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Solenoid pull-push force

  1. Mar 30, 2016 #1
    Hello, I have some problems with my solenoid calculations. I have spent many hours, on searching for Formulas, and answers, but exact answer I cannot find. I made 2 electromagnets, and then tried with solenoids.
    Last solenoid was made with 120 m long wire, 0,3 mm, center diameter was 12 mm and 4 cm long. I achieved 0,5 A on 12V. But solenoid pull force was weak.
    So here is my question: My goal is to make solenoid which can push iron bolt (center) with neodymium magnet with push force similar with when put two permament magnets with same poles.
    I planned max 1A and 12 V. How can I make solenoid with max force with such parameters, and what kind of wire I need- 0.3, 0.5, 0.6, 0.7, how long, Is it even possible, to gain such solenoid force, with such parameters.
    Thanks.
    And here is an image to show my idea.
    009b85c.png
     
  2. jcsd
  3. Mar 30, 2016 #2

    BvU

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    Hello Dave, :welcome:

    Did you also try out your solenoid with the bolt inside?
     
  4. Mar 31, 2016 #3
    Yeap, I made electromagnets with bolt inside. It was weak. I think problem is with wire, its too tight problably. And when I made my coil current was 0,5 A.
    Now I think that i have to make solenoid with 0.5 or 0.6 mm wire, with 1A current, but I dont know how to calculate force, what I can reach and will it be enough for my project.
     
  5. Mar 31, 2016 #4
    The ordinary pull force, for an iron bolt, can be estimated easily. The product of the current and the number of turns divided by the coil length gives the estimation of the magnetic field H inside the coil. The magnetic constant (1.26 microhenry/m) divided by 2 and multiplyed by the square of H gives the magnetic energy density inside the coil. Finally, the product of that energy density and the bolt cross section area gives the estimation of the magnetic pulling force (provided the end of the bolt is inside the coil).

    As for pushing a permanent magnet, it's way more complicated, and the force must be much weaker...
     
    Last edited: Mar 31, 2016
  6. Mar 31, 2016 #5
    Thanks for answer,
    My calculations: 1000 turns Current: 1A , length 0,04m
    H=25000
    Then : (0.00000126/2) * 25000^2 = 393,75

    Then: 393,75* 0.00005527000m² = 0,021 ??
    Is it correct, and is it kg or lb?? What I calculated?
     
  7. Mar 31, 2016 #6
    It's 0.021 newton

    https://en.wikipedia.org/wiki/Newton_(unit)

    Your energy density is 393.75 joule/cubic_meter ; when you multiply it by 0.000055 square_meter , it gives 0.021 joule/meter - and joule/meter is exactly newton, the SI unit for force.

    0.021 newton is the force you need to lift a 2.14 gram weight - not very much...

    To increase the force, one should use additional iron things in order to concentrate the magnetic energy near the bolt end, like here:
    s-l225.jpg 197-004-512E13B5.jpg
    - the field H (not B ! ) is concentrated between the back iron plate and the internal end of the bolt (plunger). For example, it may be 4mm between them instead of your 40 mm - and then, the field is 10 times greater and the force is 100 times stronger.
     
    Last edited: Mar 31, 2016
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