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Author: Subject: Electromagnets!
elementcollector1
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Electromagnets!

I found on Wikipedia a way to calculate the magnetic strength of an electromagnet, given number of turns of the wire, length of the core, and length of the air gap (total?). However, it also lists "H", which I don't recognize.

NI=HcoreLcore+HgapLgap

Where:
N= number of turns
I = current
Lcore = length of core
Lgap = length of air gap (how do I calculate this, estimation?)
Hcore = height of core?
Hgap = height of air gap?

This says it's derived from Ampere's Law, but I took a look at that and got even more confused.

Now on to my practical problem:
Let's say I have an electromagnet (which I'd like to use for electromagnetic levitation, with sensors and stuff). The diameter of the whole thing (wire turns and all) is 3", with the core diameter being 1-2" and the core itself being a non-hollow steel pipe, approx. 8" long. The circuit is run on 4.5 volts, and may possibly include some sort of volts->current transformer (as magnetic field strength is proportional to current). Is there a way to calculate
a) the theoretical 'strength' of the magnet,
b) whether this could levitate an object, with the appropriate sensors and such,
c) If so, how to set it up?

On c): The general basis for electromagnetic levitation is controlling the strength of the electromagnet so that it counteracts the force of gravity exactly, usually with the use of an IR LED and a sensor (I don't understand this exactly either). This can suspend it in place, but usually cannot move the object around. But what if you want to hold an object out in front of you, and wave it around like an idiot? Would a sideways electromagnet be able to do this? Or would you have to put something above the object?

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Bezaleel
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Elementcollector, I guess I could help you, but which Wiki page do you refer to?

I guess that the H you mention is the magnetic induction. B_vector := H_vector + M_vector.
elementcollector1
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http://en.wikipedia.org/wiki/Electromagnet#Magnetic_field_cr...

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Bezaleel
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Please check the definitions, further down the wiki page.

In magnetism, terminology is sometimes confusing.
When we speak of a magnetic field, we usually refer to B. H is called the magnetic induction, or (for that reason) the magnetising field. M is called the magnetisation (not in the article on wiki). H is directly proportional to your current and number of windings, or rather to the current density, or yet more accurately to the integral over all space of the current density X direction to measurement point/(square of distance to measurement point), known as the Biot Savart-law

B is what you measure, and are practically interesrted in.

When you perform your experiments in a non-magnetisable environment and with non-magnetisable substances, M will be zero, and therefore B will be equal to H.

When you add a core material to a coil, you intend to magnetise the core material by the magnetic induction H, in order to increase B, because B = H + M.

In order to keep the number of windings as small as possible, you will want to choose a material that can be magnetised strongly, so a matarial with a high value for mu (coefficient of magnetisation).

elementcollector1
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So, I'm guessing iron is so often used as the core because either it's cheap or it has a high mu-value?
Either way, I'm having trouble calculating the variables - each one leads me to another theory with a different set!
I might just have to try this out in practice, as wading through the endless computations is getting hazardous to my mental sanity.
The thing I'm hoping to achieve is a stronger version of this:
There are two ways I could accomplish this: Either put an electromagnet/sensor combo on each of the three 'claws', or place a larger one down the center. I'm probably going to pirate the sensor and such from a magnetic globe or two (did I mention those things aren't cheap?), but want the whole thing to stand out as little as possible.
What do you guys think? If this worked, it would be true 'mad science': Carrying around a levitating magnet hidden inside something while laughing maniacally?

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franklyn
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A little less theory a little more practical

Google ( "Magnetic levitation" feedback )
www.levitationfun.com

www.coilgun.info/lev_popelex1966/home.htm
http://amasci.com/maglev/magschem.html
http://web.mit.edu/6.302/www/maglev.pdf
www.arttec.net/Levitation

www.me.hawaii.edu/HRIL/publications/planmaglev-msc08.pdf

The force produced by a magnetic field

Solenoids, Electromagnets & Electromagnetic Windings , pdf
https://ia600306.us.archive.org/35/items/solenoidselectr00undegoog/solenoidselectr00undegoog.pdf
djvu format
https://ia600306.us.archive.org/35/items/solenoidselectr00undegoog/solenoidselectr00undegoog.djvu

Design Construction & Operating Principles of Electromagnets for Attracting Copper, Aluminum & Other Non-Ferrous Metals
www.rexresearch.com/mrmagnet/Non-Ferrous-Magnet.pdf

http://archive.org/details/magnet_laboratory_1959

.
elementcollector1
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http://www.levitationfun.com/globe.html
At the bottom, they mentioned there was a Hall effect sensor - is that the circuitboard? If so, I wonder if I could make this MUCH smaller (maybe pack the components together to the right shape). The components, as mentioned, look easy to find - although buying three or more of each might take a bite out of my wallet.
EDIT: Never mind - the guy stated he did not get to the Hall-effect sensor.

[Edited on 13-5-2013 by elementcollector1]

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12AX7
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Umm... I could explain a lot in depth, but it's been my experience that I can write pages and not begin to convey the concept which is in my head. I know this well, and really it's simple at heart (or at least I would like to think so), but I'm just not good enough at English (or English is an inherently poor medium) to explain. So maybe we can work through it instead.

Diagrams help -- do you, per chance, have an example of what you're looking to do? Preferably a scale drawing or something.

The first step to guessing how much force an electromagnet will exert under certain conditions is to get the magnetic path lengths correct. Once you know where the flux lines are, you can measure their lengths and plug them into the formula.

Force is given by the Maxwell Stress:
|sigma| = B^2 / (2*mu_0)
With B in tesla (or if you start with H in A/m, use B = mu_0 * H), and mu_0 in H/m, you get a pressure in Pa (which is also an energy density: the energy stored in that part of the magnetic field; the integral over all space equals the total electrical energy supplied to the coil). If you know the flux density on one side of the armature (the thingy you're pulling on) and on the other side, and its area, you can roughly calculate the force exerted. (I say 'roughly' because, if you could integrate over the geometry, you'd be golden, but good luck figuring that out without using FEA. Guesses and hand-waving are good enough for at least an order of magnitude, and as long as everything is linear and proportional, you'll only need more or less of some quantity, like coil current, or separation distance, or armature cross section.)

Tim

Seven Transistor Labs LLC http://seventransistorlabs.com/
Electronic Design, from Concept to Layout.
Need engineering assistance? Drop me a message!
Twospoons
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 Quote: Originally posted by 12AX7 but good luck figuring that out without using FEA.

But if you do want to use FEA I heartily recommend FEMM4.2. Its a free 2-d magnetic/electrostatic/heat flow/current flow finite element simulator.

Let your PC do all the tedious calculations - that's what it's for !

[Edited on 14-5-2013 by Twospoons]

Helicopter: "helico" -> spiral, "pter" -> with wings
Bezaleel
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 Quote: Originally posted by elementcollector1 So, I'm guessing iron is so often used as the core because either it's cheap or it has a high mu-value?

Correct.
 Quote: Originally posted by elementcollector1 Either way, I'm having trouble calculating the variables - each one leads me to another theory with a different set! I might just have to try this out in practice, as wading through the endless computations is getting hazardous to my mental sanity. The thing I'm hoping to achieve is a stronger version of this: http://hackaday.com/2012/10/11/hackadays-portal-gun-actually...

From the page it isn't clear to me how the device actually works. Of course, you could use a top and bottom magnet to produce a field between them. But these cannot be permanent magnets, as the cube being levitated, would fall from the region between the magnets or it would end up clinging to either of the magnets.

Conclusion: they use AC in their magnets.

Question: what is it inside the object to be levitated, that is being attracted by the magnets?
 Quote: Originally posted by elementcollector1 There are two ways I could accomplish this: Either put an electromagnet/sensor combo on each of the three 'claws', or place a larger one down the center. I'm probably going to pirate the sensor and such from a magnetic globe or two (did I mention those things aren't cheap?), but want the whole thing to stand out as little as possible. What do you guys think? If this worked, it would be true 'mad science': Carrying around a levitating magnet hidden inside something while laughing maniacally?

Please sketch a setup of what you intend to do, and add some comments in the working principle. (From the article on hackaday I can't see the 3rd magnet is located.)

Edit: the link provided by Franklyn to rexresearch, might be much worth your while to read. It wouldn't amaze me if what is descibed therein is the principle used in the levitation gun on hackaday.

[Edited on 14-5-2013 by Bezaleel]
elementcollector1
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No, there is a magnet or two hidden inside the levitating cube.

Levitating globes actually use two magnets: One at the top of the globe, and one at the bottom.

As per AC, I think you're right - I remember seeing a plug-in cord somewhere on the 'gun'.

As for a sketch... I'll see if I can put that up once I get home. I've decided to go one sensor and two infrared LED's on the three 'prongs', with a large electromagnet in the barrel (although I only have a 3" by 3" cylinder to work with if I'm to use that same toy as in the video).

For my real one, the area of the electromagnet is going to be ~8" of a 3" inner-diameter cylinder. I'm still having trouble calculating any of this, so I guess I'll skip the math and try it out.

On the upside, the larger the magnet gets, the less I have to worry about short-circuits because the copper wire itself has some resistance (which adds up the longer the wire gets).

The only real problems left are finding Hall-effect sensors, and creating a strong enough electromagnet to hold stuff sideways with reliability.

I hope to possibly combine this with a railgun in the future (to simulate the Gravity Gun from Half-Life 2, because I am a total nerd).

EDIT: Although the nonferrous electromagnetization article interests me greatly, it's not quite what I'm looking for (and uses house current).

[Edited on 14-5-2013 by elementcollector1]

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elementcollector1
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Looking back at this, this is a rather silly plan for electromagnetic levitation and projectile launching.
I found a new equation, to calculate the force an electromagnet will exert at a given distance from a magnetic material:
F=((IN)^2 kA)/(2g^2)
Where:
I is current running through the solenoid. I did my calculations for 400 feet of 32 AWG wire, which has a resistance of 0.1641 ohms per feet (taken from Wikipedia). Given a voltage of 350 V, the current is about 5.33 amps.
N is number of turns. This was calculated by dividing 400 feet of wire by (pi*1/2") the circumference of a cross section core (plus a bit for the added layers). The answer is about 255 turns, rounded up.
k is 4pi x 10^(-7), a constant.
A is the cross-sectional area of the magnet (presumably the core). This is (pi*(1/4)^2) or about 0.3927 inches squared, or about 0.0002534 meters squared.
g is the distance between the electromagnet and the metal it is acting upon, in meters. This is about 4 inches, or .1016 meters.

Plugging everything in, a magnet running at 350V with 400 ft. of 32 AWG wire at about 4 inches away from a ferrous metal will exert a force of .03 N. I interpret this as it being able to levitate 2.9 grams of metal at this distance (dividing the force by gravity equals the mass). This seems a bit low.

I did a quick recalculation for 12 feet of 12 AWG wire at 1.5V (number of turns is 72, amps is about 78, area is same, distance is same), and found that the equivalent force for this magnet would be 0.5 N, so the maximum weight it could lift is 50 grams of ferrous metal at 4" away.

My question is, does everything here make sense? The equation, the variables and the answer? The variables are all rough guesstimates to get a sense of how strong this magnet will be.

[Edited on 12-14-2013 by elementcollector1]

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IrC
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elementcollector1 that hackaday link was one depressing read. Such a nice globe destroyed merely because the guy could not wind a simple coil. If you need a Hall device (or two or three), I have a few laying around. Used to get them from keyboards circa 1980's. Each key had a tiny magnet, a slot under the spring loaded key gave room for the key to push down passing the magnet by the Hall chip which stood vertically under the key. They built everything the hard way back in those days. Oddly enough the ZX80 Timex Sinclair was using advanced membrane keys like calculators did. Although I suppose they wanted it to be super mechanically strong/reliable so office workers could hammer the crap out of the keys all day. 4 terminal type IIRC I'll have to dig the bag of them up. Haven't looked at them in years. Only used a couple to build my magnet strength meter/plus N or S pole indicator.

Looking at the page 'Magnetic Levitating World Globe', if the guy had taken a couple color pics of side views (component side) so I could read resistor values that circuit would be very easy to draw from the board pics he posted.

"Science is the belief in the ignorance of the experts" Richard Feynman
elementcollector1
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I think the point of Hackaday is to take old, "boring" stuff and make new, cool stuff with it. I'm probably going to wire my own circuit using an Arduino.
I'm actually going to need about 6 ratiometric Hall Effect Sensors, but I found a cheap site for them (Tayda Electronics).

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IrC
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http://playground.arduino.cc/Code/HallEffect

You might look at this page. Tayda Electronics is a good supplier I buy from them often.

"Science is the belief in the ignorance of the experts" Richard Feynman
elementcollector1
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Look good to me. Is that force equation even accurate?

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MrHomeScientist
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I don't see how you'd be able to levitate an object to the side of your electromagnet. The force vectors are different - gravity is pulling downward and your magnet would be pulling perpendicular to that. So your object would instead accelerate diagonally, downward and towards the magnet. You can levitate things with the magnet above or below the object because the forces line up - the magnet either pushing or pulling upwards against gravity.

The only thing I've seen where such sideways levitation is possible is called 'magnetic locking,' demonatested by the Royal Institution in one of my favorite videos: http://www.youtube.com/watch?v=zPqEEZa2Gis
It uses superconductors cooled with liquid nitrogen. You can see it go sideways about 5 minutes in, and he has a great explanation for it after that.

[Edited on 12-18-2013 by MrHomeScientist]
elementcollector1
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Hmm. In that case, I guess I'll use one particularly strong electromagnet on top. I've heard transformer wire gets pretty good results.

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Master Triangle
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Sideways levitation confirmed:
http://youtu.be/5zYqLwRMyLw

It must use a electromagnet configuration that provides a force diagonally upward towards the platform alternated with a force away from the platform resulting in a net upwards force.
elementcollector1
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Hoo boy, looking back at this gives me a rush of nostalgia.
Anyway, I just finished winding an electromagnet today. I did not measure the length of the wire, but I would put a guess at somewhere around 500 ft. of 32 AWG magnet wire, which has a resistance of 82.05 ohms. With 9V (and therefore 0.11 A) of supply voltage/current, the magnet easily picked up and held an 8g nail indefinitely. The attraction was not very strong, as the nail could be 'flung off'.
What I am wondering right now is, what if I tested the 350V camera circuit I have on hand? This would provide 4.3 A of current, and according to the relationship H=I2R (which is apparently equivalent for the relationship to power?), and my extremely approximate number of turns (I hazard a guess at 1000), this gives a(n extremely inaccurate and imprecise) value of 18490 (or rounded to significant figures, 20000) Teslas, compared to the value for 9V of only 12.1 Teslas. However, at this level, the wire might melt or the plastic could burn, and I'm not sure I want to incorporate water cooling into my device. Thoughts?

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IrC
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 Quote: Originally posted by elementcollector1 Hoo boy, looking back at this gives me a rush of nostalgia. Anyway, I just finished winding an electromagnet today. I did not measure the length of the wire, but I would put a guess at somewhere around 500 ft. of 32 AWG magnet wire, which has a resistance of 82.05 ohms. With 9V (and therefore 0.11 A) of supply voltage/current, the magnet easily picked up and held an 8g nail indefinitely. The attraction was not very strong, as the nail could be 'flung off'. What I am wondering right now is, what if I tested the 350V camera circuit I have on hand? This would provide 4.3 A of current, and according to the relationship H=I2R (which is apparently equivalent for the relationship to power?), and my extremely approximate number of turns (I hazard a guess at 1000), this gives a(n extremely inaccurate and imprecise) value of 18490 (or rounded to significant figures, 20000) Teslas, compared to the value for 9V of only 12.1 Teslas. However, at this level, the wire might melt or the plastic could burn, and I'm not sure I want to incorporate water cooling into my device. Thoughts?

Being at work and too busy to go through a bunch of calculations can I at least say something seems terribly wrong about all your calculations. Also when doing the numbers consider the flux lines added by the core materials aligned domains. Also IIRC iron can add no more than 24,000 Gauss (2.4 Tesla) no matter what the power level is. My N48 NdFeB 3" D x 2" H are about 14K Gauss or 1.4 Tesla. So saying 20,000 Teslas seems incredibly off without taking the time to do the math since I have built some serious electromagnets in my time but none of them would compare to my 'super magnets'. 350 volts at 4.3 amperes is 1,505 watts, rapidly vaporizing 32 gauge enameled wire, not to mention somehow I seriously doubt your camera supply can do 1.5 kilowatt. But that's just me, I think your number crunching has serious flaws in both Ohms law and magnetic field calculations. I guess someone can shoot me down here I don't even have a calculator laying around right now but just on the fly thinking it appears your numbers on several things are way off.

"compared to the value for 9V of only 12.1 Teslas"

One Tesla is 10,000 Gauss, one Gauss is one line per CM^2. You are saying a 9 volt battery through a little coil of fine wire is producing 121,000 Gauss? Impossible.

"Science is the belief in the ignorance of the experts" Richard Feynman
elementcollector1
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Yeah, re-took a look at that. Didn't quite understand the applications of Ohm's Law - thought that if there was a certain voltage and resistance, the current would be as such no matter what.
New plan: Get one of these http://www.amazon.com/gp/product/B00A82LBQM/ref=gno_cart_tit...
and one of these http://www.amazon.com/gp/product/B00E0NTPP4/ref=gno_cart_tit...
and couple it with my Arduino Uno. Should get significantly better results than a 9V, right? But with 12V and 4.5 amps, the same question comes into play: Would this damage the electromagnet windings?

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 Quote: Originally posted by elementcollector1 Yeah, re-took a look at that. Didn't quite understand the applications of Ohm's Law - thought that if there was a certain voltage and resistance, the current would be as such no matter what. New plan: Get one of these http://www.amazon.com/gp/product/B00A82LBQM/ref=gno_cart_tit... and one of these http://www.amazon.com/gp/product/B00E0NTPP4/ref=gno_cart_tit... and couple it with my Arduino Uno. Should get significantly better results than a 9V, right? But with 12V and 4.5 amps, the same question comes into play: Would this damage the electromagnet windings?

http://www.powerstream.com/Wire_Size.htm

Start with the specs for the wire. Yes I=E/R but that holds until the circuit vaporizes, at which point R becomes very high (think air gap). So you do not want to exceed the maximum ratings for the wire, although for a rapid pulse you could go much higher so long as the time was short enough the heat did not build up to the melting point of Cu. Below that one still must consider the melting temperature of the enamel. Double coated Formvar was an attempt to run much hotter before turns began shorting but one can only carry this so far. Pulsing at high power one must also consider the force if great can rip the turns apart.

Also consider while I=E/R, also P=EI. Merely because your camera supply does 350 volts does not mean it can do so at a low impedance (high amperes = high power). The current will fall off when the power supply's ability is exceeded. 1 amp 1 volt 1 watt 1 ohm, but what if your supply cannot deliver one amp at one volt into one ohm? I know small numbers I was just making it simple to think about. In effect even though ohms law might give a certain number, this does not mean your power supply can do it. I doubt your camera supply can supply 4.3 amperes at 350 volts. If it could think deadly dangerous supply so work carefully.

"Science is the belief in the ignorance of the experts" Richard Feynman
elementcollector1
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Have since switched to a MOSFET and mini lead-acid battery, which claims 6V and 4.5A. Should be better in terms of current/not frying the arduino...
How do I find the supply current/voltage for when my wire vaporizes? Your scale claims it should only be able to handle up to half an amp, but a different scale claims it's fine up until about 9A (http://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_A...), where it fuses in about 10 seconds.

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 Quote: Originally posted by elementcollector1 Have since switched to a MOSFET and mini lead-acid battery, which claims 6V and 4.5A. Should be better in terms of current/not frying the arduino... How do I find the supply current/voltage for when my wire vaporizes? Your scale claims it should only be able to handle up to half an amp, but a different scale claims it's fine up until about 9A (http://en.wikipedia.org/wiki/American_wire_gauge#Tables_of_A...), where it fuses in about 10 seconds.

Actually it's not my scale, just a standard wire table. I don't think you are understanding clearly what you are looking at. A typical wire table gives ratings one could say is akin to a 'steady state' condition. By this I mean the value listed is what the circuit can do being powered continuously. I have never really looked for a table giving 'exploding wire' conditions but I imagine they exist out there since the technology is common in metal forming. While I have played with the subject it was before the internet, in a day you could go check out books (if you could find one on a subject such as this in print in the 60's to 80's) but buying them was expensive. Typically I found these values by trial and error. Interesting enough I think I'll search around for awhile out of curiosity. I love that about the internet, in seconds so much information is available for free doing little work. In any case the link I gave would be more for building an electromagnet you planned on powering for long periods of time.

http://users.tm.net/lapointe/Wire_Explosions.html