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B Are electrons or protons attracted due to their magnetic moments?

  1. Dec 4, 2018 #1

    Javier Lopez

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    Does electrons or other particles attracted towards a magnet due its magnetic momentum?
     
  2. jcsd
  3. Dec 4, 2018 #2

    mfb

    Staff: Mentor

    There is a force towards regions of stronger magnetic fields. Which can be towards a magnet but doesn't have to be. The Stern-Gerlach experiment is one example.

    For free charged particles typically electric fields acting on the electric charge are more important.
     
  4. Dec 5, 2018 #3

    Javier Lopez

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    I suppose that the Lorentz force is greater also than the magnetic moment force

    At large distances I have seen that the force is:
    $$
    Fm=\mu \frac{q_{m1}*q_{m2}}{4*\pi *r^2}\\\\
    Fe=\frac{1}{4*\pi*\epsilon_0 }*\frac{Q_1*Q_2}{r^2}
    $$

    As long as magnetic moment of deuterium es 8e-24 and charge is 1.6e-18 the electrostatic force is 3.6e25 times higher.
    At short distances things changes drastically as long as magnetic moment force rises with 1/r^4 and accordingly my calculus cancells electrostatics field force for deuterium-deuterium at 2e-13 m that I supposse falls within QM equations.
    What is approximately the validity range of classical approach?. In my simulator I use classical approach to calculate magnetic fields and EM forces due coils
     
    Last edited: Dec 5, 2018
  5. Dec 5, 2018 #4

    mfb

    Staff: Mentor

    Compare the usual parameter sets (position&momentum, energy&time, spin) with the Planck constant. If they are large a classical approach might work.

    I don't see how you got 2e-13m but it looks way too large.
     
  6. Dec 6, 2018 #5

    Javier Lopez

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    I have made a more approximate calculus and obtained 4.97e-13 that is more large.
    I obtained that data using the classical aproach:
    $$
    F_e=K*\frac{Q*Q'}{2^2}\\\\
    F_m=\frac{3*10^{-7}*m1*m2}{r^4}*(-3*cos(\theta )*cos(\varphi)+sin(\theta)*sin(\varphi)*cos(\phi ))\\\\
    F_m max(\varphi,\theta,\phi=0)=\frac{3*10^{-7}*m1*m2}{r^4}*(-3)
    $$
    Where the electrostatic force at 4.97e-13 was 9.34e-4 newtons
    And magnetic force almost similar. I used for deuterium magnetic moment 7.95e-24 J/T (and all angles=0 to have maximum magnetic force)
     
    Last edited: Dec 6, 2018
  7. Dec 7, 2018 #6

    mfb

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    I don't know where you got that value from but it is three orders of magnitude too large.
     
  8. Dec 7, 2018 #7

    Javier Lopez

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    You must be true, I multiplied the deuterium magnetic moment 2.79284734 by Born magneton (9.274009994E-24 J/T) instead of nuclear magneton: 5.050783699e-27 J/T
    Thank you very much, now I can correct my equations

    Now the equilibrium point is 2.7e-16 m where both forces are 3.17 kilonewtons
    2.7 e-16 is a lot lower than the square root of the cross section of the deuterium-deuterium reaction
     
    Last edited: Dec 7, 2018
  9. Dec 7, 2018 #8

    mfb

    Staff: Mentor

    2.7e-16 m is also smaller than the size of the deuterium nucleus. They cannot come that close and the formulas you used break down before that, in addition the strong interaction is dominant at very small distances.
     
  10. Dec 7, 2018 #9

    Javier Lopez

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    I agree at all. At larger distances it gives me an idea of how an external magnetic field could help to increase the cross section, but I should be sure having the correct coefficients

    About the Planck constant I suppose that I have to use the h=1.986-25 J/m, so for 650KeV protons we should have:

    $$\frac{h}{E}=\frac{1.98644*10^{-25}J*m}{650000 eV * 1.6021*10^{-19} J/eV}=1.9*10^{-12}m$$

    That is close to the famous 2 picometers where strong forces have a minimum
     
    Last edited: Dec 7, 2018
  11. Dec 7, 2018 #10

    mfb

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    The strong force is completely irrelevant at 2 picometers. And I don't think the distance you calculated has a meaning.
     
  12. Dec 7, 2018 #11

    Vanadium 50

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    Not very famous - I never heard of such a thing. At 2000 fm the strong force is effectively zero.

    The other numbers seem to be from random equations.
     
  13. Dec 8, 2018 #12
    A magnetic pole actually repels a free charge like an electron or a proton, unless it moves precisely along a field line or is stationary.
    Magnetic momenta, such as that of a neutron, which has no charge, are attracted to magnetic poles.
     
  14. Dec 15, 2018 #13

    Javier Lopez

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    Why?, magnetic moment of loops makes them attract between them in 180º degree range and repels if in the 180º opposite directions like magnets does.
     

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    Last edited: Dec 15, 2018
  15. Dec 15, 2018 #14

    mfb

    Staff: Mentor

    Now consider where 2000 fm would be on that scale.
     
  16. Dec 15, 2018 #15

    Javier Lopez

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    You are right, I deleted the wrong part
     
  17. Dec 15, 2018 #16
    Consider a charge regardless of sign moving through uniform magnetic field with parallel field lines in direction other than directly along the magnetic field.
    If the movement of the charge is at right angle to magnetic field then Lorentz force is at right angle to direction of movement and magnetic field - to the centre of a circle, balanced by a centrifugal force. If the movement of the charge is at an acute angle to magnetic field then the Lorentz force is still at right angle, but now to axis of a helix.

    But now consider the charge moving in nonuniform magnetic field, where the field lines converge at a pole.
    The Lorentz force is still at right angle to magnetic field and direction of movement. But considering the simple case of field and motion direction at right angle: since the field lines along the circle are not parallel, the Lorentz forces along the circle no longer cancel out with each other and centrifugal force. They leave a resultant force away from the pole.
    If the charge is different or the identity of magnetic pole is different, only the direction of circling will change. The resultant repulsive force on a moving charge is the same: always from stronger magnetic field to weaker.
     
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