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# Electromagnetic and Magnetic Levitation

Hi! I am setting up a levitation system and need to estimate the amount of voltage and current that is needed for a 2" dia. coil to produce 4800 Gauss. I don't have any restrictions/requirements on the core, number of turns, etc as long as it would produce 4800 Gauss with a 2" diameter. How can this be calculated? (I have spent a lot of time googling for such information, but I am pretty confused and didn't find how to calculate this) Thanks!

This is one of the most obscure parts of conventional (ie non-quantum, etc) physics. I had to go and look it up, and on finding it I realised why.

See the equation below, which I lifted from this paper.

To find the field strength in Tesla ( = 10,000 Gauss), multiply the number of turns in the coil, radius of the coil in metres, the current flowing and the permeability of vacuum (4pi * 10

^{-7}) together, then divide by ((the radius^{2}+ the axial distance from the coil^{2})^{1.5), then halve the result.}To find the number of turns and current, you will need to set B=0.48T (=4,800 Gauss), a=0.05 (2" = 5cm = 0.05m) and z=the required distance in metres and solve the equation. This could get a bit weird as you have a fractional power on the bottom of a fraction- get a large piece of paper and put your algebra hat on.

Alternatively, if you feel like doing groundwork to make your life easier later on, write a calculator in a spreadsheet that lets you plug numbers in and calculates the field, then play with the numbers until you reach something achievable.

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This is the equation.

u

_{0}is the permeability of vacuum, 4pi*10^{-7}or about 0.0000012566N = turns

I = current (Amps)

a = radius (metres)

z = axial distance from coil (metres)

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If I remember reading the document correctly, this formula is for measuring the electromagnetic force produced

insidethe coil, as opposed to the outside for magnetic levitation.Select as Best AnswerUndo Best Answer

The z

^{2}term in the denominator is the distanceoutsidethe coil, along the polar axis. Inside the volume of the coil, you should take z=0 everywhere.Select as Best AnswerUndo Best Answer

hm, I'm confused now, especially from all the different formulas out there, like http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c3

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That's for the field in a long cylindrical (helical) coil. The equation I posted with z=0 would work in that case, but the one I posted is more general and applies to points outside the coil along its axis. Points off-axis are apparently impossible to calculate simply and presumably need calculus or other mathematical "heavy machinery", but this gives you a good estimate.

I'm still curious how you arrived at the 4800 Gauss figure- I have no idea how to equate field strength with attractive/repulsive force in simple terms.

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Well I calculated, more like estimated, the 4800 Gauss from this website Then, I looked up the magnet's real magnetic force from the same site - and arrived at 4800 Gauss for a magnet to support a certain weight.

By the way, the u0 means the permeability of a type of core, right? So if I am to put an iron core, that would produce a greater magnetic field (if I am not mistaken?). I had a look on wikipedia's permeability on common materials chart : It seems that steel or ferrites would be a good option. However, I don't quite understand what its column labeled Magnetic Field means. It sounds like the material's permeability changes according to the amount of magnetic field applied to it?

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The preprint I posted (parallel to your post) does the 3D derivations, and it is quite ugly and calculus intensive. The axial component reduces to exactly what you posted, outside the coil.

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You might find this new preprint useful. It's got the derivations, and apparently even a spreadsheet implementation.

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Haha, I looked at that, and my first thought was "Why does it have sodium iodide in it?". My second thought was "That's not the right way to write the formula for sodium iodide!".

_{My third thought was "doh!"}Select as Best AnswerUndo Best Answer

Isn't it the permittivity of the

medium, i.e. not the same if you put a lump of iron in the middle?L

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True, but this is the theoetical field strength on axis for an ideal flat circular coil anyway, so I figured approximate/order of magnitude was good enough. I still haven't worked out why the OP needs 4800 Gauss specifically for maglev. J50- any more details on what you are planning to do? If it works are you going to obey the zeroth law and write an Instructable about it?

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Well, I am basically trying to "levitate" a heavy object about (or more than) 200 kg. The way I am doing it is similar to a new type of maglev trains, where you have some magnets/electromagnets

pushingthe object upward, and otherspullingthe object upward. So, imagine a hexagon with an opening on the bottom side. The object will be put inside the hexagon and hanged down from there, thus the opening at the bottom side. -- sort of like those old tick tock clocks. Now, the bottom slant sides willpushthe object upward. The top side and slant sides willpullthe object upward. So this is in fact a 100 kg job distributed among many magnets/electromagnets. Therefore, I am estimating 4800 Gauss.Select as Best AnswerUndo Best Answer

Well I'm interested, let's wait & see. L

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