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enter image description hereI have built a toroidal inductor by wrapping an insulated copper wire over a toroidal laminated soft-iron core. The dimensions of core and copper wire are as below:

Toroidal core: Inner diameter = 20 mm Outer diameter = 70 mm Height = 15 mm

Copper wire: Total length = 150 m Diameter = 0.100 mm

My main concern is that what maximum DC current and voltage can I allow to flow through this inductor?

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    \$\begingroup\$ I'm failing to see how what you have described can be an electromagnet. Maybe draw a picture. \$\endgroup\$
    – Andy aka
    Aug 14, 2021 at 9:36
  • \$\begingroup\$ I have added one diagram for the clarification \$\endgroup\$ Aug 14, 2021 at 9:43
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    \$\begingroup\$ That isn't an electromagnet. You need an air-gap for an electromagnet. It's an inductor but it isn't an electromagnet. \$\endgroup\$
    – Andy aka
    Aug 14, 2021 at 9:47
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    \$\begingroup\$ You have not provided enough detail to answer this question. The size and characteristics of the toroid core have nothing to do with the max. DC current. That is determined by the output capacity of your power supply or the resistance of the wire you have used. Both are unspecified. \$\endgroup\$
    – jwh20
    Aug 14, 2021 at 9:57
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    \$\begingroup\$ How much DC current can be supplied by your power source and what is the resistance of the wires and how much current can those wires safely carry without burning. That is what you need to ask but, it's pointless if you think you were designing an electromagnet because... it isn't one. \$\endgroup\$
    – Andy aka
    Aug 14, 2021 at 10:04

2 Answers 2

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Your photo shows a closed torroidal core. All the flux will be contained inside the core so this will not perform as an electromagnet as you have supposed in your question title and post.

Core diameter will not affect the flow of current (except at switch on due to inductance). Your problem is simply one of wire resistance. See Current capacity of very thin (0.1mm ~ 0.5mm) copper wires? for details on that.

What maximum DC current and maximum voltage can I flow through an electromagnet?

Current flows through. Voltage is applied to the load. Once you have calculated the resistance of the wire and decided on the current required then you can calculate the required voltage from Ohm's law, V = IR.

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  • \$\begingroup\$ Your answer is very accurate to calculate the input voltage. Since the resistance of the coil will be very less, so can I use a resistor in series to avoid the heating? \$\endgroup\$ Aug 15, 2021 at 11:05
  • \$\begingroup\$ You can, but the resistor will get hot so you are wasting power and not really solving anything. It's much better to use the right voltage or use a wire with the right resistivity. What exactly are you trying to make? \$\endgroup\$
    – Transistor
    Aug 15, 2021 at 12:34
  • \$\begingroup\$ I am trying to make an inductor \$\endgroup\$ Aug 16, 2021 at 12:16
  • \$\begingroup\$ Your question title and post still refer to an electromagnet. You need to edit it. \$\endgroup\$
    – Transistor
    Aug 16, 2021 at 13:26
  • \$\begingroup\$ Yes.. I have edited it now.. \$\endgroup\$ Aug 17, 2021 at 5:33
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My main concern is that what maximum DC current and voltage can I allow to flow through this electromagnet?

The main question is 'for how long?'

The failure mode of that copper wire is it getting too hot for the insulation. If you want to run it for tens of minutes, then the heat must be dissipated continuously. While it's theoretically possible to compute a temperature rise from dimensions and fluid mechanics of air, it's far easier to just switch it on at low power and see how hot it gets. Rinse and repeat with increased power until you just can't touch the coils (will be 60 to 80 °C) which most insulation should survive.

If you want to run it for seconds or milliseconds, then heat will be absorbed into the thermal capacity of the wire rather than dissipated, and you will be able to run very large currents, with a constant value for \$I^2t\$. Reduce the operating time by 4, and you can double the current. Pulse it on for a second, estimate the temperature rise, and use \$I^2t\$ to extrapolate to other times and currents.

BTW, what you have there is not an electromagnet, it's an inductor with approximately zero external field (exactly zero if the windings are uniform). It won't attract pieces of iron, and should barely affect a compass needle.

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  • \$\begingroup\$ Your response sounds good. So can I use a resister in series with this circuit to counter the heating effect? \$\endgroup\$ Aug 15, 2021 at 11:03
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    \$\begingroup\$ @ShivamKumar A series resistor would reduce the current, which would reduce both the heat and the magnetic field in the coil. The resistor would also get hot. You would achieve the same effect by reducing the voltage from the power supply. \$\endgroup\$
    – Neil_UK
    Aug 15, 2021 at 11:09

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