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| platinum |
 Posted: Wed Apr 23, 2008 1:53 pm Post subject: question regarding magnetics |
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Forum Freshman
Joined: 23 Apr 2008
Posts: 8
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hello! first post here, and i'm happy to be part of this forum!
i'm working on a project, and i need to figure out something regarding magnets.
based on a picture i've seen. if two ring magnets each with these specifications:
Dimensions: 2" outer diameter x 1/4" inner diamiter x 3/8" thick
Tolerances: ±0.003" x ±0.002" x ±0.002"
Material: NdFeB, Grade N42
Plating/Coating: Ni-Cu-Ni (Nickel)
Magnetization Direction: Axial (Poles on Flat Ends)
Weight: 5.03 oz. (143 g)
Pull Force: 191.40 lbs
Surface Field: 5150 Gauss
Brmax: 13,200 Gauss
BHmax: 42 MGOe
are put on the same verticle plane with idenical poles facing each other, will repel almost 5 inches under normal gravity. see what i mean in this pic:
http://www.kjmagnetics.com/prodimages/RY046L.jpg
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what i want to know is if there is a formula i can use to calculate this distance accurately (or at least semi accurately).
thanks again. hope there's some smart people out there! |
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| Cold Fusion |
| Posted: Wed Apr 23, 2008 4:45 pm Post subject: |
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Forum Ph.D.
Joined: 10 Apr 2007
Posts: 793
Location: In the circuitous haze of my mind
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I think I calculated it right.
F=MA
F=(Kq1q2)/d^2
Where, F is force in newtons, M is mass, A is acceleration, K is the electric constant at 9(10^9) (NM^2)/c^2), d is distance, q is charge is coulombs for the first and second body.
143g=.143kg(9. m/s gravity= 1.4014N
[1.4014=9(10^9) (.000001717855)^2 coulombs]/d^2
1.4014=.026559/d^2
1.4014d^2=.026559
d^2=.0189519
d=.137665m
.137665/ (.0254 conversion constant to inches) =5.419 inches 
Wow science forum, its taking (9.8 )m/s as (9.m/s. Come on, we use the gravitational constant way too much for stuff like that to happen.
I know its a little off....maybe someone could find a mistake in my calculations?
I decided to calculate the force of the first magnet, and the normal force exerted by the surface on the bottom one; I then averaged them together and got around 4.7 inches.
Here is the coulomb to gauss converter I used:
http://www.convertunits.com/from/coulomb/second/to/gauss
_________________
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"Great spirits have always found violent opposition from mediocrities. The latter cannot understand it when a man does not thoughtlessly submit to hereditary prejudices but honestly and courageously uses his intelligence"
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Use your computing strength for science! |
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| platinum |
| Posted: Fri Apr 25, 2008 12:43 pm Post subject: |
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Forum Freshman
Joined: 23 Apr 2008
Posts: 8
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thanks!
i have the magnets now, and i'm going to test your math. |
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| bit4bit |
| Posted: Sat Apr 26, 2008 12:45 pm Post subject: |
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Forum Ph.D.
Joined: 14 Jul 2007
Posts: 625
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I think Coldfusion was right with the first bit....If the magnet is in static equillibrium, the nett force acting on it will be 0. The forces in this case are gravity, and the magnetic repulsion.
So you have:
-Fg=Fm
Force of gravity =ma=9.81m, m being the mass of the magnet.
The magnetic force can be trickier to calculate. The equation is actually:
F=kq1q2 /4πr2
where k is the permeability of the medium (Not the electric constant as with Coulombs law), and q represents the pole strengths of the magnets (Not charge as with Coulombs law). r is the separation between them, and the 4πr2 represents the inverse square law...same as with Coulombs law.
I think it can get much more complicated than this simple formula though, due to the angles of magnetic poles, the distribution of magetization over the surfaces of the objects, and the shape of the surfaces. I don't know what to use in that case, but if you take the formula above you have:
-9.81m=kq1q2 /4πr2
You have all the variables apart from the distance, r, which is what you want to solve for. I'm not familiar with some of the units you've given, but if you use this formula, make sure you're consistent with the units. I always work in SI units, so I would convert to those first.
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Chance favours the prepared mind. |
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