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HiI know that for a short solenoid (L<R) the magnetic field at the

  1. Feb 3, 2012 #1
    Hi

    I know that for a short solenoid (L<R) the magnetic field at the axis is (standard EM)
    [tex]
    B(z) = \frac{1}{2}\mu_0 \frac{N}{L}I(\frac{z+\frac{L}{2}}{\sqrt{(z+L/2)^2+R^2}} - \frac{z-\frac{L}{2}}{\sqrt{(z-L/2)^2+R^2}})
    [/tex]
    where R is the radius of the solenoid, N the number of turns along the axis and L the length. In this system, each vertical plane consists of a single turn, but say I am looking at a solenoid, where each vertical plane consists of e.g. 2 turns. First I thought about using the above equation twice, but that is wrong since it is not 2 independent solenoids.

    Is it correct to regard the system simply as a collection of coils with 2 turns each? I'm not quite sure how this would work out, since this way I can't take into account the widths of each individual coil. If my description is confusing, please let me know.

    Best,
    Niles.
     
    Last edited: Feb 3, 2012
  2. jcsd
  3. Feb 3, 2012 #2

    clem

    User Avatar
    Science Advisor

    Re: Coils

    Your formula is for a solenoid of any length, not just L<R.
    I don't understand what you mean be 'two turns'.
    Why would it make a difference?
     
  4. Feb 3, 2012 #3
    Re: Coils

    Hi, thanks for replying. I really appreciate it. What I mean by two turns is the following:

    OOOOOO
    OOOOOO
    ------------ (axis of solenoid)
    OOOOOO
    OOOOOO

    There I have a solenoid with 12 turns in total, i.e. a 6-turn solenoid on top of another (larger radii) 6-turn solenoid. How would one calculate the B-field for that? In principle I could just add the B-field for each one of the 12 turns, right? The reason why I am asking is because I am looking for the most general way to calculate this, because I would also have to look at e.g.


    OOOO
    OOOOOO
    OOOOOOOO
    ---------------------- (axis)
    OOOOOOOO
    OOOOOO
    OOOO
     
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