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Velocity of a magnet traveling in a magnetic field

  1. Sep 20, 2012 #1
    So if we can one large not moving magnet and a small magnet that is attracted to the large magnet we know that ( http://en.wikipedia.org/wiki/Magnet ) F = μ qm1 qm2 /(4 pi r^2). Here the small magnet is traveling in a straight line because it is attracted by the large magnet.

    Where F is force (SI unit: newton) qm1 and qm2 are the magnitudes of magnetic poles (SI unit: ampere-meter) μis the permeability of the intervening medium (SI unit: tesla meter per ampere, henry per meter or newton per ampere squared) r is the separation (SI unit: meter).

    To calculate the velocity equation can we equate mass * acceleration = F = μ qm1 qm2 /(4 pi r^2) and integrate for acceleration? this yields Velocity = constant/ r. I was wondering if this is correct. Thank you.
     
  2. jcsd
  3. Sep 20, 2012 #2

    mfb

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    The formula you copied is the force between magnetic monopoles. They do not exist in nature (or at least no monopoles were found yet), real magnets always have two opposite poles, and you have to consider this to calculate the force.
    That does not follow from the formula. Earth is attraced by the sun, but we do not fall in a straight line into sun.

    Not in the way you describe it, but it can be done. Energy conservation is easier here - find the potential, use that the sum of kinetic and potential energy is constant.
     
  4. Sep 20, 2012 #3
    thanks but the response does not answer the question and is wrong

    check the following:

    http://geophysics.ou.edu/solid_earth/notes/mag_basic/mag_basic.html [Broken]

    as you know if you keep on magnet fixed and you have another magnet some distance apart they attract if their poles are opposite

    what is the force to characterize these two in terms of distance? the further away the weaker the force and as they get closer the stronger the force

    as they more toward each other, what is the equation of acceleration and velocity?
     
    Last edited by a moderator: May 6, 2017
  5. Sep 20, 2012 #4

    pervect

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    It seems to say just what the response you said was wrong said. I don't know why you said the response was wrong, it looks OK to me.

    Specifically, from the reference you quote:

    This seems to me to say essentially the same thing MFB said to you earlier.

    I'm afraid I don't understand the question you are asking if the above doesn't answer it. Perhaps you could try clarifying what you meant to ask.
     
    Last edited by a moderator: May 6, 2017
  6. Sep 21, 2012 #5
    It seems to me that jmmy is looking for a method to calculate the excerted force on the small magnet at a certain length from the big magnet. So he can use integrals or derivatives to obtain velocity and acceleration functions.

    ....... And then he can also calculate 'work' done if he really need it for his application.

    NONE of the replies, including this one, have solved jmmy issue.

    P.S.: jmmy didn't asked for conservation of 'energy'.
     
  7. Sep 21, 2012 #6

    mfb

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    It is correct, but maybe it is not what you expected.
    Why do you think that it does not answer the question?

    It is the easiest method to get the velocity as function of distance.
     
  8. Sep 21, 2012 #7

    vanhees71

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    I also think the question has been properly answered. I'll try to clarify the issue:

    The provided link about "magnetic poles" is misleading without further explanations. There is no such thing as a magnetic monopole. That's why in nature there is nothing like a Coulomb-like interaction between magnetic poles. The most simple realistic example thus is the motion of a magnetic dipole in an (inhomogeneous) magnetic field.

    It should also be added that the rest of this page is as misleading (if not to say wrong!) as its beginning. A magnetic field usually has no scalar potential. Even if you only look at magnetostatics in the vacuum you have (in Heaviside-Lorentz units)
    [tex]\vec{\nabla} \cdot \vec{B}=0, \quad \vec{\nabla} \times \vec{B}=\vec{j}.[/tex]
    The first equation tells you that there are no magnetic charges (or monopoles as they are called in the literature), and the second equation tells you that there is no scalar potential for the magnetic field, where non-vanishing current densities are present, and somewhere there must be such currents, because otherwise there wouldn't be magnetic fields. Note that also permanent magnets are described by these Maxwell equations. You only have to set [itex]\vec{j}=c \vec{\nabla} \times \vec{M}[/itex], where [itex]\vec{M}[/itex] is the magnetization of the permanent magnet.

    So your original question doesn't make much sense, and that's why the answers you got are perfectly to the point and correct!
     
  9. Sep 21, 2012 #8
    Look what I am asking is very simple, I think many think in complected terms so forget about the theory or textbook questions for a minute,

    do you agree if you have 2 magnets they will attract each other and move toward each other (nobody mentioned monopoles) if their opposite poles face each other?

    it does not require the maxwell equations to know this

    just take 2 magnets and you will see that eventually they will stick together

    the force obeys the inverse square law (just like gravity)

    so my question is what is the equation of speed and acceleration for this type of attraction?
     
  10. Sep 21, 2012 #9

    mfb

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    Right.

    Wrong. In general, it is a complicated function of distance. With a separation much larger than the size of the objects, I think it is proportional to 1/r^4.

    For an inverse square law:
    $$a=\frac{F}{m}=\frac{c}{r^2}$$$$v^2(r)=\frac{c'}{r}-\frac{c'}{r_0}$$

    For a general 1/r^n-law: $$a=\frac{F}{m}=\frac{c}{r^n}$$$$v^2(r)=\frac{c'}{r^{n-1}}-\frac{c'}{r_0^{n-1}}$$

    Both c and c' are constants which depend on the strength of the interaction, the masses and so on. The formula for the velocity assumes that the objects are at rest at a radius r0.
     
  11. Sep 21, 2012 #10
    Expressions based on constant magnitude 'point' dipole model will be ok for relatively large separations, but breaks down otherwise, for two reasons. At close range one needs to integrate over a volume distribution of elemental dipoles, so magnet shape is important. Also for real magnets magnetization is never completely uniform or saturated. There will be some mutual change in magnetization as the fields interact strongly, though relatively small in the case of 'neo' permanent magnets which exhibit strong remanence (resistance to demagnetizing).
     
  12. Sep 21, 2012 #11
    well i did this experiments in liquid
    one magnet fixed
    one magnetized micro-particle

    turns out v= constant/r fits my data exactly
     
  13. Sep 21, 2012 #12

    mfb

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    v=constant/r would correspond to v^2=c'/r^2 or a 1/r^3-force.
    However, the liquid could change that completely.
     
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