configuration for the strongest electromagnet using these materials, I would appreciate it. (For example, is a higher permeability core good or bad? Is a longer core good or bad or neutral? etc.-- Basically, how does each property of an electromagnet effect the strength?)

If there is a maximum limit to the strength of the magnet, what is the most efficient way to achieve this maximum? (For example, maybe I only need 12 amps to achieve that)

 

[MrAl]

Hi there,


It's basically an optimization search, where you try to find the
best wire diameter based on how much of it will fit in the window
and the equations for how the current and number of turns
affects the magnetic force. More about this in a minute.


Yes, the flux density is limited in materials because when saturation happens
all of the magnetic domains in the material have been aligned with the field
and there just arent any more for the field to work with. It's like
having a whole table full of pencils where they cover the whole surface,
and one by one you rotate them until the tips all point the same way. Once
you have rotated all of them in this fashion there just arent any more
to rotate and that's the end of it.
But this doesnt mean the total force is limited in this way because
the dimension of force isnt limited to that one dimensionality.
The solution is to simply add another table. This gives us another set to
work with, and the more that are aligned the better.
To put this in proper perspective, the total flux is limited in any
material and so the force is limited, but only in that one dimensional
sense. By looking across the area, we find that the greater the area
the more flux lines we can create. In the real world this means adding
to the cross sectional area.

Just to summarize a little:
In a given material the saturation flux density is limited, but the
force is not because when we talk about flux density we havent stated
over what area this is, and although doubling the area (adding a second
core physically in parallel) does not increase the saturation level,
it does increase the total force available at the poles by 2.
Thus, the same material only two times the volume of it gives us
two times the total force available.

Now rather than go through the whole story this way, here is an equation
to look at:

F=K*I^2*N^2*A/(g+d/um)^2

This is an estimate of the force (F) available at the poles of an electromagnet.
It relates current (I), cross sectional area (A), the number of turns (N),
the length of the gap (g), the length of the magnetic core (d), and the
initial permeability of the core (um). K is a constant that never changes,
or at least not with this approximation.

To put it into words:

1. The force increases by increasing current I.
2. The force increases by increasing the number of turns N.
3. The force increases by increasing the cross sectional area A.
4. The force decreases by increasing the gap length g.
5. The force drcreases somewhat by increasing the core length d.
6. The force increases somewhat by increasing the permeability of the core.

To get max force the idea then is to try to increase everything that increases
force and decrease everything possible that decreases force, but we must
also so this in a controlled algebraic way.

There are a few other things that come into play though if we were to
want to build a coil optimized for maximum force. One would be the
way resistance increases with the number of turns and how big the
wire diameter should be. One point to consider is that the
force goes up as the square of the number of turns, but goes down
by the number of turns of the same wire diameter times a constant.
What this means is that to get a quick idea how this works we would
calculate the number of turns for a wire diameter that will fit on
the core, and then the resistance and hence the current (and we know
force goes up with the square of the current). Then calculate
for another wire diameter and see if the current went up or down.
If it went up, we try changing the wire diameter the same way we
did before and recalculate, but if it went down we reverse this
procedure. Eventually we would find a wire diameter that
is optimized for the amount of it that will fit on the core.
Of course this assumes we have enough wire to begin with if
the core is very large.



OPTIMIZATION OF FORCE

To put this into an equation we can start with this:

F=K*I^2*N^2*A/(g+d/um)^2

and since we wont be changing anything but I and N we lump everything else:

F=K2*I^2*N^2

and since I is related to the battery voltage V and the resistance R
and cell internal resistance Rb:

F=K2*(V/(R+Rb))^2*N^2

We would then calculate R knowing the wire diameter & length and
resistivity of copper, and then search for an optimum value of this
wire diameter subject to the constraint of the core window area.
We could approximate the wire diameter to be a square for the purpose
of satisfying the window area constraint, or work with layers of
circle groupings to get a little more accuracy. Since the voltage
will be low we probably dont need any tape but it helps anyway to
provide some volume for the wire to expand with temperature rise.
Approximating the wire as a square to fit the window area, we would
arrive at a total wire length and this would be part of the resistance
R calculation.

After we do this, we end up with the best wire diameter for a given
core size, in that it will provide the most force with the given
battery.

There is another point however, and that is the core saturation.
We would have to test for core saturation and subject the
optimization search to that constraint because we dont want
to wrap more turns of wire on the core if the increase in force is
going to be negligible, as that would increase the resistance
unnecessarily. This would require measuring a few points.
Be aware however that this ends up as another constraint equation
that enters in as um, unless we just want to test for some max
and then call it good enough.

There is yet another constraint, and that is the coil heating.
We would have to test for wire diameter less than say 500cm/amp
or something like that, in order to prevent the coil from overheating
and burning out unless we are going to use it intermittently.

 

[Trisorion]

Thanks MrAl, that was easily one of the best answers I have ever gotten.

Just one more thing. I dont have an osilliscope to test for exact 'hysteresis curves' (I think that is what is called). But lets say my core is average for a piece of steel. How would I calculate when I need a wider core?

F=K*I^2*N^2*A/(g+d/um)^2

What quantitiy in the equation cant be higher than 1.8 due to saturation?

 

[MrAl]

Thanks MrAl, that was easily one of the best answers I have ever gotten.

Just one more thing. I dont have an osilliscope to test for exact 'hysteresis curves' (I think that is what is called). But lets say my core is average for a piece of steel. How would I calculate when I need a wider core?
What quantitiy in the equation cant be higher than 1.8 due to saturation?

Hi again,

That 1.8 is for a particular type of steel and that is most likely the
max B (1.8T max flux density) for this type of steel. That's one
of the better steels BTW, made just for mag cores.
Since force is affected linearly by area (more or less) if you run into
saturation before you have reached the amount of force you are
looking for that is when you would increase the cross sectional
area. For example, if you reach sat and you have a force F1 and
you need a force F2 (F2>F1) then you would need to increase the
area A by a factor of F2/F1. Unfortunately, when you increase
the cross sectional area of the core you also increase winding resistance
R so you would have to do the optimization again.

I am pretty sure you can get the optimization equation into one
equation that allows a min or max to be found which will point to
the best wire diameter. I just havent gotten around to trying this
yet. Since you want to include the possibility of changing the
cross section area you would just include area A in the equation
too. Area A already indirectly has to come into play as part of
the wire resistance anyway, so no big deal there.

Actually i am happy you are interested in this. I like to see people
take an interest in these kinds of things.

 

[MrAl]

Hi again,


I felt like playing around with the equations a little so i ended up
with an optimization equation.

Combining all the equations we get the equation for force that we will
use for the optimization:

F=(H^2*K2*V^2*W^2)/(d^4*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)

where
H is core center height,
W is core center width,
V is battery open circuit voltage,
d is wire diameter,
rho is resistivity of the wire,
Wwin is the width of the window area,
Hwin is the height of the window area,
Rb is the battery internal resistance,
pi is the constant 3.1415926...
and dimensions are meters, volts, ohm meters, ohms, and later amps when needed.

To optimize all we have to do is look for a relative max.

dF/dd=(32*rho*H^3*K2*V^2*W^3*(2*W+2*H+Wwin+Hwin))/(d^9*pi^4*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))
      /(d^4*pi^2)+Rb)^3)-(4*H^2*K2*V^2*W^2)/(d^5*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)

we set this equal to zero
dF/dd=0
and we find four solutions, the one we want is:

d=(sqrt(2)*((2*rho*H*W^2)/Rb+(2*rho*H^2*W)/Rb+(rho*Wwin*H*W)/Rb+(Hwin*rho*H*W)/Rb)^(1/4))/sqrt(pi)

and this leads directly to the wire diameter in meters of about 0.0031
meters, which is roughly a AWG #8 wire gauge.

A simpler way to avoid all the math however is to just plot the equation for
F between wire diameters say 0.0005 meters (AWG=24 approx) to 0.005
(AWG=3 approx) and see where the max occurs graphically. This has been
done in this drawing:

**broken link removed**


Now we must find the current so we can calculate the wire current density
to make sure it doesnt overheat. We will use a 600 cmil/amp criterion for
this test because this is known to work well with transformers which have
the same basic construction.

Calculating the total resistance from:

R=(4*Awin*mwl*rho)/(d^4*pi^2)

we get R=0.0888 approx.

Now we calculate I from

I=V/(0.0888+Rb)

and we get I=15.9 amps and this allows us to check the wire diameter for
heating.

We know the current is about 16 amps now, and since the wire diameter is
0.0033 meters (the actual number 8 gauge wire) we can calculate the
circular mils from:

Asm=pi*(d/2)^2 (d is diameter in meters)
cmils=Asm/5.067e-10

Doing this, we get
Asm=8.55e-6 square meters
cmils=16880

Now that we know the circular mils, we can calculate the current density:

density=cmils/amps
so
density=16880/16=1055,

so the current density is 1055 cmils per amp which is better than
600 cmils per amp so this construction should not overheat.

The other thing we have to do however is check the current draw to make sure
our voltage source can handle the operating level of current. If not, we may
end up decreasing the wire gauge to increase resistance to make it easier
on the source, or we have to think about getting a higher power supply to
drive the coil with.

 

[Trisorion]

Wow, thanks for doing all of that. However, now I'm really confused.

How did you find the best wire diameter without the length or cross sectional area of the core? (or maybe im just missing something) Is the core square or circular in this model?

Wwin is the width of the window area,
Hwin is the height of the window area,

What is "window area"?

R=(4*Awin*mwl*rho)/(d^4*pi^2)

What does the first half of this equation mean?

We know the current is about 16 amps now

My batteries could do 2.4 volts, 20 amps with .06 ohm Rb in series or 1.2 volts, 40 amps with .015 ohm Rb in parallel. 16 amps is not pushing it, even for series. Do I understand this correctly? Are you saying there is no benefit to have more than 16 amps?



Additional Information if it helps:

• I would buy the magnet wire from here:magnet wire at PlanetEngineers.com
• I have my own horseshoe-shaped, low carbon steel round bar that I could use as the core. It has a .5 inch diameter and is 15 inches long (meaning it was 15 inches long before being bent into a horseshoe shape.
• I could buy any of the cores on this website (except the very largest, because of cost issues). I would then cut the bottom off of a C type model to make it U shaped. **broken link removed**
• The competition will be conducted in two parts. 1)A speed challenge to move the most steel paperclips from one bin to another in 30 seconds. 2)A challenge to lift as many paperclips at once as you can. No time limit. I don't know if the fact that the goal is to pick up many small things (opposed to one heavy thing) changes anything.

 

[Willbe]

Hi again,


I felt like playing around with the equations a little so i ended up
with an optimization equation.

Combining all the equations we get the equation for force that we will
use for the optimization:

F=(H^2*K2*V^2*W^2)/(d^4*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)

where
H is core center height,
W is core center width,
V is battery open circuit voltage,
d is wire diameter,
rho is resistivity of the wire,
Wwin is the width of the window area,
Hwin is the height of the window area,
Rb is the battery internal resistance,
pi is the constant 3.1415926...
and dimensions are meters, volts, ohm meters, ohms, and later amps when needed.

To optimize all we have to do is look for a relative max.

dF/dd=(32*rho*H^3*K2*V^2*W^3*(2*W+2*H+Wwin+Hwin))/(d^9*pi^4*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))
      /(d^4*pi^2)+Rb)^3)-(4*H^2*K2*V^2*W^2)/(d^5*pi^2*((4*rho*H*W*(2*W+2*H+Wwin+Hwin))/(d^4*pi^2)+Rb)^2)

we set this equal to zero
dF/dd=0
and we find four solutions, the one we want is:

d=(sqrt(2)*((2*rho*H*W^2)/Rb+(2*rho*H^2*W)/Rb+(rho*Wwin*H*W)/Rb+(Hwin*rho*H*W)/Rb)^(1/4))/sqrt(pi)

and this leads directly to the wire diameter in meters of about 0.0031
meters, which is roughly a AWG #8 wire gauge.

A simpler way to avoid all the math however is to just plot the equation for
F between wire diameters say 0.0005 meters (AWG=24 approx) to 0.005
(AWG=3 approx) and see where the max occurs graphically. This has been
done in this drawing:

**broken link removed**


Now we must find the current so we can calculate the wire current density
to make sure it doesnt overheat. We will use a 600 cmil/amp criterion for
this test because this is known to work well with transformers which have
the same basic construction.

Calculating the total resistance from:

R=(4*Awin*mwl*rho)/(d^4*pi^2)

we get R=0.0888 approx.

Now we calculate I from

I=V/(0.0888+Rb)

and we get I=15.9 amps and this allows us to check the wire diameter for
heating.

We know the current is about 16 amps now, and since the wire diameter is
0.0033 meters (the actual number 8 gauge wire) we can calculate the
circular mils from:

Asm=pi*(d/2)^2 (d is diameter in meters)
cmils=Asm/5.067e-10

Doing this, we get
Asm=8.55e-6 square meters
cmils=16880

Now that we know the circular mils, we can calculate the current density:

density=cmils/amps
so
density=16880/16=1055,

so the current density is 1055 cmils per amp which is better than
600 cmils per amp so this construction should not overheat.

The other thing we have to do however is check the current draw to make sure
our voltage source can handle the operating level of current. If not, we may
end up decreasing the wire gauge to increase resistance to make it easier
on the source, or we have to think about getting a higher power supply to
drive the coil with.

Holy Cow, Mr. Al (It sort of rhymes).

And all this time I was blaming that North Jersery air pollution for my chronic laziness!
That's the end of that excuse. . .

 

[MrAl]

Hi again,



Willbe:
He he



Trisorion:

Wow, thanks for doing all of that. However, now I'm really confused.

How did you find the best wire diameter without the length or cross sectional area of the core? (or maybe im just missing something) Is the core square or circular in this model?

I am glad that you picked that up. In my haste i did not enter
in all the values for the example core, which includes the window
area and other things. We have to know the window height
and width as well as the core height and width to get the answer
of course.
The window height is the same as the length of the winding.
The window width is the same as the coil 'build'.
The core width is the center width of the core material (the center
width of one lamination for example).
The core height is the core stack height.


Im sorry i didnt make this clear, but i was in a hurry yesterday and
rushed to get that all posted. Since i made a few mistakes, i'll
provide the equations now and all the input data necessary
to do these calculations, and we come out with an interesting
result after all this.

Units: meters, ohms, ohm meters, volts, amps.


The force at the poles:

F=K*I^2*N^2*A/(g+d/um)^2


Area of wire
Awire=pi*(d/2)^2

Area of rectangular window cross section:
Awin=Hwin*Wwin

Circular to square wire area loss factor:
cslf=pi/4

Number of turns that fit into window area:
N=(Awin*cslf)/Awire

The mean wire length:
mwl=2*(Wcore+Wwin)+2*(Hcore+Hwin)

Total wire length:
twl=N*mwl

Wire resistivity:
rho=1.7241e-8*(1+.00393*(Ta)) ohm meters

Total wire resistance:
R=twl*rho/Awire

The force including volts and wire resistance:
F=K2*(V/(R+Rb))^2*N^2

Combining all the equations into one force equation:
F=(Hcore*Hwin^2*Wcore*Wwin^2*K2*V^2)/(d^4*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^2)


Finding the max:

dF/dd=(32*Hcore*Hwin^3*rho*Wcore*Wwin^3*(2*(Wwin+Wcore)+2*(Hwin+Hcore))*K2*V^2)/(d^9*pi*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^3)-(4*Hcore*Hwin^2*Wcore*Wwin^2*K2*V^2)/(d^5*((4*Hwin*rho*Wwin*(2*(Wwin+Wcore)+2*(Hwin+Hcore)))/(d^4*pi)+Rb)^2)

Set that equal to zero to find relative max:
dF/dd=0

Find all solutions to dF/dd=0 ,

and the solution we want is:
d=8^(1/4)*((Hwin*rho*Wwin^2)/(pi*Rb)+(Hwin*rho*Wcore*Wwin)/(pi*Rb)+(Hwin^2*rho*Wwin)/(pi*Rb)+(Hcore*Hwin*rho*Wwin)/(pi*Rb))^(1/4)

Just for reference, at d=optimum:
F=(Hcore*Hwin*pi*Wcore*Wwin*K2*V^2)/(32*rho*Rb*Wwin+32*rho*Rb*Wcore+(32*Hwin+32*Hcore)*rho*Rb)


Now plug in all the values for the equation for d above:

K2=1 (constant for optimization)
V=3 (battery open circuit voltage)
Rb=0.1 (battery internal resistance)
Ta=25 (temperature in degrees C)
Hcore=2/39 (height of core approx 2 inches)
Wcore=1/39 (width of core approx 1 inch)
Hwin=1/39 (height of window approx 1 inch)
Wwin=1/39 (width of window approx 1 inch)
rho=1.7241e-8*(1+.00393*(Ta)) resistivity of wire at ambient temperature Ta

Now graph or use the dF equation to find d, and we get:
d=0.002478 meters (and use a wire gauge chart to find the wire gauge number)

Now from:
Awire=pi*(d/2)^2
N=(Awin*cslf)/Awire
we get:
N=107 turns

and from:
mwl=2*(Wcore+Wwin)+2*(Hcore+Hwin)
twl=N*mwl
R=twl*rho/Awire
Im=V/(R+Rb)
we get:
Im=15 amps

and from:
Asm=pi*(d/2)^2
cmils=Asm/5.067e-10
we get:
cmils=9521

and calculating the current density:
CurrentDensity=cmils/I=635 cmils/amp

so this shouldnt overheat.


Now we take a quick look at what R came out to:
R=twl*rho/Awire
so
R=0.1

Interesting, isnt it?

So we found that the coil resistance R should match the
battery internal resistance.
I'll leave it up to you to prove if we get this result no matter
what the construction is.

My batteries could do 2.4 volts, 20 amps with .06 ohm Rb in series or 1.2 volts, 40 amps with .015 ohm Rb in parallel. 16 amps is not pushing it, even for series. Do I understand this correctly? Are you saying there is no benefit to have more than 16 amps?

Well, it's not quite that simple. We can not simply increase current
because current is dependent on the battery internal resistance,
and because we have optimized for force there should be nothing
we can do to get more force unless we change one of the optimizing
parameters (like the battery internal resistance). But we have to
remember that we are attempting to optimize the wire gauge (via
wire diameter) for a given construction which makes certain things
constant:

Window area
Core area
Battery internal resistance

We have to keep some of these things constant because if we
didnt we could not optimize because after all, if we were able to
increase current indefinitely (without overheating of course) we
would be able to increase the force infinitely even with a core
that saturates. We cant do this though, because for one we
are limited as to what the battery internal resistance is and the
open circuit voltage.

This doesnt mean we cant find out what the effects are when
we increase some parameter, such as core area, but to do
this we would really have to enter in all new parameters and
do a new optimization for wire diameter.

It may also be possible to find other interesting facts such as
maximum force possible for a given construction given all of
the other parameters. Solving for wire diameter and then
inserting that into the equation for force for example provides
for an equation that shows the max force possible for the
best wire diameter, without actually calculating the wire diameter.
This would show the difference between various cores and
window areas because the force for each construction could
be compared.
Since we are talking about optimization processes here, perhaps
you should look into other processes before attempting this one
because after all this construction has a lot of dimensions to
consider. If you dont want to do that then you are just going
to have to keep trying to understand the current process we
are talking about. After that you should be able to factor in
things we havent looked at yet, such as core saturation and
its effect on the optimized wire diameter.

One last note:
A circular cross section for the core is the optimum core shape
because this shape minimizes the wire resistance per turn,
and a square cross section comes in second. Rectangular
cores are common also however, but they wont be as good
as a square cross section. It's easier to get more core area
however with a rectangular cross section because you can
keep adding laminations to make the core height larger
and thus increase the core area.