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Is 1/R^3 descriptive of magnetic force drop off?

  1. Jul 22, 2009 #1
    Someone on YoutTube stated...

    Each magnetic pole drops off by 1/R^2. Once you get away from both poles (Cavendish setup) they they start canceling each other out and it ends up as 1/R^3.

    Comments?

    Here's the link if you need backup info
    http://www.youtube.com/comment_servlet?all_comments&v=euvWU-4_B5Y [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 22, 2009 #2

    negitron

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    Yes, that's correct. Far field magnetic strength decreases as the cube of the radius.
     
  4. Jul 22, 2009 #3
    What do you mean by far field magnetic strength? And can the setup on the youtube page work like it depticts?
     
  5. Jul 22, 2009 #4
    I've never seen this equation ever!
     
  6. Jul 22, 2009 #5

    diazona

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    The field strength of a magnetic dipole drops off as 1/r^3 when you're relatively far away from both poles.

    A magnetic monopole's field strength would drop off as 1/r^2, but there are no true magnetic monopoles (as far as anyone knows), so the leading-order contribution is the 1/r^3 term.
     
  7. Jul 22, 2009 #6
    So, I guess I have to forget what I've read that magnetism of a metal object near a magnet (always a dipole) fades off with the square of the distance. Have they started changing the text books yet? lol
     
  8. Jul 22, 2009 #7
    Relatively far away from both poles. Like a million miles? How do I know the tranistion point? So the force varies like this?

    R being 1 inch = 1 N, 2 inch =1/4 N, 3 inch = 1/9 N...(somewhere it changes)...1 mile = 1/R^3
     
  9. Jul 22, 2009 #8
    Can someone look at the YouTube video and indicate whether this is a faked setup or not?

    http://www.youtube.com/comment_servl...&v=euvWU-4_B5Y [Broken]
     
    Last edited by a moderator: May 4, 2017
  10. Jul 22, 2009 #9

    diazona

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    Ah, but you're talking about near a magnet. When you're much closer to one pole of a magnet than the other pole, then it is a 1/r^2 force (or, well, very nearly so) because the effect of the one pole you're close to is so much larger than the effect of the other pole. The 1/r^3 behavior is only far away from the dipole.

    If you were to start next to one pole of a bar magnet and move away in any direction, measuring the magnetic force as you go, you'd find that it starts out looking like 1/r^2 and then gradually transitions to 1/r^3.
     
  11. Jul 22, 2009 #10
    Wow, I have never ever read this anywhere and even in the MIT Walter Lewin lectures either. I'm amazed. Please check out the YouTube video and comment. Thanks. I was wrong.
     
  12. Jul 22, 2009 #11

    diazona

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    I made a little graph to show the magnetic force dropoff from a dipole, as you start from one pole and move off to the side. The red line is the actual dipole force, the green line is 1/r^2, and the blue one is 1/r^3, and you can see that the red line starts out following the green line then "moves over" to the blue line. As I said, it's a gradual transition, but the center of sorts is at r=1 on the graph, which represents the length of the dipole. So if the two poles are a distance d apart, roughly speaking, the force is like 1/r^2 if you're closer than d, and like 1/r^3 if you're further away than d.

    Technical details: I actually used the formula for an electric dipole,
    [tex]\vec{E} = \frac{q}{4\pi\epsilon_0}\left[\left(\frac{1}{x^2}-\frac{x}{(x^2+d^2)^{3/2}}\right)\hat{x} - \frac{d}{(x^2+d^2)^{3/2}}\hat{y}\right][/tex]
    but a magnetic dipole behaves the same way with respect to r.
     

    Attached Files:

  13. Jul 22, 2009 #12
    Haha that youtube video is definitely faked somehow. The gravitational constant is far far far too weak to have such an effect.
     
  14. Jul 22, 2009 #13
    So why does not gravity do the same thing??
     
  15. Jul 22, 2009 #14

    negitron

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    Because gravity is universally attractive. There are no gravitational poles. However, note that tides, which are ultimately caused by gravity, do also decrease with the radius cubed because tides are a differential force.
     
  16. Jul 22, 2009 #15
  17. Jul 23, 2009 #16

    jtbell

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    Which textbooks say this? Last semester I taught intermediate level E&M from https://www.amazon.com/Introduction...=sr_1_1?ie=UTF8&s=books&qid=1248326246&sr=8-1 which derives the far field from both electric and magnetic dipoles as [itex]1/r^3[/itex].

    See also for example Wikipedia's article on dipoles.
     
    Last edited by a moderator: May 4, 2017
  18. Jul 23, 2009 #17
    What's the equation to determine how far away the far field is?
     
  19. Jul 23, 2009 #18

    jtbell

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    See Diazona's answer in post #11. There's no sharp "boundary" between the near field and the far field. The far-field formula is always an approximation to the true field, which gets better and better as you get further and further away from the dipole.

    How far away you have to get before the far-field formula is "good enough," depends on your definition of "good enough."
     
  20. Jul 23, 2009 #19
    Is it like this? Since gravity is totally attractive and magnetism has attractive and repulsive attributes, the attractive and repulsive tend to cancel each other?
    Also can you check the link to YouTube and tell me if this setup if faked????
     
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