1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Forces on multiple cylindrical magnets stacked

  1. Dec 28, 2011 #1
    I am trying to calculate the forces between ten small cylindrical magnets stacked (inside a tube so they cannot flip/shift) so that consecutive magnets are repelling. The bottom magnet is attatched to the bottom of the tube with, lets say, the positive pole facing up. Then the second has a positive pole facing down, third has it facing up.. etc. So they appear to be floating above each other. I have a stick which i can use to push on the top magnet and compress the space between each magnet.
    If anyone could give some insight on how to calculate the force between the magnets or how to find the maximum force I can push down on them with, it would be really helpful because I can't seem to find much online with this kind of set-up.
    Thanks :)
     
  2. jcsd
  3. Dec 28, 2011 #2

    I like Serena

    User Avatar
    Homework Helper

    Welcome to PF, julie827! :smile:

    The force between magnets is not so straight forward as the force between electric charges I'm afraid.
    Here's the wikipedia article that describes the force between magnets:
    http://en.wikipedia.org/wiki/Force_between_magnets

    Specifically it has a section named "Force between two cylindrical magnets".
    From it you can deduce that for cylindrical magnets of height t and separation x in extreme cases you get:
    $$F \approx {C \over x^4}\qquad\textrm{if x >> t}$$
    $$F \approx {C \over x^2}\qquad\textrm{if x << t}$$
     
    Last edited: Dec 28, 2011
  4. Dec 28, 2011 #3
    would this mean that I would have to find the force between each possible set of two magnets? so 1 with 2, 1 with 3, 1 with 4.. etc?
     
  5. Dec 28, 2011 #4

    I like Serena

    User Avatar
    Homework Helper

    If you keep the tube horizontal, the forces and distances between the magnets will be the same. They will balance out.

    If you keep the tube vertical and if you cannot neglect the weight of the magnets themselves, yes, you will have to find the force between each set of 2 magnets.
     
  6. Dec 28, 2011 #5
    For my purposes, the tube will be verticle.
    thanks for the help! I guess it's just a lot of calculations now...
     
  7. Dec 28, 2011 #6

    I like Serena

    User Avatar
    Homework Helper

    Good luck!
     
  8. Dec 28, 2011 #7

    rcgldr

    User Avatar
    Homework Helper

    For the vertical tube, the downwards force from above on any magnet in the tube will equal the weight of all the magnets above that magnet, plus any force you apply to the top magnet. The magnets will space themselves so that the magnetic forces cancel the gravitational and any external force applied to the top magnet. The downwards force at the bottom the tube will equal the weight of all the magnets plus any force applied to the top magnet.
     
  9. Dec 29, 2011 #8
    I don’t blame you but I’ve found a number of problems with that Wiki page.

    (1) If you do the maths then when x>>t, F goes to zero.
    (2) In the formula just below the heading you mentioned, there’s a term R^6/t^4 missing.
    (3) But look at this statement: (at the end of that same chapter)
    F in that formula is proportional to x^-4 whereas earlier on F was worked out to be proportional to r^-2. ( x and r are both representing the same distance).
     
  10. Dec 29, 2011 #9

    I like Serena

    User Avatar
    Homework Helper

    Huh? :confused:
    How do you get that?
    The first term at x=∞ of the Taylor expansion is:
    $$\frac 3 2 \pi\mu_0 M^2 R^4 t^2 {1 \over x^4}$$
    As you can see here.
    Just like the article says.


    I tried to find another reference to compare the formula but couldn't find one.
    Do you have a reference?


    Erm... earlier two poles were mentioned, not two dipoles.

    As you can see here, in a magnetic dipole ##F \propto r^{-4}##.
     
  11. Dec 29, 2011 #10
    Ok I slipped up badly. Also I need to make mare use of Wolfram to do my maths. Thanks anyway.
     
  12. Jan 6, 2012 #11
    To determine the force without doing a billion calculations, I used a scale to measure the force (in kg's). I put the magnets (arranged so they repel: north-south then south-north then north-south etc) in a tube and then capped one end. I put a dowel in the other end and pushed the magnets together so that they were almost all touching (there was a tiny gap between at least two of the magnets). Then, I used the scale to push the dowel and thus the magnets together and recorded the force in kg's. What I found was that no matter how many magnets I had in the tube (two through to nine), the force was about the same. Does this make sense? or have I done something wrong?
     
  13. Jan 6, 2012 #12

    rcgldr

    User Avatar
    Homework Helper

    Yes, it makes sense. The forces on the first magnet must cancel or else the magnet would be accelerating. The forces on the first magnet are the force from the dowel on one side and the repelling force from the second magnet on the other side. The force that the first magnet exerts on the second magnet is equal and opposing to the repelling force the second magnet exerts on the first magnet, which was equal and opposing to the force on the first magnet from the dowel. So the forces between magnets, dowel and the other end of the tube have the same magnitude, regardless of the number of magnets. The gap between magnets vary a bit with the number of magnets due to the attractive force between every other magnet, such as a 2 magnet case versus a 3 magnet case, but this may be too small a difference to notice.

    If you oriented the tube vertically, then the force on each magnet would be the weight of the remaining magnets above, and the gaps would be smaller at the bottom than at the top.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook