I have been trying to create an air-core electromagnet with no luck

[davenn]

, and I want to learn what gauge of wire I need

reread what I wrote

you should be aiming for a coil resistance of around 100 Ohms
A 12V relay coil is typically around 250 - 300 Ohms and more than 1000 turns

 

[NathanSM]

reread what I wrote

you should be aiming for a coil resistance of around 100 Ohms
A 12V relay coil is typically around 250 - 300 Ohms and more than 1000 turns

I meant "I want to learn how to determine what gauge I need" Does this mean I could use any combination of length/gauge that would reach 100 ohms? How do I figure out how many ohms are required?

 

[Tom.G]

The field strength of a solenoid is proportional to (N×I)², where
N is the number of turns
I is the current in Amps.

So you want to get the product of N and I as high as your physical design will allow. For instance that may be 10 turns at 100A or 2000 turns at 0.5A.

Of course the wire size needed depends on your choice of current.

And the voltage needed depends on the chosen current and on the resistance (length) of wire in the winding.

Also keep in mind that the magnetic field strength decreases with the square of the distance from the coil.

A quick search found some calculators and a discussion related to the above. They have 'Area' as a variable, which seems to be the cross section area of what you are trying to pick up/move (as long as it fits in the core of the solenoid), and a variable 'g' which is the distance from the solenoid to what you are trying to attract.

Note: My knowledge of this subject is limited, so these calculators could be completely wrong too! Hopefully others here can chime in on their relevance.

https://daycounter.com/Calculators/Magnets/Solenoid-Force-Calculator.phtml
http://onlinecalculators.brainmeasures.com/Electric/SolenoidCoil.aspx
https://www.eevblog.com/forum/beginners/solenoid-force-calculations/

For further information, above found with:
https://www.google.com/search?&q=solenoid+force+calculator

Hopes this helps!

Cheers,
Tom

p.s. I can see how that external core that @Rive suggested would help, alot, when the target object is already inside the coil by concentrating the field there. I believe however that this would reduce/eliminate the external field of the solenoid. From your brief description, it sounds like this is not what you are after.

 

[NathanSM]

The field strength of a solenoid is proportional to (N×I)², where
N is the number of turns
I is the current in Amps.

So you want to get the product of N and I as high as your physical design will allow. For instance that may be 10 turns at 100A or 2000 turns at 0.5A.

Of course the wire size needed depends on your choice of current.

And the voltage needed depends on the chosen current and on the resistance (length) of wire in the winding.

Also keep in mind that the magnetic field strength decreases with the square of the distance from the coil.

A quick search found some calculators and a discussion related to the above. They have 'Area' as a variable, which seems to be the cross section area of what you are trying to pick up/move (as long as it fits in the core of the solenoid), and a variable 'g' which is the distance from the solenoid to what you are trying to attract.

Note: My knowledge of this subject is limited, so these calculators could be completely wrong too! Hopefully others here can chime in on their relevance.

https://daycounter.com/Calculators/Magnets/Solenoid-Force-Calculator.phtml
http://onlinecalculators.brainmeasures.com/Electric/SolenoidCoil.aspx
https://www.eevblog.com/forum/beginners/solenoid-force-calculations/

For further information, above found with:
https://www.google.com/search?&q=solenoid+force+calculator

Hopes this helps!

Cheers,
Tom

p.s. I can see how that external core that @Rive suggested would help, alot, when the target object is already inside the coil by concentrating the field there. I believe however that this would reduce/eliminate the external field of the solenoid. From your brief description, it sounds like this is not what you are after.

Wow this really helps a lot. I should clarify that the object attracted to the magnet is passing through the center of the coils, so yes, I could surround the coil in iron to contain the field. What is the field strenght of the coil measured in? Tesla? Also I'm am struggling with which gauge of wire to select as I all really know is the voltage of the battery I have.

 

[davenn]

I meant "I want to learn how to determine what gauge I need" Does this mean I could use any combination of length/gauge that would reach 100 ohms?

That would be a minimum and it is more to stop the coil getting too hot because of high current flow
and draining the battery(s)

How do I figure out how many ohms are required?

The Ohms required isn't your main requirement

In post #33, @Tom.G gave you the main requirements

The field strength of a solenoid is proportional to (N×I)2, where
N is the number of turns
I is the current in Amps.

I could surround the coil in iron to contain the field.

the physics of the coil will already contain the magnetic field reasonably well ( sort of)
putting the coil in a metal box isn't going to help concentrate the magnetic field.Only an iron core will do that

What is the field strenght of the coil measured in? Tesla?

absolutely tiny ... microTesla or less

 

[NathanSM]

That would be a minimum and it is more to stop the coil getting too hot because of high current flow
and draining the battery(s)
The Ohms required isn't your main requirement

In post #33, @Tom.G gave you the main requirements
the physics of the coil will already contain the magnetic field reasonably well ( sort of)
putting the coil in a metal box isn't going to help concentrate the magnetic field.Only an iron core will do that
absolutely tiny ... microTesla or less

That is what I thought initially, that the outside of the magnet really doesn't matter. I noticed that only the area inside of the coil was creating a field, which is what I wanted. I have found a way to use about 1300 turns of wire will pull 7 amps. The only thing that worries me is that my current magnet was suppose draw 17 amps but i only read 6

 

[Baluncore]

You maximise the amp·turns by filling the largest possible winding cross-section.
If the winding is 75 mm long and 5 mm deep it has a section of 5 * 75 = 375 mm².
You can probably fill π/4 ≈ 80% of that section with round copper, the rest being insulation and air.
So the section of copper is 375 * 0.8 = 300 mm².
The ampacity of copper magnet wire is about 10 A/mm².
For a one turn winding that makes 300 * 10 = 3000 amp·turns.
It remains 3 kA·turns for a full section of any diameter copper wire.

If you know the DC voltage available you can select wire that is thin enough to limit the current by resistance, to the ampacity of that sized wire.

The attached file contains the calculations needed to evaluate different copper wire sizes.
Adjust the wire diameter to have actual current less than ampacity.

 

[sophiecentaur]

ampacity.

Now that's a lovely word. It could be straight out of the lexicon of George W Bush. But it's meaning is very clear (clearer than most of Dubbya's utterances).

 

[Baluncore]

Here is the direct way of finding the wire diameter.
The wire_diam in mm = √ ( Cu_resistivity * Cu_ampacity * volume /voltage /100000 )
Where;
Cu_resistivity = 1.715 ' The resistivity of copper in ohms·metres·10^-8.
Cu_ampacity = 10 ' The maximum current in amps per mm² of copper.
volume = π * ( Rout² - Rin² ) * length ' The volume of the winding in mm³.
voltage = the DC voltage connected to the coil.

 

[alan123hk]

What is the field strenght of the coil measured in? Tesla?

The magnetic flux density is measured in Gauss or Tesla units.
I think you may need a Gauss meter.

In any case, it seems that you are working on a fairly complex and difficult engineering project, and some scientific instruments should be inevitable.

https://www.alphalabinc.com/product-category/gaussmeters/

 

[NathanSM]

You maximise the amp·turns by filling the largest possible winding cross-section.
If the winding is 75 mm long and 5 mm deep it has a section of 5 * 75 = 375 mm².
You can probably fill π/4 ≈ 80% of that section with round copper, the rest being insulation and air.
So the section of copper is 375 * 0.8 = 300 mm².
The ampacity of copper magnet wire is about 10 A/mm².
For a one turn winding that makes 300 * 10 = 3000 amp·turns.
It remains 3 kA·turns for a full section of any diameter copper wire.

If you know the DC voltage available you can select wire that is thin enough to limit the current by resistance, to the ampacity of that sized wire.

The attached file contains the calculations needed to evaluate different copper wire sizes.
Adjust the wire diameter to have actual current less than ampacity.

Could one of the reasons of my magnet being much weaker that projected was that it was winded poorly? The magnet made almost no force. I winded all 160 feet around a small spool and I did not do a very good job. All of the winding was clockwise but it is a little uneven. Thank you for the detailed explanation of ampacity and how I can use it.

 

[Baluncore]

If you scramble wind the coil, the same length of wire will have a few less turns, so the amp·turns will be slightly less. I do not think that is the real problem.

I think you need many more amp·turns, or a magnetic core.

If the former was aluminium it will act like a shorted secondary winding. You should have an insulated gap along the wall of the tube to prevent the shorted turn.
If the former was an iron tube it would greatly reduce flux through the air core because the flux would follow the iron path rather than the air path.
What material is the former?

Can you reduce the inner diameter by 1%, and so reduce the area of the former by 2%, to get a greater field density?

 

[NathanSM]

If you scramble wind the coil, the same length of wire will have a few less turns, so the amp·turns will be slightly less. I do not think that is the real problem.

I think you need many more amp·turns, or a magnetic core.

If the former was aluminium it will act like a shorted secondary winding. You should have an insulated gap along the wall of the tube to prevent the shorted turn.
If the former was an iron tube it would greatly reduce flux through the air core because the flux would follow the iron path rather than the air path.
What material is the former?

Can you reduce the inner diameter by 1%, and so reduce the area of the former by 2%, to get a greater field density?

The wire is wrapped around an aluminum tube. This tube has a 1/2 diameter and is hollow. Ferrous items are supposed to be able to move through this tube and be affected by the magnet, if that makes sense. If i were using an iron tube would it not just turn the entire tube into a large magnet? The force of the magnet is supposed to be concentrated in one section of the tube.

 

[Baluncore]

You have a long magnetic coil. You will only notice a change when the magnetic material is near the ends of the coil.

The thick aluminium tube wall is a shield that will slow any changes within the air core. The presence of ferrous material in the tube will distort the field, but that change will be delayed by the aluminium tube.
Maybe change to a plastic tube, or cut an insulating slot along the aluminium tube in the region of the coil.

We do not know what you are trying to do so it is difficult to suggest more practical alternatives to your concept.

 

[NathanSM]

You have a long magnetic coil. You will only notice a change when the magnetic material is near the ends of the coil.

The thick aluminium tube wall is a shield that will slow any changes within the air core. The presence of ferrous material in the tube will distort the field, but that change will be delayed by the aluminium tube.
Maybe change to a plastic tube, or cut an insulating slot along the aluminium tube in the region of the coil.

We do not know what you are trying to do so it is difficult to suggest more practical alternatives to your concept.

So are you saying that the aluminum tube is hindering the force of magnet inside the tube. As for the project imagine having a piece of metal inside the tube i want to toggle different magnets to hold the metal at different positions within the tube. I can certainly cut gashes into the tube on either end of the coil. The item only travels inside of the tube that the coil is wrapped around though.

 

[Baluncore]

I can certainly cut gashes into the tube on either end of the coil.

That will make no difference. You must cut the aluminium for the length of the coil and insulate the cut to make sure there is no circulating secondary current. You do not seem to understand that your coil is the primary of a transformer. The aluminium tube is a short circuited secondary of that same transformer.

The item only travels inside of the tube that the coil is wrapped around though.

So it is big and heavy to fit the tube, while the magnet is weak. Is the tube horizontal?

 

[NathanSM]

I see what you mean. The tube is 2 1/2' but I am only coiling around about 6' of it. Is that what is causing the issue? As for the item, consider something small like a ball bearing. Any small item about the weight of a screw or nail. The tube is horizontal, shaped like a thin PVC pipe. I thought this would not be an issue as aluminum is non-ferrous

 

[JBA]

I see what you mean. The tube is 2 1/2' but I am only coiling around about 6' of it.

Do you mean 6" rather than 6'?

A small ball bearing (or item) has very little volume and will quickly become saturated regardless of the strength of the coil field.

 

[Tom.G]

You must cut the aluminium[/color] for the length of the coil and insulate the cut to make sure there is no circulating secondary current. You do not seem to understand that your coil is the primary of a transformer. The aluminium tube is a short circuited secondary of that same transformer.

This ONLY makes a difference when the current is changing. If AC (Alternating Current, maybe from a transformer) was used, their would be a problem. With DC from a battery, the only effect is when the current is actually changing, the magnetic field would change slightly slower than if the tube was an insulator.

For these first experiments, I suggest you continue with what you have and make changes later if needed for speed or other reasons. Right now your main problem is to make a magnet strong enough to move your ball bearing.

 

[Charles Link]

This ONLY makes a difference when the current is changing. If AC (Alternating Current, maybe from a transformer) was used, their would be a problem. With DC from a battery, the only effect is when the current is actually changing, the magnetic field would change slightly slower than if the tube was an insulator.

There is an additional effect=the iron piece that is being attracted produces a changing magnetic field, so there is the possibility of Faraday EMF's in the aluminum.

 

[NathanSM]

This ONLY makes a difference when the current is changing. If AC (Alternating Current, maybe from a transformer) was used, their would be a problem. With DC from a battery, the only effect is when the current is actually changing, the magnetic field would change slightly slower than if the tube was an insulator.

For these first experiments, I suggest you continue with what you have and make changes later if needed for speed or other reasons. Right now your main problem is to make a magnet strong enough to move your ball bearing.

That is what I thought. Today I am doubling the length of wire on the coil, so hopefully that will help.

 

[NathanSM]

Do you mean 6" rather than 6'?

A small ball bearing (or item) has very little volume and will quickly become saturated regardless of the strength of the coil field.

I did mean 6". I am not really farmiliar with saturation. Is this like a limit as to how fast I can make a ball bearing can go or with how much strenght it can be held within the field?

 

[NathanSM]

There is an additional effect=the iron piece that is being attracted produces a changing magnetic field, so there is the possibility of Faraday EMF's in the aluminum.

That shouldn't be too signifigant should it?

 

[Charles Link]

That shouldn't be too signifigant should it?

It could be very significant, particularly if the iron ball is moving fast. It might be an interesting experiment to have one aluminum tube with a slot and another without a slot and compare results. Faraday EMF's in the aluminum would have currents whose magnetic fields would be opposite the magnetic field of your coil. The slot should reduce these "eddy" currents considerably.

 

[Tom.G]

That shouldn't be too signifigant should it?

Try dropping a permanent magnet thru the tube. You will get an idea of how much of a problem it can be. Of course your ball bearing will not be magnetized nearly as much as the permanent magnet, but there will be some effect.

Oh, I see @Charles Link hit Send while I was still typing.

 

[Charles Link]

@NathanSM See https://www.physicsforums.com/threads/spherical-magnet-dropped-through-aluminum-pole-rotates.924489/
along with other threads on Physics Forums of Lenz's law with a tube. You may want to do a search of the topic. There are a number of posts that you might find good reading.

 

[NathanSM]

@NathanSM See https://www.physicsforums.com/threads/spherical-magnet-dropped-through-aluminum-pole-rotates.924489/
along with other threads on Physics Forums of Lenz's law with a tube. You may want to do a search of the topic. There are a number of posts that you might find good reading.

Ill check those out, I will have to cut slots anyway for some sensors.

 

[Baluncore]

If you attempt to control the position of balls in the tube by turning different coils on or off, then what advantage can there be in wasting energy from the battery to heat the aluminium tube every time you change the coil current?

The coil and the ball interact through the aluminium tube wall. The field of the coil will be affected by the presence of the ball. Skin effect in aluminium limits the rate of interaction. Cutting a full length slot in the aluminium, or using insulated tube will eliminate the waste of energy, and speed the interaction by a factor of close to a million, to the speed of light. The changing field conditions between coil and ball will communicate through the slot, or the wall of an insulated tube.

The insulation between transformer laminations provide a rapid path for flux to reach the magnetic material deep in the core. Likewise, the slot in an aluminium tube provides the path for the external coil to interact with the surface of an internal ball.

If a ball rolls along inside the tube it's induced magnetic polar axis will tumble, end-over-end and so induce currents in the top and bottom of the aluminium tube. That will not be reduced by a single slot cut in the wall of a conductive tube.

... The tube is 2 1/2' but I am only coiling around about 6' of it. ...

Please use the SI unit metre, m or mm. They are useful and can be understood. The fractional foot' and inch" just lead to confusion.

 

[Charles Link]

The insulation between transformer laminations provide a rapid path for flux to reach the magnetic material deep in the core. Likewise, the slot in an aluminium tube provides the path for the external coil to interact with the surface of an internal ball.

This, I believe, is somewhat inaccurate. The laminations and the slot block eddy currents. Upon blocking most of the eddy currents, there is very little magnetic field generated in the opposite direction. In the case of the laminated transformer, the magnetic field is that of the applied field from the current in the primary coil, $\\$
[considering the case here of no current in the secondary=otherwise, the result is the primary current increases to offset the field in the opposite direction that would be caused by the secondary=see https://www.physicsforums.com/threads/magnetic-flux-is-the-same-if-we-apply-the-biot-savart.927681/ especially posts 15-20=perhaps getting off on a tangent here, but it may be of interest], $\\$
along with what can be considered to be (bound) magnetic surface currents in the iron that enhance the magnetic field from the primary coil by a factor of 500 or more. The laminations do not block the magnetic surface currents,(which give rise to the enhanced magnetic field), because there is no actual charge transport for these currents. The laminations do block the eddy currents from the Faraday EMF in the iron, which would generate an opposing magnetic field, if they were left unblocked. $\\$ With a completely DC system, there would be no need for laminations or slots, because there would be no Faraday EMF's to generate eddy currents.

 

[Baluncore]

The laminations and the slot block eddy currents.

If that is all they do then why is the thickness of a lamination related to the frequency and skin effect in the lamination. The magnetic flux that enters the magnetic material does it via the insulation between the laminations. The eddy current excuse is given to pacify trainee technicians.