# Strongest Electromagnet using 2 AA bateries

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by Trisorion, Aug 17, 2008.

1. ### TrisorionNew Member

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Strongest Electromagnet using 2 AA batteries

I am currently competing in a voluntary class contest to build the strongest electromagnet possible under these guidelines.

Objective: To construct an electromagnet that can pick up as much weight as possible.
Construction: Any construction may be used but the maximum power source is two (2) 1.5 volt batteries.
Competition: There are 2 segments to the competition:
1. Power: The greatest number of paper clips that can be transported from one bin to another in 20 seconds. Note: You may not touch the paper clips with your hand or any other apparatus.
2. Energy: The greatest number of paper clips that can be held at one time.
Score will be determined in each segment by the following formula:
Score = # Paper Clips (by mass​

Any and all feedback is welcome. Please feel free to share any general tips, suggestions for materials or thoughts on the shape of the electromagnet. However, I do have some more specific questions. I understand these are not really electronics questions, but it does not hurt to ask. I will edit the answers into this thread after they are given for future readers.

1)Does wrapping a larger wire one layer thick have the same effect as wrapping many layers of a small wire to achieve the same thickness?

2)Can you ever wind too much wire into the electromagnet? Or to put it another way, is there any downside to winding more wire on?

3)What dictates the thinest gauge wire you can use? current? amperage? voltage?

4)What effects the strength of the magnetic force generated the most: the voltage, amperage or total current running through the wire? Another way to ask is, should I place the batteries in series or in parallel?

5)Would it be better to add a large capacitor, drain the batteries to it, then use it for the heavy lifting. I might have to wire this up with some switches because there is no time limit on the heavy lifting, but i cant waste time charging in the 20 second power round. I think 10-15 seconds of heavy lifting would be much more useful than 3 minutes of slow battery drain.

6)How dangerous is a large capacitor?

Thank you very much for the help

Edit: I just realized I missed a "t" in batteries in the title, unfortunately I cant change it.

Last edited: Aug 17, 2008
2. ### Hero999Banned

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Generally the more turns the better but you need to consider that more turns will have a higher resistance so not as much current will be able to flow at a given voltage.

Thicker gauge wire is also better as it has a lower resistance.

You also need a good core, does it matter if it stays magnetised when the power is removed? If not then use a 'hard' iron core that acts as a permanent magnet when the external field is removed. If you need it to loose its magnetism you could degauss it be applying an alternating voltage to the coil.

3. ### WillbeNew Member

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I'm going to be a little abstract here.

Magnetomotiveforce is proportional to the product of amp-turns, so 10 A flowing through 10 turns [= 100 a-t] equals the force exerted by 1A flowing through 100 t [= 100 a-t].

You should model the battery by measuring its open circuit voltage and maximum current delivered, let's say into a load that gives 0.2 volts across it and use this to figure out the Thevenin equivalent internal battery resistance.
Leave the load on just long enough to make the measurements.

Then run through tradeoffs with this model, the number of turns and wire size and wire resistance on a spreadsheet. Use a graph to find the optimum point.

To approximate the number of turns in the coil, figure out the cross sectional area of the windings using the formula for a cylinder of max winding diameter and then of min winding diameter, then subtract the two volumes, then use a correction factor for the wire having a round cross section [the entire area is not filled by copper, it has spaces between the wires].

#30 AWG wire in air fuses at about 10A, but it will carry more current if the heat is spread out through the volume of the windings.

If it doesn't work on paper it certainly will not work in reality.
If it does work on paper it still might not work in reality.

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5. ### TrisorionNew Member

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How quickly would a hard iron core degauss? In the heavy lifting section I would want to use the additional magnetism, but in the speed challange I would want it to release the paper clips quickly. Also, does the voltage need to be alternating in the low hertz scale, or could i just reverse the flow in the circuit for 1/4 of a second?

6. ### WillbeNew Member

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If I made no derivation error

mmf = (Vb*t)/[Ri + PI*D*t*k]

where Vb is the battery voltage = 3v, t is the number of coil turns, Ri is the internal resistance of two cells in series, D is the coil diameter in inches and k is the ohms/inch of the wire.

Typically
#30 AWG = 1 ohm/120 inches
#20 = 1 ohm/1200 inches
#10 = 1 ohm/12000 inches

You might be able to reduce Ri by heating the battery.

With Ri of 0.3 ohms and a PI*D*t*k product of 0.3, the Ri term will begin to predominate the mmf value, so I'd keep this product between 0.3 and 0.06.

Last edited: Aug 17, 2008
7. ### GlyphNew Member

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Will you be allowed to use superconducting wire?

If so, and if you can acquire the wire and the liquid nitrogen to cool it, you could create a very powerful magnet. Since it would give you alot more turns for the same resistance (in theory infinite, but you still have to consider the resistances in the battery itself and the non-idealities in the wire and the contacts).

8. ### MikebitsWell-Known Member

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It's probably hard to source the stuff, but it's not like black magic or anything. Just find an intelligent person at a hospital to find out what local company they source the liquid nitrogen/helium for their MRE machines. Long term storage isn't really practical but short term storage for transport is as simple as a ventilated vacuum carafe. As long as the materials don't crack you could use an off the shelf Thermos.

10. ### MikebitsWell-Known Member

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Properties: Liquid Nitrogen

Liquid Nitrogen has a boiling point of -195.8°C
Volume of expansion liquid to gas (at 15°C, 1 atm.) = 682.1
Sg = 0.808 (at -195.8°C).
Density of liquid (normal boiling point, 1 atm.) = 0.807 g/cc
Colourless, odourless liquid similar in appearance to water.
Known or Expected Hazards

a) Temperature Related

The extremely low temperature of the liquid can cause severe burn-like damage to the skin either by contact with the fluid, surfaces cooled by the fluid or evolving gases. The hazard level is comparable to that of handling boiling water.
The low temperature of the vapour can cause damage to softer tissues e.g. eyes and lungs but may not affect the skin during short exposure.
Skin can freeze and adhere to liquid nitrogen cooled surfaces causing tearing on removal.
Soft materials e.g. rubber and plastics become brittle when cooled by liquid nitrogen and may shatter unexpectedly.
Liquid oxygen may condense in containers of liquid nitrogen or vessels cooled by liquid nitrogen. This can be extremely hazardous because of the pressure rise on the slightest degree of warming above the boiling point of oxygen ( -183°C ) and the possibility of explosive reaction with oxidisable material.
Thermal stress damage can be caused to containers because of large, rapid changes of temperature.
b) Vapour Related

Large volumes of nitrogen gas are evolved from small volumes of liquid nitrogen (1 litre of liquid giving 0.7 m3 of vapour) and this can easily replace normal air in poorly ventilated areas leading to the danger of asphyxiation. It should be noted that oxygen normally constitutes 21% of air. Atmospheres containing less than 10% oxygen can result in brain damage and death (the gasping reflex is triggered by excess carbon dioxide and not by shortage of oxygen), levels of 18% or less are dangerous and entry into regions with levels less than 20% is not recommended.
Oxygen condensed into leaking containers can explode on heating following resealing or blockage with ice
-------------------------------------------

Is it just me, or does this sound like a bad idea?

Last edited: Aug 18, 2008
11. ### jpanhaltWell-Known MemberMost Helpful Member

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Back to the batteries... Will you supply your own or will the school provide a "standard" battery? If you supply your own, find one with a low internal resistance. That is, one that can supply the most current (see: Willbe post#3). Batteries with a larger physical size and capacity often have lower internal resistances compared to the same chemistry with a small physical size.

NiCd's can have very low internal resistance, even though their nominal voltage may be less than a non-rechargeable alkaline battery. Thus, two large NiCds may give more magnetic force than
to alkaline batteries with a higher internal resistance.

Here are some (of many) sites that show internal resistance under various conditions:

Internal resistance for alkaline batteries (click on datasheet): http://www.duracell.com/oem/productdata/default.asp

NiMH(scroll down to 5.7.1):
http://www.duracell.com/oem/Pdf/others/nimh_5.pdf

NiCd lowest internal resistance under conditions of test:
http://www.batteryuniversity.com/partone-22.htm

Good luck. John

12. ### mashersmasherNew Member

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make sure to cut 2 of those huge 1.5v cells out of a 6v square battery. give you a bit of current

13. ### TrisorionNew Member

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I dont think I am going to be using liquid nitrogen, but it was an interesting idea.

I will be supplying my own batteries, and they must be AA, I don't think i can cut cells out of other batteries.

I am sure Willbe derived it correctly, but I have practically no knowledge of how this works. Could someone explain that in layman's terms? I was following him until "so I'd keep this product between 0.3 and 0.06."

Thanks for all the help everyone.

Last edited: Aug 18, 2008
14. ### TrisorionNew Member

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Last edited: Aug 18, 2008
15. ### WillbeNew Member

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Here's entries cut and pasted from a spreadsheet.

mmf = (Vb*t)/[Ri + PI*D*t*k]

vb= 3
Ri= 0.3
t= 1000
PI= 3.14
D= 0.3
inch= 1200
k= 0.000833333

PI*D*t*k= 0.785

mmf= 2764.976959

Your mission, should you decide to accept it, is to change D, t, and k until you maximize mmf while staying within your practical, budgetary and time constraints.

Anyway, in this example, I fussed with t and inch [which is inches of wire to get one ohm] to maximize mmf.

Going from #20 wire to #10 wire didn't have that much effect so I stayed with #20. Winding #30 wire around a form will probably be easier than winding #20.

I didn't try changing D.
You can see from the formula that if I doubled D and halved k it would not change the mmf, so you can already see a tradeoff between D and k.

Since the PI*D*t*k product is more than Ri, you can lower this product and still get a sizeable increase in mmf. I would lower this product by changing only D or k or both.

If you want to go to 1000 turns, chuck the coil form in a variable speed drill and wind it that way.

The problem is, I've never actually tried this!

You may want to come up with a way of measuring the mmf and try it for some simple examples to confirm the formula, or find a commercially published formula for a coil, or experimental data that could confirm/deny whatever formula you use.

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You have to punch in the numbers of what you have available. Wire thickness length etc.. into that equation, he was saying if the internal cell resistance is .3 ohms then you'll get maximum magnetic force by keeping the product of the PI*D*t*k part of the equation between .3 and .06

17. ### jpanhaltWell-Known MemberMost Helpful Member

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On one of the sites I referenced (as I recall) it shows that the internal resistance of NiMH increases when discharged at high currents. The implication I got was that that was a permanent effect. So for the "finals" you might want to use NiMH that had not been abused.

It is also something you could add to your documentation. You know, max. number of paper clips vs. number of charge/discharge cycles. John

edit: NiCd's, if you use them, can have extremely high discharge rates -- at least once in their lives. I have seen them actually smoke when shorted.

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NiCads are still used in RC cars frequently because of their extreme ability to discharge current. I think they even beat primary alkaline cells. The 1.2 volts vs 1.5 volts is completely over shadowed by the lower internal impedance, as the voltage on a NiMH or alkaline cell will drop well bellow a NiCads at the same current. The only thing better is lead acid, even then cells in parallel have such low equivalent resistance you may want to think about running then in parallel instead of series for the increased current, you just have to wind your coil to keep in mind the 1.2 volts. The current will be very large.

19. ### TrisorionNew Member

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So you derived this with:

mmf= maximum magnetic force
I= current
V= voltage
R= resistance
Rb= internal resistance of batteries
Rw= resistance of wire
T= number of coil turns
D= diameter of wire
K= resistance per inch
Ct= circumference of a turn

Where Vb is the battery voltage = 3v, t is the number of coil turns, Ri is the internal resistance of two cells in series, D is the coil diameter in inches and k is the ohms/inch of the wire

mmf= I*T
I=V/R
mmf= (V/R)*T
mmf= V*T/R
R= Rb + Rw
mmf= V*T/(Rb + Rw)
Rb= Rb1 + Rb2 (since they are in series, right?)
mmf= V*T/(Rb1 + Rb2 + Rw)
Rw= K*Ct*T
mmf= V*T/(Rb1 + Rb2 + (K*Ct*T)

20. ### jpanhaltWell-Known MemberMost Helpful Member

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Sceadwain makes a good point about paralleling the cells. Two cells in series will have twice the voltage and twice the resistance. Two in parallel will have half the voltage as a series connection, but one-forth the internal resistance and be able for a short time to provide more current to a very low resistance coil.

Since it is for school, you may want to include that concept too, regardless of what type of cell chemistry you use.

John

21. ### WillbeNew Member

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To measure the battery internal resistance, you can use
Ri = (Voc-V)(R/V)
where Voc is the open circuit voltage of the battery, V is the voltage across a load resistor and R is the value of the load resistor in ohms.

So, for Voc = 3.1v, R = 1 Ω, V = 2v, then Ri = 0.55 Ω.

Ri probably changes with the current drawn out of the battery so you might try different load resistors, like 10, 3, 1, 0.3, 0.1 Ω.

If you can post some measurements we can crank out some mmf values.

Last edited: Aug 18, 2008