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Thread: question regarding magnetics

  1. #1 question regarding magnetics 
    Forum Freshman
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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

    --------------

    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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  3. #2  
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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(9.8)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 (9.8)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/cou...econd/to/gauss


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  4. #3  
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    thanks!

    i have the magnets now, and i'm going to test your math.
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  5. #4  
    Forum Masters Degree bit4bit's Avatar
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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:

    -F<sub>g</sub>=F<sub>m</sub>

    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=kq<sub>1</sub>q<sub>2</sub> /4πr<sup>2</sup>

    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 4πr<sup>2</sup> 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=kq<sub>1</sub>q<sub>2</sub> /4πr<sup>2</sup>

    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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