Electromagnet Coil Performing Poorly

In summary: V 2A then the current would be 400A. So it would be about the same as if you were to apply 12V 2A.The coil was more than half the diameter of the tube, filled most of it up. 1200 turns. the coil is six inches long, the rod was maybe a foot. I am not sure how powerful I am expecting it to be but I would hope 1200 of feet of 17AWG being powered by well over 30V would be at least more powerful than a fridge magnet, if not much more.The electromagnet will not be very powerful. You need two co-planar poles close to each other to create a strong magnetic field. The field
  • #1
NathanSM
37
5
I recently posted about an air cored electromagnet I was designing. It currently uses 300' of 17 AWG magnet wire, coiled around a .5'' diameter tube. I purchased a DC to DC step up to test different voltages and currents. I noticed while using it that regardless of what current I supplied, the magnet got no stronger. I also tried to increase the voltage and there was little result. I got up to 40V 7A. Preformed similarly to 12V 2A.

Honestly hit a wall here. At this point I'm about to take apart a microwave to try to make a magnet out of the transformer's wires. I have no clue why my coil will not work. Some have suggested thinner wire with more turns but that still doesn't explain why increasing the current on my current build isn't working. I am also not sure what gauge wire to select if I were to switch.
Edit- I will be uploading photo of the set up shortly.
 
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  • #2
NathanSM said:
Preformed similarly to 12V 2A.

How are you measuring the magnetic field? How similarly? FYI the B field outside a solenoid along its length is negligible. It is strongest at the end opening. The strength of a magnetic field is determined by the concentration of the magnetic field lines. Since the field lines are closed, the concentrated field line inside the solenoid are spread out over all of the exterior space thus reducing the field strength outside significantly.
 
  • #3
NathanSM said:
I recently posted about an air cored electromagnet I was designing. It currently uses 300' of 17 AWG magnet wire, coiled around a .5'' diameter tube. I purchased a DC to DC step up to test different voltages and currents. I noticed while using it that regardless of what current I supplied, the magnet got no stronger.
I don't remember if I already posted the following comments in your first thread or not:

https://www.physicsforums.com/threa...n-air-core-electromagnet-with-no-luck.981479/

but you are not going to make a strong electromagnet with an air-core solenoid. Why are you using an air core if you want a strong electromagnet?

And to get a strong electromagnet, you need the two poles of your electromagnet close and co-planar. That minimizes the air gap from your ferrous core to the ferrous metal piece you are trying to attract/hold.

https://imgaz.staticbg.com/thumb/large/upload/2012/lidanpo/SKU092627/SKU092627 (1).JPG

1578951648766.png
 
  • #4
gleem said:
How are you measuring the magnetic field? How similarly? FYI the B field outside a solenoid along its length is negligible. It is strongest at the end opening. The strength of a magnetic field is determined by the concentration of the magnetic field lines. Since the field lines are closed, the concentrated field line inside the solenoid are spread out over all of the exterior space thus reducing the field strength outside significantly.
I wasn't measuring anything in an intellegent way. I had a thin steel rod in the tube that I was holding with my hand, while I increased the amperage
 
  • #5
NathanSM said:
I wasn't measuring anything in an intellegent way. I had a thin steel rod in the tube that I was holding with my hand, while I increased the amperage

Could you please provide the following data?

How much force do you actually expect from the electromagnetic coil? 10 or 20 or more grams force?

How thin is the steel rod you use? What is its diameter?

Do you know how many turns have actually been wound on the tube?

What are the lengths of the coils and the steel rod?

...etc
 
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  • #6
alan123hk said:
Could you please provide the following data?

How much force do you actually expect from the electromagnetic coil? 10 or 20 or more grams force?

How thin is the steel rod you use? What is its diameter?

Do you know how many turns have actually been wound on the tube?

What are the lengths of the coils and the steel rod?

...etc
The rod was more than half the diameter of the tube, filled most of it up. 1200 turns. the coil is six inches long, the rod was maybe a foot. I am not sure how powerful I am expecting it to be but I would hope 1200 of feet of 17AWG being powered by well over 30V would be at least more powerful than a fridge magnet, if not much more.
 
  • #7
My humble opinion is the same as it was way back: an external 'core' would help you a lot.
Try to put the coil inside a steel tube (of similar length as the coil).
 
  • #8
Because the electromagnetic force of the electromagnetic coil is proportional to the square of the current, it is inexplicable that the force remained almost the same when the current increased from 2A to 7A, that is, (7/2) ^ 2 = 12 times greater than before. 🤔

If you applied 40V 7A which is equal to 280W into the coil, it is very unreasonable if the device doesn't even produce a force greater than a mini fridge magnet, especially the size of your solenoid actuator seems to be already comparatively large.🤔

Since the force should change with the depth of the steel rod pushed into the hollow center of the aluminum tube, have you found out the depth that produces the greatest force?

Have you verified that the magnetic properties of the steel rod are good enough?
You might just use a fridge magnet to test and compare the forces produced by the steel rod and iron, which may help ensure that the steel rod can conduct magnetic flux no less than iron.

Besides, using a magnetically permeable frame to reduce external reluctance or increase the cross-sectional area of the steel rod should also help.:smile:
 
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  • #9
alan123hk said:
Because the electromagnetic force of the electromagnetic coil is proportional to the square of the current, it is inexplicable that the force remained almost the same when the current increased from 2A to 7A, that is, (7/2) ^ 2 = 12 times greater than before. 🤔

If you applied 40V 7A which is equal to 280W into the coil, it is very unreasonable if the device doesn't even produce a force greater than a mini fridge magnet, especially the size of your solenoid actuator seems to be already comparatively large.🤔

Since the force should change with the depth of the steel rod pushed into the hollow center of the aluminum tube, have you found out the depth that produces the greatest force?

Have you verified that the magnetic properties of the steel rod are good enough?
You might just use a fridge magnet to test and compare the forces produced by the steel rod and iron, which may help ensure that the steel rod can conduct magnetic flux no less than iron.

Besides, using a magnetically permeable frame to reduce external reluctance or increase the cross-sectional area of the steel rod should also help.:smile:
I will try to test with different materials. I really am unsure why the magnet isn't getting strong. I don't think I wound it wrong, all the layers are wound clockwise. One thing that may affect it is that I am coiling around a 6 inch section of the tube, while the aluminum tube is 2 feet long.
 
  • #10
NathanSM said:
I will try to test with different materials. I really am unsure why the magnet isn't getting strong..

Let's roughly estimate how much force the device will generate.

Coil.png

Equation : F = ((NI)^2 u0 A) / (2g)

Assumptions : -
N = Nos. of turns = 1200
I = Current = 2A
uo = permeability of free space = 1.257 *10^-6
A = cross-sectional area of the steel rod = 50mm^2 = 5*10^-5 m^2
g = air gap as show on the image = 20mm = 0.02m
The magnetic reluctance in the space outside the coil is assumed to be small, so its effect can be ignored

Result : -
F = 0.455N = 46 gram-force

By the way, I hope that the structure shown on the image correctly describes the device you are using.
 
  • #11
Now we can see the shortcoming of not installing an external magnetically permeable return path, normally the force should be much and much stronger when the gap (g) length is shortening. However, the effective magnetic reluctance in the outside space is actually not so small when there is no dedicated return path, therefore, the outside space magnetic reluctance will become comparable to the gap magnetic reluctance as the gap length reduced to a certain extent, and as a result, the expected force at this time will be reduced.
 
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  • #12
alan123hk said:
Let's roughly estimate how much force the device will generate.

View attachment 255725
Equation : F = ((NI)^2 u0 A) / (2g)

Assumptions : -
N = Nos. of turns = 1200
I = Current = 2A
uo = permeability of free space = 1.257 *10^-6
A = cross-sectional area of the steel rod = 50mm^2 = 5*10^-5 m^2
g = air gap as show on the image = 20mm = 0.02m
The magnetic reluctance in the space outside the coil is assumed to be small, so its effect can be ignored

Result : -
F = 0.455N = 46 gram-force

By the way, I hope that the structure shown on the image correctly describes the device you are using.
That looks exactly like my unit. The aluminum tube is .125 inches thick. What do you think I could do to safely make this magnet a lot stronger? Different wire size, amperage, or materials?
 
  • #13
alan123hk said:
Now we can see the shortcoming of not installing an external magnetically permeable return path, normally the force should be much and much stronger when the gap (g) length is shortening. However, the effective magnetic reluctance in the outside space is actually not so small when there is no dedicated return path, therefore, the outside space magnetic reluctance will become comparable to the gap magnetic reluctance as the gap length reduced to a certain extent, and as a result, the expected force at this time will be reduced.
How do I construct a return path
 
  • #14
NathanSM said:
How do I construct a return path
Find an iron tube that fits neatly over the coil, cut it to the length of coil. Cut a slot the full length of the iron tube to prevent eddy currents around the tube.

Get two iron washers that fit neatly on the aluminium tube. Cut slots in the two washers.

Assemble tube onto coil, then slide a washer on each end with slots lined up with the slot in the tube, outside of the washer in contact with the iron tube.

The outer return magnetic path is through first washer, through tube wall, and then through second washer. The sliding rod is the internal magnetic path.
 
  • #15
NathanSM said:
That looks exactly like my unit. The aluminum tube is .125 inches thick. What do you think I could do to safely make this magnet a lot stronger? Different wire size, amperage, or materials?

NathanSM said:
How do I construct a return path

Since I am neither a magnetic design engineer nor an expert in the field, I may not be able to tell you how to make the device in detail, especially if the design conditions and requirements are not clearly stipulated. Therefore, I just want to share some of my own understanding of related principles or theories, not detailed design.

Maybe you should check out the datasheets of commercial off-the-shelf linear solenoid actuators from various manufacturers.

You will then notice that force is proportional to power input and inversely proportional to stroke. In addition, with the same power input, a larger size unit produces more force.

So for a coil with fixed size, you need to increase the power to generate more force. On the other hand, you can make another coil of a larger size to increase the force without costing more for electricity, and the trade-off between them is up to you.

The relationships are easy to understand. As an example, for fixed power input to the coil, this means that the voltage, current, and resistance of the coil are fixed, we can then double the number of turns and the cross-sectional area of the magnetic wire at the same time to keep the original resistance unchanged, so the power remains the same but at the cost of increasing the coil size to get more turns.

By the way, there is another benefit of studying the specifications of commercial off-the-shelf products, we can at least compare and realize how bad our design is.

For the return path of the magnetic field, you can find vast amount of information online. All in all, its role is to reduce external reluctance, reshape the intensity and direction distribution of magnetic field in the space. To increase force, this measure is necessary in high-performance designs

I recommend that you spend some time studying magnetic circuit theory, which is an interesting subject and will also help you design electromagnetic equipment.
 
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  • #16
NathanSM said:
How do I construct a return path
The 'return path' is the external core what was already mentioned above and also in your previous topic.
 
  • #17
alan123hk said:
Since I am neither a magnetic design engineer nor an expert in the field, I may not be able to tell you how to make the device in detail, especially if the design conditions and requirements are not clearly stipulated. Therefore, I just want to share some of my own understanding of related principles or theories, not detailed design.

Maybe you should check out the datasheets of commercial off-the-shelf linear solenoid actuators from various manufacturers.

You will then notice that force is proportional to power input and inversely proportional to stroke. In addition, with the same power input, a larger size unit produces more force.

So for a coil with fixed size, you need to increase the power to generate more force. On the other hand, you can make another coil of a larger size to increase the force without costing more for electricity, and the trade-off between them is up to you.

The relationships are easy to understand. As an example, for fixed power input to the coil, this means that the voltage, current, and resistance of the coil are fixed, we can then double the number of turns and the cross-sectional area of the magnetic wire at the same time to keep the original resistance unchanged, so the power remains the same but at the cost of increasing the coil size to get more turns.

By the way, there is another benefit of studying the specifications of commercial off-the-shelf products, we can at least compare and realize how bad our design is.

For the return path of the magnetic field, you can find vast amount of information online. All in all, its role is to reduce external reluctance, reshape the intensity and direction distribution of magnetic field in the space. To increase force, this measure is necessary in high-performance designs

I recommend that you spend some time studying magnetic circuit theory, which is an interesting subject and will also help you design electromagnetic equipment.
I think I will try to find some actual resources that can help me study electromagnets and their construction. The main reason I was so confused is that I've got a literal pound of copper on that coil and I'n not even able to retain a two pound weight in the tube. Hopefully I can find some sort of pdf online that at least goes through the basics of exactly how these things work and how I can improve mine.
 
  • #18
One thought. The rod that you were using to test the solenoid strength wasn't stainless steel was it? SS tend weakly or none magnetic. Try a plain iron nail or spike available at any hardware store. The galvanized ones are magnetic too.
 

1. Why is my electromagnet coil performing poorly?

There could be several reasons for this. One possibility is that the coil is not wound tightly enough, causing a weak magnetic field. Another possibility is that the wire used for the coil is not the appropriate gauge or material for the intended purpose. Additionally, if the coil is overheating, it could be due to excessive current or insufficient cooling. It is important to check all of these factors and make any necessary adjustments to improve performance.

2. Can the shape of the coil affect its performance?

Yes, the shape of the coil can have a significant impact on its performance. For example, a tightly wound coil with a smaller diameter will produce a stronger magnetic field compared to a loosely wound coil with a larger diameter. The shape of the core material used within the coil can also affect its performance. It is important to carefully consider the design and shape of the coil for optimal performance.

3. How can I increase the strength of my electromagnet coil?

There are a few ways to increase the strength of an electromagnet coil. One method is to increase the number of turns in the coil, which will increase the magnetic field. Another way is to use a stronger core material, such as iron or steel, which can enhance the magnetic field. Additionally, increasing the current flowing through the coil can also increase its strength, but be careful not to exceed the coil's maximum capacity.

4. Why is my electromagnet coil attracting some objects but not others?

This could be due to the type of material that the objects are made of. Some materials, such as iron, nickel, and cobalt, are highly magnetic and will be attracted to the coil. Other materials, such as aluminum or plastic, are not magnetic and will not be attracted. The strength of the magnetic field and the distance between the coil and the object can also affect the level of attraction.

5. Can external factors affect the performance of an electromagnet coil?

Yes, external factors such as temperature, humidity, and nearby electromagnetic fields can all impact the performance of an electromagnet coil. For example, extreme temperatures can cause the wire to expand or contract, affecting the tightness of the coil and potentially weakening the magnetic field. It is important to consider these external factors and make any necessary adjustments to maintain optimal performance.

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