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Calculating lifting force of a magnet

  1. Apr 14, 2015 #1
    If the magnetic field strength is measured to be X Tesla on a large flat permanent magnet is there a standard formula for determining the force the magnet will supply as a function of distance from a piece of metal with well defined properties.

    I see the problem of different metals are going to respond differently to same magnet and they will themselves become magnetised etc but for simplicity use some uniform, constant grade of common metal.


    thanks any ideas.
     
  2. jcsd
  3. Apr 14, 2015 #2

    CWatters

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    Looks like you may have to experiment...

    https://www.kjmagnetics.com/calculator.asp

     
  4. Apr 14, 2015 #3
    thanks will check out the link.

    I have looked through a few engineering handbooks for a formula or general rule of thumb to no avail. I thought it would have been a common and rather mundane application problem with a simple approximate algebraic formula to plug in numbers like what type / grade metal, distance and magnetic field strength, evidently not.

    I would settle for a formula to calculate just the maximum lifting force.
     
  5. Apr 14, 2015 #4
    hey that link is cool, seems hyperbolic (or something) with distance as I would roughly expect.

    love to know how they calculated it, does not appear to be any links to the source.
     
  6. Apr 14, 2015 #5

    Hesch

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    If you separate the magnet from the metal, you create an airgap in the magnetic circuit. In this airgap the density of the magnetic energy will be:

    Edens = ½*B*H [ J/m3].

    So the total energy in the airgap will be: E = ½*B*H*A*ds, A=surface area, ds=width of airgap.

    The force of attraction will be: F = dE/ds.

    But the real problem is to calculate B(s) and H(s), when the width (s) of the airgap becomes "some distance". A numerical calculation is suggested.
     
  7. Apr 14, 2015 #6
    A is the surface area of the magnet or steel?

    B is not the magnetic field strength of the magnet away from any magnetic material? if no what is it?

    and H is ???
     
  8. Apr 14, 2015 #7

    Hesch

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    The B-field is the magnetic induction ( unit: Tesla ). It may be compared to electrical current.
    The H-field is the magnetic field strength ( unit: A/m ). It may be compared to electrical voltage.

    Ohms law: I = U / R
    "Magnetic" law: B = H * μ ( μ is the permeability (magnetic conductivity) of some material ).
     
  9. Apr 14, 2015 #8
    so which one, B or H is the actual measured field strength in Tesla I could measure with a Gauss meter placed near the magnet?
     
  10. Apr 14, 2015 #9

    Hesch

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    With a Gauss meter, you are measuring the B-field. (As I remember: 1 Tesla = 10000 gauss ).

    μ0 ( permeability in vacuum ) = 4π*10-7.

    μr (relative permeability) in iron ≈ 1000.

    μ = μ0r.
     
  11. Apr 14, 2015 #10
    so if I measure B I can easy calculate H ie B= uH from above post and given Edens = ½*B*H [ J/m3] and E = ½*B*H*A*ds and F = dE/ds = d(B^2uA)ds/ds therefore
    F = B^2uA, seems to simple, what have I messed up?
     
  12. Apr 14, 2015 #11

    Hesch

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    No, it's not that easy. Say you have a cylindric magnet. Between the magnet and the metal, you will have an (small) airgap, and you can approximately calculate the volume of this airgap to be: V = A*ds. But on the other side of the magnet you have another airgap, and what is the dimensions of this airgap?

    Remember that magnetic fields circulates. They must create curves that return into themselves. What is the volume/length of such a field? It's correct, that the magnetic flux is constant in a magnetic circuit, but the H-field is not. ( The current is constant in an electric loop, but the voltage drop in a resistor in the loop depends on if the resistor is 10Ω or 1kΩ ).

    That's why I suggest a numerical calculation (computer calculation).
     
  13. Apr 14, 2015 #12
    "But on the other side of the magnet you have another airgap"

    I wonder how much damage if I assumed there is no other side of the air gap or it just had a value of zero or 1, just kidding.

    thanks to all those that replied, I got this now.

    appreciate it.
     
  14. Apr 14, 2015 #13

    Hesch

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    You may substitute your permanent magnet with an electromagnet (iron cylinder with some coil added).

    Computer calculation may be done, using Biot-Savart law with respect to this coil. (Complicated, not kidding).
     
  15. Jun 4, 2015 #14
    Simple Formula Derived by my lecturer.

    F= B^2*A / 2μ0

    B - Flux Density (Squared)
    A - Area of the parts in contact (or in close proximity)
    μ0 - Constant 4*pie 10^-7
     
  16. Jun 4, 2015 #15

    Hesch

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    Yes. More generally:

    The energy density of a magnetic field: Emagn = ½*B*H [ J/m3 ] →

    F = dE / ds, ( s is the distance, E = magnetic energy. )
     
    Last edited: Jun 4, 2015
  17. Jun 5, 2015 #16
    thanks that is a simple enough formula.

    in the second formual above F = dE/ds, there is no distance variable, how do you differentiate it WRT s? or are you saying you have to first find an expression for B or H as a function of s?
     
  18. Jun 5, 2015 #17

    Hesch

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    Yes, B and H becomes a function of s when s is "some distance". So: Emagn = ½*B(s)*H(s).

    When you separate magnet and iron, you change the magnetic circuit. In electrical terms you cut some wire in the circuit and insert a resistor as connection between the two pieces of wire. If current flows in the wire, you will have a voltage drop across the resistor ( you will have a strong H-field in the magnetic circuit ). If the electrical circuit is supplied by a fixed voltage, the voltage drop across the resistor will tend to decrease the current ( tend to decrease the B-field ).
    So cutting the wire and inserting a resistor (say 1kΩ) will not change the current very much if there is another resistor in the circuit in series
    ( 100kΩ ), representing the other airgap. So the calculations are rather complicated when the airgap becomes bigger than tiny.
     
    Last edited: Jun 5, 2015
  19. Jun 5, 2015 #18

    Hesch

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    Another thing:

    E = Emagn * V (V is the volume of the airgap ) →
    E = Emagn * A * s (A is the cross section area of the airgap) →
    F = dE/ds = Emagn * A (even if H(s) and B(s) are regarded as constants )

    So in the expression for E, there is the variable s "built in".
     
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