Sciencemadness Discussion Board

Electromagnets!

elementcollector1 - 12-5-2013 at 22:02

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?

I know I have a lot of questions, but I would like to learn more about this - electromagnets are a fun project.

Bezaleel - 13-5-2013 at 01:53

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 - 13-5-2013 at 05:23

http://en.wikipedia.org/wiki/Electromagnet#Magnetic_field_cr...

Bezaleel - 13-5-2013 at 06:22

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 - 13-5-2013 at 08:50

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. :D
The thing I'm hoping to achieve is a stronger version of this:
http://hackaday.com/2012/10/11/hackadays-portal-gun-actually...
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?

A little less theory a little more practical

franklyn - 13-5-2013 at 10:04

Google ( "Magnetic levitation" feedback )
www.google.com/search?q=magnetic+levitation+feedback&tbm...
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
www.crealev.com - www.youtube.com/watch?v=OEExd0wlsPo

flying_discs_0.jpg - 213kB flying_discs_1.jpg - 228kB

The force produced by a magnetic field
http://info.ee.surrey.ac.uk/Workshop/advice/coils/force.html

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 - 13-5-2013 at 10:38

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]

12AX7 - 13-5-2013 at 15:33

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

Twospoons - 13-5-2013 at 20:01

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 !

FEMM4.2 Download page

[Edited on 14-5-2013 by Twospoons]

Bezaleel - 14-5-2013 at 01:54

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. :D
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 - 14-5-2013 at 11:52

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]

elementcollector1 - 13-12-2013 at 20:03

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]

IrC - 14-12-2013 at 00:45

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.

elementcollector1 - 14-12-2013 at 10:36

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).

IrC - 14-12-2013 at 13:34

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

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

http://www.youtube.com/watch?v=EHM3ahJ8FYQ&feature=playe...

Another link.

elementcollector1 - 14-12-2013 at 18:56

Look good to me. Is that force equation even accurate?

MrHomeScientist - 18-12-2013 at 09:38

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 - 18-12-2013 at 09:55

Hmm. In that case, I guess I'll use one particularly strong electromagnet on top. I've heard transformer wire gets pretty good results.

Master Triangle - 25-12-2013 at 00:20

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 - 11-1-2014 at 13:27

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?

IrC - 11-1-2014 at 15:08

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.



elementcollector1 - 11-1-2014 at 16:09

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?

IrC - 11-1-2014 at 16:57

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.

elementcollector1 - 11-1-2014 at 18:22

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.

IrC - 11-1-2014 at 20:30

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

This link is interesting.


[Edited on 1-12-2014 by IrC]

elementcollector1 - 18-1-2014 at 23:06

I know it's not actually your table, that was just for reference.
Anyway, the lead-acid battery, disappointingly enough, does not actually output 4 amps as I thought it claimed to. Instead, it does output 4 Ah for 20h, which I think means that the battery would run for 20 hours at a drain of 4 amps before being fully discharged.
As it stands, I did use the amperage and voltage measured off the battery (0.056 A and 6.52 V respectively), as well as a resistance value for 1 ft. of 32 AWG wire from Wikipedia of .1641 ohms to calculate the length of wire currently in the electromagnet, which is about 710 ft. This will come in handy for some of my later calculations, I suppose.

So, back to my original plan, with a few change-ups. What if I planned to use a boost converter to give a voltage output of roughly 350V, making the current (by Ohm's Law) 3 A?
This would make the battery life 30 hours, assuming that kind of basic math even works here.

IrC - 19-1-2014 at 00:47

Quote: Originally posted by elementcollector1  
I know it's not actually your table, that was just for reference.
Anyway, the lead-acid battery, disappointingly enough, does not actually output 4 amps as I thought it claimed to. Instead, it does output 4 Ah for 20h, which I think means that the battery would run for 20 hours at a drain of 4 amps before being fully discharged.
As it stands, I did use the amperage and voltage measured off the battery (0.056 A and 6.52 V respectively), as well as a resistance value for 1 ft. of 32 AWG wire from Wikipedia of .1641 ohms to calculate the length of wire currently in the electromagnet, which is about 710 ft. This will come in handy for some of my later calculations, I suppose.

So, back to my original plan, with a few change-ups. What if I planned to use a boost converter to give a voltage output of roughly 350V, making the current (by Ohm's Law) 3 A?
This would make the battery life 30 hours, assuming that kind of basic math even works here.


350 volts times 3 amperes is 1,050 watts per second into 116.66 ohms. 32 gauge wire will vaporize. Say your battery is rated for 20 amp/hr at 6 volts. First off the internal impedance would make it hard to get 20 amps from it. So a more realistic way to look at it's maximum power delivery is say 10 amperes for 2 hours. 10 amperes at 6 volts is 60 watts per second. In other words in your thoughts when designing your circuits not only should you be giving consideration to the internal impedance of your power source (generator), you need to be considering power P = EI. While never forgetting your 32 gauge wire is going to vaporize anyway. Thinking in terms of power, what would it take at 6 volts to get 1,050 watts per second. 1050/6 = 175 amperes. Of course this assumes your inverter is perfect. Say it is 70 percent efficient. 1/0.7 = 1.42857. 175 amperes times this = 250 amperes at 6 volts in your case here. So now we have 25 6 volt batteries in parallel. 1,500 watts per second for 2 hours. Next consider the battery voltage is dropping as it runs down meaning your not going to get a steady 1,500 watts per second for 2 hours. Also consider things melting and smoke filling the room means a fire extinguisher and O2 mask is not a bad idea either. You are not looking at the problem correctly is what I am trying to say. You should be considering things also in terms of power, what materials can handle, and safety. Being nearly the Hour of the Wolf I may have made an error somewhere in my thoughts in this post but I don't think so. What you are considering doing is not reasonable.

I am assuming since you omitted units you meant 0.1641 ohms per foot for 710 feet giving 116.51 ohms. I=E/R, at 350 volts I = 3 amperes at 350 volts. 1,051.4 watts per second. So my original thoughts are close. Think about how much heat a steady kilowatt represents. I imagine you may need more than one fire extinguisher but that's just me. Having been hurt badly many times in 60 years of mad science I try to look at things from a safety viewpoint. OK your right not 60, but I swear at 5 years of age my dad caught me sitting on the sidewalk in front of the house with a hammer and a box of .22 long rifle shells. I beat all but 2 or 3 flat laughing with glee at the pop and fizzle. Yes that is what they did, nothing went flying anywhere. More than a few I hit just right and they were loud enough it drew him out from his morning coffee. I consider that the beginning of my foray into mad science so being more precise in 55 years of mad science I have learned through pain and injury to look at problems with a more analytical mind never forgetting safety. What you are trying to do is not safe and your design thinking is not correct. But you have to start somewhere, hopefully not the way I did. My first rocket set an entire field ablaze bringing out two fire trucks. I'm hoping your electromagnet does not do the same thing to you but it sure sounds like this will be the end result.

elementcollector1 - 19-1-2014 at 10:35

I see. In that case, what about running at just 1 amp? This is 116 V, and 116 W/s as well. Is this a more reasonable goal? This is about the input for the average lightbulb, so the magnet should heat up considerably (it already warms to the touch at the previous 0.056 A). If no, I could work with half an amp as well. The point is something signficantly stronger than what I'm working with now - 1 amp represents a multiplication in strength of x18, 1/2 amp represents a multiplication of x9, etc.

The main purpose of all this is to electromagnetically levitate a neodymium magnet from 6" away, which (by my thinking) can only be done by a particularly strong electromagnet and a particularly strong Nd magnet.

IrC - 19-1-2014 at 13:18

More reasonable yes although I'm confused as to how the 350 volt power supply lowered to 116 volts, unless it is adjustable. Having done no calculations I cannot say if it will or will not work. However I don't think your electromagnet will be strong enough. Go another way. Weigh your magnet, measure the field strength. Calculate what field the electromagnet will require to do what you want. I see another flaw in your approach. One dipole field is not going to levitate another easily since it will tend to shift off to the side. Try it with two magnets to simulate your experiment you will see what I mean. I have a sheet of graphene which requires 4 N50 magnets to levitate and sit stationary in zero wind. Better yet study the Levitron, it must have perfect weight, balance, rate of spin, and proper height above the magnetic base. Those floating globes use a dynamically adjusted field with feedback from a Hall signal. They are also far lighter than a large super-magnet. I have never tried your experiment so I am not the best one to help you in this but just from experience I think what you wish is going to be much harder and require much stronger fields than you realize. I also think it makes sense to have a soft Iron core for your electromagnet which diverges, along with far more turns than you are talking about. In any case I would begin by calculating the field strength your electromagnet will need to work, then design the electromagnet based upon that.

Going back to my concern about levitating with two magnets, those globes would also shift off to the side if the field were steady. In reality it lifts, shuts down field, falls, re-powers the field, lifts, and so on very rapidly many times each second. So that it appears to float stationary. The reason a superconductor levitates is the opposing field is dynamic, every microscopic movement alters the opposing field to yet again oppose. Two simple magnets in your case, the light one will never stop trying to flip over sticking to the other one. As the Levitron inventor discovered it must be rotating rapidly in a field very carefully shaped. I would search and study magnetic levitation before I went further.

elementcollector1 - 19-1-2014 at 17:26

The boost converter would have supplied 116 volts or possibly more if I built it in the way mentioned here (http://www.instructables.com/id/DC-Boost-Converter/). Multiplying that voltage would have been subject to being figured out later, but if 110 volts is fine then that doesn't need to even happen.

Similarly to how you mentioned the globe circuit works, I will be using the same type of stabilizing feedback using an Arduino Uno and two Hall Effect Sensors, both of which are ratiometric. These will likely be placed some distance from the coil itself (about 1 inch) to avoid maxing out the readings, and make for better sensory capability. This leaves a space of about 5 inches between the magnet and the HES.

Flipping over might be a problem - but for all the research I've done (this video especially: http://www.youtube.com/watch?v=LaGv2FHS5zg), disc magnets appear to be resistant, if not immune to such an occurrence. Noticeably, the magnet appears to spin very slightly, but spinning rapidly causes it to destabilize. I've been researching this for quite a while, and disc magnets are always the magnet of choice.

IrC - 19-1-2014 at 19:33

The levitation idea will work but your choice of power supply design I fear will be severely lacking in power output on the level you need for your project. Maybe I missed it but how heavy is the magnet you wish to levitate and what is its strength. Sounds like a fun project. You should try experimenting with building the electromagnet and see how much it can lift. I had a thought, study some typical inverter circuits using higher power levels than the boost converter you mentioned with an actual step up output transformer with full wave rectification plus filtering for a steady source to power the electromagnet. Using a TL494 you can modulate the output with a voltage from your Arduino. Look at this example:

arc-speaker8.jpg - 51kB

Image from: http://hackerfriendly.com/wp-content/uploads/2008/12/arc-spe...

Change the flyback to a transformer that will work for your electromagnet (rectify/filter output). Omit the capacitor to pin 4 and feed a DC control voltage which you can find by experimenting if your not far along enough in electronics to calculate it. Just be sure the control input does not go below zero or above the DC supply used for the IC. Also do not go too high quickly until you know what current the power mosfet will see or you may blow it. Study how hot it gets not forgetting a proper heat sink. At least this circuit approach has a better chance working than merely ringing a low value inductor as the link you posted does. Whether it will provide enough power I do not know as you really have not given specific enough information. I do know this circuit can be scaled up to be the heart of a 1,500 Watt inverter for what it's worth assuming you had the DC source to run it. Obviously you may end up building your own output transformer but this is far easier than most think it is. Many is the time I used an oven (not too hot) to aid in pulling all the E's from an old power transformer. A bobbin you can buy or build or if careful salvage from the source (note they melt easy). My very first HeNe laser was powered with one I built from the bad transformer from a CB base station (was around 4A/18Vac originally). Of course if going for a high frequency from the TL494 think big ferrite transformers. Best source here is from you guessed it, a dead old 1,500 Watt inverter. You can rewind the transformer secondary for a higher voltage but if your going to work around 110 volts it seems the transformer from the inverter would already be perfect.

I have to wonder though why you are focused on high voltage high current circuits when something better exists. First off forget 32 gauge, go lower voltage higher currents (heavier wire less turns). Have you looked at this video?

http://www.youtube.com/watch?v=nZL6mLISxeo

Seems like it works very well at far lower voltages meaning safer and less problematic in other areas such as heat.

Or: http://www.youtube.com/watch?v=zCsg0phiY0c



[Edited on 1-20-2014 by IrC]

elementcollector1 - 19-1-2014 at 23:29

The magnet I wish to levitate can be of any size, so long as it can be levitated from over 6" away... In that case, I'm not sure whether lighter or heavier is good.

Problem with lower gauge wire is way too high current - for example, 10 ft. of 12 AWG wire and 1.5V leads to 94 amps, something I definitely don't want to mess with.

A major concern is size restraints - given that the end objective is a sort of literal realization of a video game prop, consider the model below:



See that claw-shaped thing on the upper left? It's about 6 inches long, and that is where the magnet has to fit (without being too visibly intrusive - I'm fine with a bit of iron core sticking out the end, but not a 2" fat solenoid...). Add that with the fact that the object should be levitated about 6" below (the center of the barrel), and this complicates things severely. Of course, I'm not going to give up - that would be no fun.

In the center, where the barrel is, I'm planning to have another electromagnet - this one attracts a spring and an aluminum plate. When the button is pressed, the magnet is turned off, the spring flies forward, and assuming there's an object floating there, Newton's 3rd law takes care of the rest. For this magnet, I'm probably going to use thick wire and low voltage.

Now, what I could do is wind about 1000 ft. of 32 AWG wire onto a very small nail. Most of this nail would be hidden from sight under that claw thing, but a bit would be sticking out (the nail would be vertical, for the most part, in compliance with what you said about the field earlier). Then, I would build a boost converter (shown here: http://www.instructables.com/id/DC-Boost-Converter/) to bring the voltage up to 110 V. Assuming the power supply is decent (i.e. lasts a hell of a lot longer than a 1.5V battery would), the amperage would now be 0.67 A and the power would be 73.7 watts. Would this be a more reasonable power?

(And yes, I am such a nerd.)

[Edited on 1-20-2014 by elementcollector1]

IrC - 19-1-2014 at 23:46

It would take many calculations to answer. More than I have time to do especially by guessing about details and parameters from a distance. I think you need to be building at least the coils and circuitry and experimenting.

elementcollector1 - 24-1-2014 at 22:40

Hmm. I have 50 feet of that same 32 AWG wire on my coil, and I was wondering. If I cut this off now, the current would be 0.73 amps, and the power would be 4.4 W. Is this any better than scaling up to 1000 feet, using a boost converter to increase the voltage to 110 V, only to get the same current (with much more turns and a much fatter solenoid)?

I guess what I'm asking is, which is more important? Current or number of turns? I know the magnetic field is proportional to both, but I'd like to ask those who have experimented.

IrC - 25-1-2014 at 15:48

If I had to make a guess without doing any research I would say go with more turns higher voltage. The simple reason being at fewer turns the required current is much greater and it is the current and resistance giving rise to heating and materials problems. For the very same reason large energy levels are transmitted long distances by upping the voltage/lowering the current and therefore reducing both heating and loss effects. Of course depending upon the field strength you require as well as considering various limitations involved. Possibly you could experiment with both approaches on a small scale just to get an idea which route would be better. Make a couple test coils and measure field strength, energy input, heating and so on. Unless of course you do not have access to the wire to wind from several hundred to a few thousand turns.

elementcollector1 - 26-1-2014 at 23:21

Tested boost converter circuit today. Appears to work fine, with the drawback that the MOSFET heats up severely - 5-10 seconds, and it burns to the touch. Should I use aluminum for a large heat sink for this? (large being a 3"-by-1"-by-1" rectangular prism.) I am really worried about the MOSFET burning out from such heat, as it would be hard to replace - both in terms of the circuit and the cost.

The electromagnet seems to be quite a bit stronger. Not sure if this is quite enough to levitate objects, and I haven't tested the voltage or current or anything. A cursory test with picking up LEDs (comparing to counts from earlier variations of the magnet) showed that it's at least as good as before, if not better.

IrC - 27-1-2014 at 00:59

Adequate size heat sink is mandatory. If you paint it flat black (not glossy) it will radiate around 25 percent more heat which helps quite a bit. Anodized black is better but if it is not handy the flat black trick works fine. Along with a thin coat of thermal compound of course. Rule of thumb is if sink is adequate you can hold your palm against it for 30 seconds without discomfort although I do not recommend this test if powered by dangerous voltages. Really I am just giving you an idea how hot is too hot meaning bigger sink and/or reduced duty cycle to the waveform, also reduced voltages helps. Just mentioning the common sense ideas.

elementcollector1 - 27-1-2014 at 08:41

Voltage confirmed at 150 volts, but it doesn't last long - it quickly drops back down to about 6V or so. Why is this? Is it the MOSFET overheating?

IrC - 27-1-2014 at 09:12

Insufficient data for a meaningful reply. Sounds like a reasonable assumption.

I should probably add you know there is no way someone from a distance can answer anything without an exact circuit with parts list, including what it's powered by and loaded with. All measurements and so on. Measuring field could be a simple compass mounted a distance away comparing deflection. Assuming you do not lay a pile of bolts between the coil and compass or make other alterations.


[Edited on 1-27-2014 by IrC]

elementcollector1 - 7-2-2014 at 15:24

(Questions and concerns are bolded so they're easier to find.)

Have since wired a much better coil:



This one is as strong as it's predecessors, with much less size and length of wire - and that's *without* the boost converter. *With* the boost converter, this could be the lucky combination of amps and wire turns I've been looking for. Of course, the heat produced would be severe unless I used some sort of heat sink - 2 A of current running through about 460 feet of magnet wire cannot be good for it.

This comes into my next question/problem: I talked to a physics professor, and he said the system with the large electromagnet and the spring/aluminum rod would *not* work, because the spring (being welded to the core) would now have a pole of the magnet at its tip, and that pole would want to distance itself from the other pole at the end. He suggested instead to use a railgun as the launcher of choice.

So, I would have to construct a railgun with an H-Bridge circuit. The aluminum rod would now be attached to the 'roller' (an iron nail and two disc-shaped Nd magnets as wheels), and allowed to rotate (this way it would roll forward while still keeping the aluminum rod pointed forward). The railgun would naturally require a capacitor bank, and a high source of current.

I am hoping that the use of the railgun, and the subsequent charging of the capacitors will use enough energy to lower the current through the first set of electromagnets (as there are now 3, explained below) to a more heat-tolerable level. The railgun and electromagnets would all run off the same power source, and the railgun would possibly run off the same boost converter as the electromagnets, depending on what power consumption is needed.

The reason there are now 3 electromagnets out in the front is because I realized that vertical levitation will not be enough to keep this in the air, as the slightest oscillation or gust of wind would cause it to leave the affecting field. To fix this, I would put 2 more electromagnets on the two other 'claws' (see the image a few posts up). These would face the center of the barrel, same as the first electromagnet (keep in mind that these are exactly 120 degrees apart), and have the same configurations (Hall Effect sensors, logic-level MOSFET controls and all), and the same basic purpose: If the electromagnet moves too far away from either of them, they will pull it back towards the center. This ought to lessen the effects of horizontal interference, right? The magnet can still escape, but it is harder to do so. Hopefully, I'll be able to pick this thing up without the magnet escaping - that is the goal, after all.