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DIY electromagnet under £100: Disappointing results

Hi everyone,

I took part in a competition of my college's electronics club to build an electromagnet under =A3100, with the strongest and cheapest one winning.

Approach #1: I bought a 500W ATX power supply and used the 12V 35A lead to energise an electromagnet. I used a full 100m length of 0.8mm thick enameled copper wire - that length had a resistance of about 5 ohms. The wire reel in the as-bought state conveniently had both leads exposed.

I put a 10mm steel hex bolt through the reel's centre hole as the core material. The electromagnet was feeble and was no where even close enough to carrying its own weight :(

Approach #2: I also tried attaching a 6kV Cockroft-Walton voltage multiplier (as in instructables.com) to the same 100m 0.8mm thick enameled copper wire reel, but although the current was presumably very high it was of such short duration that some 5mm chrome steel bearings I had nearby didn't even budge :(

So I'm back to square 1 and am open to suggestions on how to create the strongest possible electromagnet under a tight budget.

Incidentally, am I correct in concluding that the length of the wire is not relevant to the magnet's strength as the resistance is proportional to the length but the strength is proportional to the number of turns (hence length) so both terms cancel out?

So presumably the only electrical considerations as far as a resistive electromagnet is concerned is how many amps you can put through it before you burn it out. Does this imply that thick copper wire (a few mm at least) with a large thermal mass and a very high voltage, high capacitance power supply is the way to go?

Thanks, Jessie xx

Reply to
Jessie
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--- For a solenoid, its internal magnetic field is described by:

µ N I B = ------- l where µ is the permeability of the core, N is the number of turns surrounding the core, I is the current in the coil, and l is the lenghth of the core

so, you can see that B will increase if the permeability of the core increases, the number of turns increases, the current increases, or the length of the core decreases.

To get the maximum strength, then, you want to wind a short solenoid with a lot of low resistance wire because, as the resistance increases the current will decrease for a given supply voltage.

Also, you'd like the external field to add to the internal field, so you'd want the solenoid to be encased by a high-permeabilty core.

To do that you'd want the core to be shaped like a bundt cake pan, with the coil nested inside of it, like this side view: (View in Courier)

. +---+ +---+ +---+ . | |ooooooooo| |ooooooooo| | . | |ooooooooo| |ooooooooo| | . | |ooooooooo| |ooooooooo| | . | |ooooooooo| |ooooooooo| | . | +---------+ +---------+ | . | | . +-------------------------------+

That way, the part of the magnetic field which would be lost will be captured in the walls of the "core" and will add to the pull of the center leg.

That's how junkyard electromagnets are made, BTW.

---

--- Both lengths _don't_ cancel out since one is the length of the wire used to wind the coil and the other is the length of the wound solenoid.

---

--- In a sense, yes, but you also need to be concerned with how many turn of wire you can get on the thing because that, and the current in the coil (ampere - turns) will determine the strength of the magnet.

---

--- Not necessarily.

If you want to get a feel for the numbers, here ya go:

Let's say we have a core that looks like this:

. |

Reply to
John Fields

If an electromagnet is thermally limited, and you plan to fill the available winding window, I think it doesn't matter what size wire you use. So size the wire to match whatever power supply you have, to get as many watts into the coil as it can stand.

It's like a transformer. A 100 VA transformer is the same size whether it has a 120 volt primary, a 240 volt primary, or a dual primary.

Square wire *is* better for a couple of reasons.

John

Reply to
John Larkin

snip

A persuasive argument but would it not be the case that if you wound with wire half the diameter of #10 say #16 that there would be 400 turns with .16 ohms resistance (both four times as much as with #10) but half as much current for the same power level since 25*25*.16 = 100 so there are 10000 ampere turns which is twice as much as before.

Reply to
joe

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