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

[Charles Link]

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.

In the link that I gave above, see post 20, especially the second part. The post is by Jim Hardy, and I think he gives a very good and accurate explanation.

 

[Baluncore]

Reading that reference, I see we agree. It is just that you approach from the direction of eddy current reduction, while I approach from the direction of skin effect and lamination orientation.

The primary coil induces a counter current in the conductive tube (secondary), which cancels the internal field due to the coil, so significant energy is reflected back out from the tube and does not initially penetrate the air core. That reflection must be prevented by elimination of the circular current path. That is why a longitudinal slot must be cut in the conductive tube, or the tube must be made from an insulator. Do you disagree with that conclusion.

 

[NathanSM]

Currently I am having an issue I do not understand. When plug the 14V battery into a length of wire with 1.6 ohms of resistance I only read 7 amps.

 

[zoki85]

Currently I am having an issue I do not understand. When plug the 14V battery into a length of wire with 1.6 ohms of resistance I only read 7 amps.

There is a voltage drop of your battery . Use a voltmeter across the terminals of the battery while the current is flowing through the circuit to see it.

 

[NathanSM]

There is a voltage drop of your battery . Use a voltmeter across the terminals of the battery while the current is flowing through the circuit to see it.

Got not even a volt.

 

[Charles Link]

Got not even a volt.

Then for 1.6 ohms resistance, you should read about .7 amps of current. Did you get the decimal place correct in your current reading? Perhaps another factor is the resistance of the wire will increase if it heats up. Something is inconsistent here.

 

[NathanSM]

Then for 1.6 ohms resistance, you should read about .7 amps of current. Did you get the decimal place correct in your current reading? Perhaps another factor is the resistance of the wire will increase if it heats up. Something is inconsistent here.

How did you get .7 amps? I am rather confused, a lot have people said use thinner wire as my setup has too little turns/resistance. I double the amount of wire used instead. This still increases resistance and turns but I've got a weak magnet that is drawing 7 amps, and heating up.

 

[NathanSM]

Im beginning to believe that since doubling the wire length did not affect strength, voltage drop is the issue. I may purchase DC-DC boost converter and try to run the circuit at a much higher voltage and see how this affects the strength.

 

[Baluncore]

I may purchase DC-DC boost converter and try to run the circuit at a much higher voltage and see how this affects the strength.

It is easier to calculate the best solution than to buy something, then wait to find it does not seem to work. Before you go looking for another power supply you should identify all the parameters for your coil.

You know that you need to maximise the product of the current and the number of turns.
If you know the geometry and voltage you can calculate the optimum wire diameter. If you know the geometry of your coil and wire diameter, you can calculate the optimum voltage.

 

[Charles Link]

You may need a power supply that is designed to deliver higher currents. (You don't necessarily need higher voltage, but simply a lower internal resistance). When you read less than one volt across a 14 volt supply, you clearly have a supply that has a much larger internal resistance than the resistance in the wires. $\\$ Meanwhile, when you have less than one volt, and a resistance that exceeds 1.5 ohms, a current of 7 amps is inconsistent. Perhaps it is 0.7 amps.

 

[davenn]

Currently I am having an issue I do not understand. When plug the 14V battery into a length of wire with 1.6 ohms of resistance I only read 7 amps.

You have been told multiple times and several times by me
GET MUCH MORE turns = more resistance

You don't seem to listen to what you are being told

Higher voltage isn't the answer, 5V generates a nice electromagnet.
When you have only a few ohms of resistance, you are effectively short circuiting the
battery/power supply. That's why you don't read any voltage across the battery when
it is connected.

Again ...
If you are not using AWG/SWG of around 26 or finer and 500 - 1000 or more turns
you WILL NOT achieve you goal

a bobbin of 28 SWG wire as purchased roughly 2.5" long 1.5" diameter
bare the 2 ends and measure the resistance and see what it is

 

[bobob]

I don't think you've though this through enough to know what you need. I have wound a magnet similar to yours except a lot heftier (aluminum spool wound with copper tubing for cooling water flow), so I know the general idea works and all of the stuff about eddy currents is nonsense. Also, the voltage is irrelevant. The field depends on the current and the number of turns, so you need to figure out how many amp-turns you need. If you can't use a power supply operating in current mode, then you need to know the resistance of the wire so you can figure out what the current will be for whatever batteries you will be using. Don't be surprised if the current you need is pretty substantial.

 

[davenn]

Don't be surprised if the current you need is pretty substantial.

again ... It doesn't need to be. electromagnets in relay/other solenoid coils
work just sweet on lower current

 

[Charles Link]

again ... It doesn't need to be. electromagnets in relay/other solenoid coils
work just sweet on lower current

@NathanSM The relevant equation is $B=\mu_o n I$, for the magnetic field inside the cylinder. If you have a lot of turns per unit length $n$, (turns per meter), then you don't need a very large current $I$ to get a magnetic field that at least has a little strength. Note: $\mu_o=4 \pi \cdot 10^{-7}$. You might get $.001< B<.005$, if you do a good job with the number $n$. You can't expect much more with an air core. If your current $I=.1$ amperes, that should be ok, if your $n$ is large enough. $\\$
With an iron core, a typical $B$ you could expect to achieve would be about $1.0$. With an iron core, the equation is $B=\mu_o \mu_r n I$, where $\mu_r \approx 500$ for iron. One other note: The magnetic field will saturate for iron for $B$ slightly greater than $1.0$, so that with an iron core, increasing $n$ to a large number doesn't offer the same gain that it does with an air core.

 

[Baluncore]

I am afraid the relevant equation won't help as NathanSM has not yet done the wire resistance calculation.

The orientation of the long winding field in the tube will not interact with a steel ball in the way NathanSM expects. That would require a horizontal field across the tube which could use a magnetic core.

 

[alan123hk]

Im beginning to believe that since doubling the wire length did not affect strength, voltage drop is the issue. I may purchase DC-DC boost converter and try to run the circuit at a much higher voltage and see how this affects the strength.

I believe you need to maximize the magnetomotive force (MMF) in order to obtain more mechanical force in your solenoid design, and I think MMF is proportional to the electrical power input to the coil, so it needs to provide more electrical power to generate a larger MMF.For a coils with limited winding space, you can choose thin wire or thick wire to make the coil. Thin wire results in more turns and high resistance, which in turn requires high voltage and low current inputs. On the other hand, thick wire results in fewer turns and low resistance, which in turn require low voltage and high current inputs.

However, if the input electrical power is the same, the MMF produced in the two cases will be roughly the same.

So the important factor is how much power you are willing to provide, but, of course, if too much power is applied, you also need to consider overheating.

Moreover, just increasing the coil winding space can increase MMF without adding additional power input.

Remark :
MMF = NI, where N is the number of turns in the coil and I is the electric current through the circuit

 

[alan123hk]

Some other interesting cases about wires and power supplies under the same conditions of limited coil winding space are the following:

Case 1, If the power supply has a fixed voltage ouput and infinite current capability, thicker copper wire will produce a larger MMF

Case 2, If the power supply has fixed current output and infinite voltage capability, thinner copper wire will produce larger MMF

Case 3, if the voltage across and the current through the coil are fixed values, then the resistance, input power and generated MMF of the coil are also fixed values. We can continue to calculate the corresponding number of turns and cross-sectional area of the wire, and then fill up the limited winding space.

Note on case 1 and 2 : -
Assumes the coil will not be destroyed by excessive voltage, current and power input..Typical coil data of relay for reference : -