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Can anyone please let me know how to generate kilo ampere current in a given circuit? I am thinking of the possibility of creating a magnetic field in a large solenoid, say of radius 5 m & length 15 m. There cannot be iron core inside the solenoid to enhance the field. So, to generate 1-2 tesla field, I will need quite large current and the wire should have many turns. It looks like the standard 1 mm wire can carry only 10 A which is not sufficient. But is it impossible to generate kilo ampere level current in non- superconducting wires? If so, what will be the dimensions of the wire? Is there any example to learn from? Any help is appreciated.

Regards, Kolahal

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  • $\begingroup$ Thank you all for your responses. I am working in simulation and data analysis and did not have enough exposure to magnet design hardwares. Thus, I was not sure if making such solenoid is feasible. I am working in an particle physics experiment (MicroBooNE) that is a very large rectangular parallellopiped: 3 m x 3 m x 10 m and this ground based detector is always showered by the cosmic muons. I am thinking of the possibility of trapping these cosmic background by using a magnetic mirror configuration:(en.wikipedia.org/wiki/Magnetic_mirror), (cont.) $\endgroup$
    – kolahalb
    Commented Jun 22, 2018 at 14:49
  • $\begingroup$ (cont.) by placing it on top of the parallellopiped. This will deflect the muons and prevent them from reaching the detector. As you can see, I shall not need much uniformity. Using a superconductor will be impractical in this case: creating the equipments on top of the experimental hall, it seems. I was looking for a crude working solution. $\endgroup$
    – kolahalb
    Commented Jun 22, 2018 at 14:49

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A wire of 12mm diameter has a max rated current of about 700A. If you wound that over 15m, you'd have 1250 turns. The field strength is then: $$ B = \mu_0 {NI\over{L}} $$ Plugging in the numbers comes to 0.07 Tesla, which is a bit short on what you want.

Your solenoid is the size of a bus and the copper wire, as thick as your thumb, carrying 700A would be running at about $200^oC$. That's a lot of heat to get rid of. If you increase the current, you get more heat until the wire melts - long before you get to the field strength you need.

Having said that, such magnets do exist! They are used to bend charged particles coming out of interactions in particle accelerators. But I'm afraid they all use superconducting coils...

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  • $\begingroup$ Thank you, for your response. Is not it possible to have multiple layers of turns of the wire? Say, we take the length of the solenoid to be 12 m (for ease of calculation with the wires of 12 mm diameter). Then, over 12 m, we will need 1000 turns. Say, we wrap it up 15 times: i.e. 15000 turns in total over 12 m. This should result in: $4\pi\ 10^{-7}\ (1000*15/12) * 700$ tesla ~ 1 tesla. $\endgroup$
    – kolahalb
    Commented Jun 22, 2018 at 16:06
  • $\begingroup$ The problem is the heat dissipation - with one layer of wire you already have over 30MW of heat. If you add more layers it goes up accordingly. You are basically making a giant electric oven. $\endgroup$
    – Oscar Bravo
    Commented Jun 25, 2018 at 9:21
  • $\begingroup$ Yes, this indeed looks too ambitious to be feasible. I may need to resort to inserting a ferromagnetic material of high permeability (like MuMetal bar) on top of the experimental hall. $\endgroup$
    – kolahalb
    Commented Jun 26, 2018 at 17:54
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I did an experiment where we had a normal-conducting, iron-free, toroidal magnet operating between 1kA and 10kA, driven by a megawatt (in this case, 100V) power supply. So yes, it's possible. The conductors were hollow copper channels, exterior cross section square roughly 5cm on a side, water flowing in the interior channel to carry away the heat. The power supply electronics enclosure was about the size of an RV, and became somewhat unreliable when the room was irradiated. It was a major engineering project and there was internal debate during the experiment about whether a superconducting magnet would have been simpler.

This was the "QTor" magnet for the "QWeak" experiment, and there are some published technical reports that I can link when I'm not posting from mobile.

I have also heard of a large solenoid which needed to be operated with low inductance, to turn the field on and off rapidly, whose final design (and working implementation) was something like the geometry you describe and operated at 3kA. Unfortunately I don't recall the name of that project but I am sure there are published technical papers describing it as well.

(Wait, did you really mean to write "five meter radius"? That's enormous! QTor's enclosure was a cube-ish frame five or six meters on a side, and that was enormous. You won't get much uniformity from a fifteen meter solenoid in that case. Now I am terribly curious about why you want to produce this field.)

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  • $\begingroup$ Hello Rob, it will be great if you can refer me to the note where you were engaged in the design of kA current. I also checked with Bravo's suggestion of 12 mm wire and that is also coming to be megaWatt level - I think that is perhaps unavoidable. $\endgroup$
    – kolahalb
    Commented Jun 22, 2018 at 15:59
  • $\begingroup$ Since you mention elsewhere that you're working on MicroBooNE, I'll just point you towards the magnet section of the Qweak public document database and invite you to poke around. $\endgroup$
    – rob ♦
    Commented Jun 22, 2018 at 16:20
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Current, I, is given by the formula:

$$ {I=\frac{V}{R}} $$

So, for a current of 1000 A, you need:

$$ {1000 A = \frac{volts}{resistance}} $$

Resistance is proportional to the length of a wire, and inversely proportional to the cross-section of a wire. If you have more turns, you increase the resistance, which reduces the current at any given voltage. As you're interested in making an electromagnet, it's worth saying that this reduces the current by exactly the same ratio that increasing the number of turns strengthens the magnetic field.

So, for your example, you either need a thicker wire, or you need more volts. Copper has a resistivity of $1.68\cdot10^{-8}\Omega/m$, so a wire with a cross section of $1mm^2$ and a length of $1m$ would have a resistance of $0.01678 \Omega$, and a 1kA current flowing through that 1m wire would require a voltage of

$$ V = 1000A \cdot 0.01678 \Omega = 16.78 volts $$

However, it would get very hot. Power is $P=VI$, here $16.78V \cdot 1000A = 16.78 kilowatts$. Normally this sort of power demand is engineered away by using low current and high voltage, but you don't want that, so instead you will need to lower the voltage by making the conductor thicker. Make it twice the cross-section to halve the resistance, and halve your voltage, and halve your power use. A solid copper bar with a $1cm^2$ cross section, being 100 times the cross section of your wire, would reduce the power use to 167.8 W, which is far less likely to melt when you switch it on.

And remember, with these numbers, this is what it looks like for each $1m$ long section of the coil. A 15m long, 1m radius cylinder with 2 turns/meter would be close to 190 such sections, needing 190 times the voltage and dissipating 190 times the power. For the $1cm^2$ copper bar example, that's ~31.9 volts and 31,882 watts for 1kA. (Which wouldn't be anything like enough for 1-2 T field, but that scales up in a nice clean way so I won't go into it).

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    $\begingroup$ It will probably be hard to find a power source that can supply 1000A at 17V. $\endgroup$
    – Jasper
    Commented Jun 22, 2018 at 12:43
  • $\begingroup$ @Jasper Oh, totally. But at this scale, you'd be building everything from scratch anyway, so I chose to ignore that problem. $\endgroup$
    – BenRW
    Commented Jun 22, 2018 at 23:18

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