I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

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    \$\begingroup\$ Welcome to EE.SE. I would say both, but it takes voltage to push amps through the resistance of the wire. If taken to extremes, the resistance of the wire will limit the amp, hence the field strength. Some iron and ferrite cores are referred to by max ampere-turns, beyond which the core saturates. \$\endgroup\$ – Sparky256 Mar 6 '18 at 3:11
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    \$\begingroup\$ it appears that you do not have a clear understanding of the relation between voltage, current and resistance ... please explore ohm's law \$\endgroup\$ – jsotola Mar 6 '18 at 3:12
  • \$\begingroup\$ For high intensity magnetic fields, only short pulses of current will work, or the wire will start to burn. \$\endgroup\$ – Sparky256 Mar 6 '18 at 3:13
  • \$\begingroup\$ I would research into electromagnets as well. \$\endgroup\$ – Sparky256 Mar 6 '18 at 3:19
  • \$\begingroup\$ @jsotola. How do you get that gray 'hi-light' effect? \$\endgroup\$ – Sparky256 Mar 6 '18 at 3:21

Firstly, voltage plays no part in the strength of an electromagnet, it's only the current through the windings that generate the field. Consider a super-conducting magnet with zero resistance windings. There's no voltage, no power dissipation, and a large magnetic field.

However, most of us don't have the luxury of using a super-conductor to wind our magnets, we have to make do with a good-conductor, like copper. As it has resistance, we need to apply a voltage across it to push a current through it. This results in bad things like heating in the coil, whose temperature needs to be limited if it's to keep working.

For any given wound magnet, we do not have independent control over voltage and current. The magnet coil defines the ratio of those, it's called its resistance. If we apply a voltage, then (voltage/resistance) current flows. If we connect a current source, then the voltage across the coil responds, and becomes (resistance * current).

The limit for both voltage and current is given by the coil's cooling. If we operate the coil for a very short time, we can increase the power supply, and allow the heat to be stored as a temperature rise in the windings, as long as we switch off before the coil overheats, to let it cool down before the next pulse.

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  • \$\begingroup\$ What frequency do you recommend for ~17mΩ? I'm thinking it should at least be less than 2Hz... \$\endgroup\$ – Sir Wumpus IV Mar 6 '18 at 7:02
  • \$\begingroup\$ @RubiQuikRQK What frequency do you recommend for ~17mΩ? I'm thinking it should at least be less than 2Hz... Your question makes no sense to me at all, ohms and Hz are not compatible units. Would you clarify or expand your question please. \$\endgroup\$ – Neil_UK Mar 6 '18 at 8:44
  • \$\begingroup\$ What I meant was with a magnet wire that has a resistance of ~17mΩ, what frequency of bursts of electricity with what cooldown time would you recommend to keep the magnet at a reasonable temperature? Sorry, I realize now that my question was somewhat confusing. \$\endgroup\$ – Sir Wumpus IV Mar 8 '18 at 4:53
  • \$\begingroup\$ depends on thermal resistance ['C/W] * I^2R [W] = Temp rise ['C] \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Mar 8 '18 at 4:58
  • \$\begingroup\$ If the pulse frequency is more than a frequency that depends on the size of the magnet, and for one I could hold in my hand, I'd say in the 10mHz to 100mHz range (one or a few per minute), then the frequency is irrelevant, you're only looking at the duty cycle as a way to reduce the average power dissipation. Once the frequency is less than perhaps 1mHz (3.6 per hour), then the frequency is irrelevant, you're only looking at the pulse length which controls the total energy delivered to the coil in a single pulse. \$\endgroup\$ – Neil_UK Mar 8 '18 at 6:36

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