# Need simple math done for my updated project?

Discussion in 'General Electronics Chat' started by ElectromagnetNewbee, Sep 5, 2014.

1. ### ElectromagnetNewbee Thread Starter Member

Jul 13, 2014
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For this project I have 1.5 pounds of 24 GA copper magnet wire. I have a 25 pound iron cast dumbell wieght that is 8.2 inches by 1 inch. I want to use as the core for a new strong electromagnet. I have only 30 volt and 20 amp variable dc power supply. I want the MOST amps i can get out of this configuration!!!?urn

How many 'turns' around this 25 pound wieght do I need to get MAX amps I possibly can?

Also, please provide the math so I can do it myself next time.

2. ### wmodavis Well-Known Member

Oct 23, 2010
739
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Since your power supply will provide 30 V @ 20 A that means that the minimum load resistance on the power supply would be calculated using Ohms Law, R=V/I =30V/20A= 1.5Ohms. So you need a length of your 24AWG wire that will be equal to that resistance (1.5Ohms).

Get a copy of the wire table that gives the resistance per ft (or per 100ft or per 1000ft) of 24AWG wire and determine from that what length of 24AWG wire you need. That is very simple math so I will leave it to you.

Measure out and cut off that length of wire and see how many turns you can wind around the dumbell. Be sure to count them so you will know the answer to your question (How many 'turns' around this 25 pound wieght do I need to get MAX amps I possibly can?).

P.S. The proper spelling is WEIGHT not weight.

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3. ### #12 Expert

Nov 30, 2010
17,840
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*not wieght

I have that chart hanging on my wall. 25.67 ohms per thousand feet.

math math math

58.4 feet.

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4. ### crutschow Expert

Mar 14, 2008
16,215
4,334
Note that 20A at 30V is 600W. So you likely will be able to apply that to the coil of wire for only a few seconds before the coil overheats.

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5. ### #12 Expert

Nov 30, 2010
17,840
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So you have 1210 feet of wire. What if you used 584 feet to achieve 15 ohms and drove 2 amps through the coil with 30 volts? The result would be the same amp turns but with only 60 watts of heat.

The point is, play with the math. You can discover different ways to do the same thing with less risk to your components.

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6. ### MrCarlos Active Member

Jan 2, 2010
400
135
Hello ElectromagnetNewbee

If your power supply provides Direct Current, You can calculate the coil as mentioned below.
but if not then the calculation should be done differently.

Let me talk as I do:
You say:
I have 1.5 pounds of 24 GA copper magnet wire.
Looking at the data sheets for wire AWG #24 meeting that supports a maximum current of 808 mA.
This is the basis for calculating Your coil: The maximum current that supports the wire.

You say:
I have only 30 volt. and 20 Amp. Variable dc power supply.
With this source we will circulate, through the coil, say, 808 mA.
Arbitrarily we will adjust our source to 10 V.
So 10 / 0.808 = 12.38 Ohms.
Looking at the data sheets wire AWG #24, found that has a resistance of 20.9870 Ohms per pound.
Therefore: 1 / 20.9870 = 0.0476.
0.0476 X 12.38 = 0.589288 Pounds. a little over half a pound.
Remember that #24 AWG only supports current: 808 mA.
if you apply more, you will begin to heat the coil and burn.
Since V / I = R, if you increase the voltage maintaining the current value, the resistance value is increased with the consequence that you will require more wire.

Now: Your power supply has the following electrical characteristics:
It is Variable.
From 0 to 30 V.
It can provide 20 A.
So it is 30 X 20 = 600 Watts.
Asked to feed a circuit (Our coil) with 10V @ 808 mA.
We will be requiring to the power supply, a power: (30-10) X 0.808 = 16.16 Watts.
What if we increase the voltage to 20V ?.
We will be requiring to the power supply, a power: (30-20) X 0.808 = 08.08 Watts.

Make an exercise:
Based on when the above,
The data sheets of the AWG wires
And the attached image Ohm's Law.
How much voltage you would have to apply to your coil made with all the wire you say have ?.

Now, speaking of the core You say have:
I have a 25 pound cast iron dumbell wieght That is 8.2 inches by 1 inch.
I want to use as the core for a new strong electromagnet.

Question: how much weight does each sheet of 8.2 for 1 inch?
or, many sheet up that core?

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7. ### ElectromagnetNewbee Thread Starter Member

Jul 13, 2014
69
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Just to verify what you are saying... 58.4 feet of wire around the iron core will give me the MAX amps possible, around 8 amps? Is this correct?

This would work great for the project!

8. ### tom_s Member

Jun 27, 2014
301
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MrCarlos typed - Looking at the data sheets for wire AWG #24 meeting that supports a maximum current of 808 mA.

this is perfect scenario for a kettle or am i missing something here?

9. ### #12 Expert

Nov 30, 2010
17,840
9,178
No. You said your power supply would provide 20 amps at 30 volts.
wmodavis said the lowest resistance to use up that power is 1.5 ohms.
I said 58.4 feet will provide 1.5 ohms.
30 volts will drive 20 amps through 1.5 ohms with a resulting heat of 600 watts.

If you want 8 amps, use Ohm's Law and 25.67 ohms per thousand feet.

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10. ### wayneh Expert

Sep 9, 2010
13,440
4,274
Why? Don't you really want the strongest magnet you can make with your wire and core? If you optimize magnet strength, the solution is not likely to be the same as maximum current.

As noted, 20A will likely turn your coil into a fuse. Poof. As #12 noted (in #5), more turns will also increase magnet strength, will be far gentler on your power supply, and will have a lower risk of burning up.

Are you wanting a strictly DC magnet? If you want to use AC, there are other considerations, namely inductance.

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11. ### ElectromagnetNewbee Thread Starter Member

Jul 13, 2014
69
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Yes, I want the strongest electromagnet I can get. I thought that meant 'amps'?

Please tell me how to get create the strongest electromagnet possible with the supplies I have? So more turns? how many feet then?

12. ### wayneh Expert

Sep 9, 2010
13,440
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Play with the math. Magnet strength will depend on amp•turns. The amps will depend on the DC resistance of the wire you add, and of course so will the number of turns. At the optimum, you will not gain or lose amp•turns by adding or removing one turn. If adding a turn increases the field more than the extra resistance you are adding, then you gain from adding that extra turn. Note that your turns will get a bit larger as the coil grows. Each layer will have a radius larger than the previous by the width of your wire and insulation.

My hunch is that you'll want to use all the wire you have, but that's just a hunch.

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13. ### MrCarlos Active Member

Jan 2, 2010
400
135
Hello ElectromagnetNewbee

Well, let's look at this issue from your point of view. I hope I have understood correctly.
As you mention in your original message.
You have a power supply with the following features:
Variable, 0 to 30V which can provide up to 20 Amp.

The materials you have available are:
1.5 pounds of 24 gauge wire.
25 pounds of cast iron leaves 8.2 inches by 1 inch.

And what you intend to do is:
How many 'turns' around This 25 pound weight do I need to get MAX amps I possibly can?

Also want the mathematical development to do it yourself next time.
Also, Please provide the math so I can do it myself next time.

First of all You should keep in mind that the wire You have You can only apply a current flow of 808 mAmp. maximum.
If you begin to apply more current, the coil would act as a fuse. it would burn immediately.

For the development of the coil in that core, and that wire, We need to know what current supports the wire, ohms either by weight or by length.
Since we do not know this information is necessary to rely on data provided by manufacturers. These data are contained in the data sheet in Excel that I have attached to my message #6.
But if you don’t want to taking the time to open that type ZIP file and unpack the data sheet, I attached a picture where you can see these characteristics for the 24 gauge wire.
By the way these data are official and are recognized worldwide

Given the fact that You have 1.5 pounds of 24 gauge wire.
We can calculate how many ohms is 1.5 pounds such as:
Ohms per pound equals 20.9870.
But as You have 1.5 pounds then it would be 20.9870 + (20.9870 / 2) = 31.4805 Ohms. The easiest way.
How much voltage we can apply to these 31.4805 Ohms to flow a current of 808 mAmp. ??
If you look at the picture I attached you my message #6 you can discover It. (Ohm's Law.PNG)
The formula is: V = R x I = 31.4805 x 0.808 = 25.436244 Volts. Easy. . . Do not ?.
They could be 25 or 25.5 Volts.

To calculate how many turns would be in your core you should take into account the value in the named columns:
*Ft/Lb and **TPI. And, of course, the dimensions you gave in your post #1 of that core.

It is very easy to play with those numbers. !try it.
If you want to have more strength electromagnet, You need to increase the thickness of the wire.
You can divide by 2 the weight of the wire you have and coiling 2 parallel sections. This way you can increase the current flow doubled. 1,616 Amp.

*Feet per Pound.
**Turns Per Inch.

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14. ### wayneh Expert

Sep 9, 2010
13,440
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I have the math worked out for relating the length of wire to the number of turns around a cylinder. You can input any wire gauge and the spreadsheet has the all the wire data built into it. (It also computes the expected inductance.) Might take a bit of digging for me to find it, if you want it.

15. ### shortbus AAC Fanatic!

Sep 30, 2009
4,520
2,556
You don't say how or what the magnet will be used for. A straight/bar magnet will be strong on both ends, but not as strong as a "U"/horseshoe shaped magnet is on one single end. Given the same volts -amps turns on the coil.

And cast iron isn't a good material for an electromagnet core. It will stay magnetized when the power is turned off. If cast iron was good for electromagnetic things, they wouldn't make motors, transformers, and other stuff using steel for the cores. The cast iron is cheaper than steel, so there's a reason they don't use it.

16. ### ElectromagnetNewbee Thread Starter Member

Jul 13, 2014
69
0
So i sat down and worked through the math for some different coils. One of them was 5V/3.33A = 1.50Ohms.
Looked at the data sheet and got 0.0476 * 1.50 = 0.0714 Pounds of copper coil.

How do I convert pounds into feet?

17. ### wayneh Expert

Sep 9, 2010
13,440
4,274
You look up the properties of your wire in a table, and find the feet per pound, or lbs per 1000ft value.