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!

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PKM8 years ago
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 radius2 + the axial distance from the coil2) 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.
PKM PKM8 years ago
This is the equation.

u0 is the permeability of vacuum, 4pi*10-7 or about 0.0000012566
N = turns
I = current (Amps)
a = radius (metres)
z = axial distance from coil (metres)
J50Nunlimited (author)  PKM8 years ago
If I remember reading the document correctly, this formula is for measuring the electromagnetic force produced inside the coil, as opposed to the outside for magnetic levitation.
The z2 term in the denominator is the distance outside the coil, along the polar axis. Inside the volume of the coil, you should take z=0 everywhere.
J50Nunlimited (author)  kelseymh8 years ago
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
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.
J50Nunlimited (author)  PKM8 years ago
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?
kelseymh PKM8 years ago
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.
You might find this new preprint useful. It's got the derivations, and apparently even a spreadsheet implementation.
Kiteman PKM8 years ago
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!"
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