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I’m trying to make a device that can propel rings of conductive material upwards and make them hover, like the one seen in this video at 6:24, albeit on a smaller scale. I got a stepdown transformer to 24-volt AC and was having trouble getting this effect to work and realised that dealing with AC will be complicating things beyond the simple \$ B = \frac{\mu I N}{l}\$ because, assuming a purely inductive load, \$ I = \frac{V}{2\pi f L} \$ and \$ L = \frac{\mu N^2 A}{l}\$, which has some of the same variables as \$B\$.

I can insert an iron core into my coil to increase the magnetic field strength, but that will also increase the inductance, which will decrease the current, which will decrease the magnetic field strength, possibly leading to zero or negative change in the final magnetic field (so it seems to me that the only reason for the core is to help bring the magnetic field up higher from the coil). I decided to do some maths to figure out which variables I should change to increase my magnetic field:Maths relating Magnetic Flux to AC coil dimensions that doesn't make much sense

I substituted values in 2 different ways to see if I had made a mistake but the result both times was that if I increase the number of turns in the coil, I'll decrease the magnetic field strength (and also that if I use a really low frequency, I'll get a massive magnetic field???), which sounds like nonsense to me. Clearly I am approaching this the wrong way. My first instinct is that there is some effect that I have not heard of that affects the magnetic field of AC coils much differently to DC ones, but also, for the purposes of getting the eddy currents required to overcome gravity in the aluminium ring, is the B-field even the most important number to be looking at? Any help with this will be greatly appreciated.

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  • \$\begingroup\$ Perhaps this will answer your question: \$\endgroup\$
    – qrk
    Jun 19, 2022 at 2:25
  • \$\begingroup\$ Nice catch by @qrk. Another way to see your question, Architect, is that \$B\propto\frac{N}{L}\$ but \$L\propto N^2\$ so \$B\propto\frac1{N}\$ once you cancel out \$L\$. Yet another: as you increase the turns, the impedance goes by the square and therefore the current declines also by the square (fixed frequency) and therefore the magnetic charge (Webers, i.e. volt-seconds) can't build up as much. \$\endgroup\$
    – jonk
    Jun 19, 2022 at 2:41
  • \$\begingroup\$ @jonk I understand, that's the problem that I have; I don't understand how this can be the case, because it suggests that one loop of wire will produce a much larger magnetic field than a more substantial number of turns would produce. How can I possibly produce a stronger magnetic field if the equation I derived is true? \$\endgroup\$
    – Architect
    Jun 19, 2022 at 2:59

2 Answers 2

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the result both times was that if I increase the number of turns in the coil, I'll decrease the magnetic field strength (and also that if I use a really low frequency, I'll get a massive magnetic field???), which sounds like nonsense to me.

No, it's correct.

Maximum field strength is attained by pushing the most power into the coil that you can without vaporizing it. This value is determined by the size of the coil only. The number of turns (for a particular coil size) determines the voltage. The more turns you have the more voltage you need, because while more turns produce more magnetism at the same current, the longer thinner wire has higher resistance which reduces the current and magnetic field more.

So you choose the coil size needed to produce the magnetic field strength you want, and then the number of turns and wire size to match the voltage you want to use.

If you have a certain size coil that isn't producing enough magnetism on the voltage are using, either raise the voltage, or reduce the number of turns while increasing wire size to fill the same volume. One way to do this is split a single winding into two halves and then wire them in parallel. This will give you half the turns but 1/4 the resistance for 4 times the current and double the magnetic field strength with the same amount of copper. Power consumption increases by 4 times too, just like it would if you doubled the voltage on the original coil to get double the current and magnetic field strength.

The highest current (and therefore highest magnetic field strength) is achieved at DC, where the inductance has no effect. If the coil has high inductance when reducing the frequency will lead to increasing current as the inductive reactance decreases.

If you have an iron core and reduce the frequency it may saturate at some point as the current increases, reducing inductance dramatically and causing a huge increase in current (but not magnetism). The same thing will happen if you raise the voltage.

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  • \$\begingroup\$ Thanks for this. You mention choosing the right coil size for the magnetic field I want. How do I find the optimal coil size for my desired magnetic field? Also, What if current is the bottleneck I am coming up against, rather than voltage? My power supply can only do 6.25 amps. Can I change the dimensions of my coil to use this current more effectively? \$\endgroup\$
    – Architect
    Jun 19, 2022 at 3:37
  • \$\begingroup\$ @Architect If you go back to your video, you will see that the professor is using AC, not DC. There is a phase shift in the four quadrants of AC operation, two of which attract and two of which repel, when the current induced in the aluminum (note the use of aluminum, by the way, and not copper) via non-Coulomb EMF generates its own field. By phase shifting (and this is a design issue which is complex as nature is working to oppose change but that opposition "takes time") in just the right way, the repulsion effect dominates the attraction effect. This ONLY works just that way, with AC. \$\endgroup\$
    – jonk
    Jun 19, 2022 at 4:52
  • \$\begingroup\$ @jonk I know that Professor Laithwaite is using AC. That’s why I used an AC supply and talked about AC effects in my question. I also generally understand how eddy currents work, and that is why I mentioned them in my question. I don’t see how your above comment addresses the new question I have under this answer. \$\endgroup\$
    – Architect
    Jun 19, 2022 at 5:25
  • \$\begingroup\$ Make the coil long and thin rather than short and fat. Use silicon steel laminations in the core. Examine the current waveform with a 'scope to make sure the core doesn't saturate (waveform should be a sine wave, not spiking at the peaks). here's an example running on 24VAC and 1.5A. 576 turns of 095mm dia wire 18.6mm id, 31.8mm od. Laminated core is 200mm long. hrcak.srce.hr/file/359719 \$\endgroup\$ Jun 19, 2022 at 8:37
  • \$\begingroup\$ @BruceAbbott you appear to have missed the very significant point of inductance being proportional to turns squared and kind of fudged your answer by introducing thinner wire with higher resistance. You also say that power produces magnetic field and that is also largely untrue. \$\endgroup\$
    – Andy aka
    Jun 19, 2022 at 8:45
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Why is my maths saying that increasing turn count will decrease magnetic field strength in an AC coil?...........which sounds like nonsense to me.

A coil's magnetic field strength is related to its current and the number of turns. It's called MMF (magneto-motive-force) and, it's measured in units of ampere-turns (not surprisingly) .

So, increasing current or turns (but not at the expense of each other) increases magnetic field strength. But, here's the problem with AC magnetics; if you (say) double the turns, the inductance rises by 4 times (it's a square thing) and, for the same applied voltage, the current drops by 4 times.

Hence, doubling the turns means a quarter of the current and, inevitably, the MMF falls to half. More turns means less magnetic field.

Also, halving the frequency means that the inductive reactance halves and, as a result, twice the current flows for the same applied voltage and turns. Take the frequency lower still and you will reach core saturation.

$$\color{red}{\boxed{\text{None of the above is nonsense}}}$$

it seems to me that the only reason for the core is to help bring the magnetic field up higher from the coil

You can use ferrous cores (laminated or ferrite at AC) to extend the magnetic field quite significantly but, to impact a magnetic field on to an object (the thing you want to hover) still requires an air-gap. The air-gap should be designed to match the size and shape of the object you want to hover in order to maximize the eddy currents in said object.

But, as you say, the added core (and object) will increase the coil's inductance and that appears counter-productive to your aims. However, the air-gap will dominate the effective inductance of the coil so it's not that big of a deal for a moderate air-gap. You can get a feel for this numerically if you looked at the data sheet for a ferrite transformer core with various air-gap distances: -

enter image description here

Image modified from here.

The ungapped core length for this product is 43.7 mm and, with a 1.1 mm air-gap (about 2.5% of the core length) the inductance has dropped to about 2.7% of the ungapped inductor. OK, looking at \$\mu_e\$ tells us that the permeability (and inductance) is still 45 times that of air-cored inductor but, with a much bigger air-gap (maybe 10 mm) it's only going to be maybe 3 times bigger. However, we should consider the benefits...

Without a core, the distance that the magnetic field travels in air is quite long and that means true H-field (magnetic field strength) reduces considerably due to widespread fringing (the "average" path for a line of flux is much longer).

H equals MMF/distance and, if you make the path more constrained (by using a core to "concentrate" the field), you will have more magnetic field for levitation than if you had a long solenoid without a core: -

enter image description here

Image from Hyperphysics with the air-core on the left.

Even so, it's likely that you should also extend the core from the bottom of the right-hand coil so that the south pole is vertically extended in parallel with the central core towards the north pole. This will reduce the air-gap even more and, although inductance may increase by (say) another 50%, the H field will still increase considerably at the newly created north and south (at the top).

After all, consider how electromagnets do this: -

enter image description here

Image from here.

The outer metal casing is ferrous material and is an extension of the south pole I mentioned above. The middle part of the casing is the north pole. See how much closer the north and south poles are now; this creates a significantly higher magnetic field strength (for a given excitation current) despite the moderate increase in inductance that the ferrous metal core brings.

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