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I Cylindrical magnet

  1. May 25, 2018 #1
    Hi, I am looking for the formula of the magneti field along the axis of a axially magnetized cylindrical magnet.
    Unfortunately, there are quite different answers on Internet.
    Is the uploaded formula (where R is the magnet radius and L its length) correct?
    upload_2018-5-25_12-26-25.png
     
  2. jcsd
  3. May 25, 2018 #2

    kuruman

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    It looks right. If I were you, I would go along the more complete equation (8) given here
    http://web.mit.edu/6.013_book/www/chapter9/9.3.html
    because it provides the field inside the magnet as well. Note that this equation has the mid point of the magnet as the origin, unlike yours where the origin is at one of the faces.
     
  4. May 25, 2018 #3
    Thank you very much. However, eq. (8) does not have an anlytical solution, does it? You need to use numerical integration to solve it.
     
  5. May 25, 2018 #4

    kuruman

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    It is an analytical solution. Perhaps you confused the ##d~M_0## up front for a differential. It is not; ##d## is the length of the cylinder as shown in the drawing and ##M_0## is the magnitude of the magnetization.
     
  6. May 28, 2018 #5
    Ohh, I did!
    Thank you so much.
    Is a 1/2 the value subtracted to z/d?
     
  7. May 28, 2018 #6
    Actually I see that the two formulae are different. Indeed, replacing d with L and changing the origing z = z1-d/2 in eq.(8), you get:
    upload_2018-5-28_11-11-42.png
     
  8. May 28, 2018 #7

    kuruman

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    Right.
     
  9. May 28, 2018 #8
    So μ0 is replaced by d in your formula.
    eqn-9.14.gif
    IS M0 measured in A/m?
     
  10. May 28, 2018 #9

    kuruman

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    No. If you follow the proof in the MIT link, you will see that μ0 drops out in equation (6). The symbol ##d## is the half-length of the cylinder. In other words ##2d=L##.
    Yes.
     
  11. May 28, 2018 #10
    I have a doubt: going back to eq. (8), the term in the square brackets is adimensional, M0 is [A/m] times d[m], you get a Hz in [A].
     
  12. May 28, 2018 #11

    kuruman

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    You are correct. I re-derived the whole thing and equation (8) should have ##\frac{M_0}{2}## as the overall factor without the ##d##. The ##d## in the numerator cancels out when one makes the denominator dimensionless. Whoever posted this probably missed the cancellation, but this goes to show that you should distrust what you see on the web, even if it's from MIT.
     
  13. May 28, 2018 #12
    Thank you very much. Not distrust, but re-think critically.
    Finally I have the field thanks to your help.
    Last question: I can swop from B to H with μ0, but if I need the force on a particle in the field of the magnet, I have to use B or H?
     
  14. May 28, 2018 #13

    kuruman

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    I would say use B. What equation do you plan to use to find the force?
     
  15. May 28, 2018 #14
    The classical derivation from the analogy with the force on an electric dipole: (m*Del) B.

    Kuruman, please, do you know a complete expression for the field of a cylindrical magnet, not only along the axis of the magnet?
     
  16. May 28, 2018 #15

    kuruman

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    An analytical expression for that involves elliptic integrals so I don't think there is one. If the force is to be calculated very near the z-axis you may do a series expansion of the radial field component. If not, then a numerical calculation may be appropriate.
     
  17. May 28, 2018 #16
    And and expression to calculate numerically it? Do you know a link in the web?
     
  18. May 28, 2018 #17

    kuruman

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    Sorry, I don't know of any links. I am sure you can research the web as well as me, if not better.
     
  19. May 29, 2018 #18
    Applying the two formulae, which we have discussed so far, it is easy to calculate the z-component of field at the top of the magnet along its axis:
    upload_2018-5-29_13-5-43.png
    Now, I found a page of Mathematica by Wolfram, which reports an expression, to be computed numerically, for the whole field:
    http://demonstrations.wolfram.com/MagneticFieldOfACylindricalBarMagnet
    However, this expression does not converge to the above one for ρ -> 0 and z ->L/2.
    Do you have the same result?
     
  20. May 29, 2018 #19

    kuruman

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    To me the Wolfram expression looks like the same as the integral leading to the corrected MIT expression. I don't understand what you mean by ρ→0. ρ is an integration variable. If you want to see what happens when you have a long thin cylindrical magnet, first you integrate then you look what happens in the limit R/L <<1.
     
  21. May 29, 2018 #20
    The integration variables are R and Φ. R integrated over [0, a] and Φ over [0, 2 Pi].
    After the integration, you have the field in generic ρ and z, namely Bz(ρ,z).
    I want the field in (ρ = 0, z =L/2), i.e. Bz(0,L/2).
     
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