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I want to make for some demonstration (caching some paper clips on ON and doping them on OFF) electromagnet made from copper wire (should it be single wire?) with ferromagnetic inside coil: enter image description here

As power source I am going to use Panasonic POLV19CE AC ADAPTOR INPUT: 220-240V ~ 50/60Hz 40mA OUTPUT: 6V DC 500mA

Is it safe to use this adapter as power source of the? Is the power of the adapter enough? The copper wire is enough or it is necessary to add some resistance?

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  • \$\begingroup\$ What's the resistance of the wire? \$\endgroup\$ – immibis Sep 19 '18 at 23:48
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    \$\begingroup\$ Your picture shows a short length of wire that is thicker than it should be. With 6 volts, connecting the wire will be like a short circuit. The power supply will shut itself off of fail. Use a lot longer piece and more turns of thinner wire and a single D cell battery (1.5 volts). Learn how to calculate current from voltage and wire resistance. Learn about wire size and resistance per length. \$\endgroup\$ – Charles Cowie Sep 20 '18 at 0:00
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    \$\begingroup\$ Wrenches are a bad choice for such a demonstration because their alloy is selected so they are hardly magnetizeable —because that could give you problems with metal saw dust in your workshop. Try cast iron instead. \$\endgroup\$ – Janka Sep 20 '18 at 0:14
  • \$\begingroup\$ @Janka Thanks, I will use just a big nail or something like this. \$\endgroup\$ – XuMuK Sep 20 '18 at 0:24
  • \$\begingroup\$ @CharlesCowie Yes, I know that it is too short. Is this calculator OK? \$\endgroup\$ – XuMuK Sep 20 '18 at 0:29
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To pick up some paperclips, your gigantic wire in the photo needs a current of roughly ten to thirty amperes. (You need 100 or 200 amp-turns.)

Much easier: don't wind eleven turns at 10 amps, wind 110 turns at 1 amp, or 1,100 turns at 100mA. All these give the same field-strength (the same value of amp-turns.) How many turns can you stand to wind, without going crazy? Maybe 500 turns?

Once you decide on the number of turns, then you can use your 6V and 500mA power-supply values.

For maximum magnetic field (max watts,) your total wire resistance needs to be 6V/0.5A = 12 ohms. If each turn of wire is 1-2 inches diameter, or 3-6 inches of wire per turn, then from the number of turns, add up the total wire length needed. Use a wire-gauge table to figure out what thickness of wire needed to give 12 ohms for the length you've chosen. Then, figure out the total weight (like, a half pound?) Buy more than that amount (buy enameled motor-winding wire, lots of it on Amazon.)

Either that, or just buy a half-pound of #30-gauge enameled wire. Wind a few hundred turns on an iron object. Hook it to a variable power supply, and crank it up until it gets somewhat warm, like five or ten watts. That's the simple way to produce a maximum field. To get higher field, you need to run hotter (with cooling fans.) That, or add more pounds of wire at the same wattage. Heavier total copper gives stronger fields, since it gives more amp-turns for each watt of resistive heating.

The rules: all electromagnets are limited by how hot they can get without being ruined (charred insulation down inside.) That sets a certain max wattage. Then, when operating at the max wattage, the total volume of the metal wire-windings determines the field strength. So, bigger electromagnets can produce stronger fields (without catching fire like smaller electromagnets would.) To double the field, just double the pounds of copper used.

For any given power, ten pounds of copper wire will quite literally give you ten times stronger fields than one pound of copper. The magic rule! Also it's the fundamental reason why a 10-horsepower motor is bigger, it's why a ten-kilowatt transformer is bigger. (However, you can cheat by using iron core designs, or liquid cooling pipes.)

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  • \$\begingroup\$ So, I am planning to get something like 6 meters of ⌀2cm for coil (~100 turns) and ⌀0.1mm wire. It should be enough. Or it is better to get more turns with smaller diameter (longer coil)? \$\endgroup\$ – XuMuK Sep 20 '18 at 8:53
  • \$\begingroup\$ @XuMuK more total copper is always best (more grams.) 0.1mm wire is far too thin! A good coil would be 100M long, not 6M, and using thicker wire than 0.1mm. The wire chart says that 0.1mm wire is 0.66ohms per foot, so 6M is correct, 13 ohms, but gives a small weak coil, see amasci.com/tesla/wire1.html#awg. Instead try this: use a new fresh 9V alkaline battery to power your wrench in the photo, but only connect it for very short times (three seconds.) It will provide about ten amps! But the battery will greatly heat, and rapidly drain. So only perform very brief tests. \$\endgroup\$ – wbeaty Sep 21 '18 at 2:40
  • \$\begingroup\$ I will try this as best and fastest option. And it takes some time to wait when the wire will arrive. \$\endgroup\$ – XuMuK Sep 21 '18 at 12:39
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A wrench is bad — an Iron core material has 10x more permeability than that wrench.

Also if you use that small of a wire (don’t forget insulation has its ratings, along with wire) you’re going to start a fire — if you don’t get shocked first lol

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  • \$\begingroup\$ I totally agree that the wire is too wide and the length is too short. It was just demonstration purposes. I won't ever try to connect AC adapter if I am not sure. I have some bad experience. \$\endgroup\$ – XuMuK Sep 20 '18 at 8:01

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