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I Magnetic field of circular loops and solenoid

  1. Feb 13, 2018 #1
    I don't understand something. At the center of N circular loops, the magnetic field is μ_0NI/2a. And that for a solenoid is μ_0nI. Why are they not the same when the number of loops is large and the length for the solenoid is long?
     
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  3. Feb 13, 2018 #2

    Charles Link

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    It's two very different geometries that you are trying to compare. Note: ## n=\frac{N}{L} ##, where ## L ## is the length of the solenoid.## \\ ## In the first case, ##a ## is the radius of the ring(s), and it has a very short length. Essentially, ## a>> L ##. ## \\ ## For the second case,=the solenoid, its radius doesn't matter, so long as it is fairly long compared to its radius. For the solenoid formula to be accurate, ## L >> a ##.
     
    Last edited: Feb 13, 2018
  4. Feb 13, 2018 #3
    If we place many rings side by side, it looks just like a solenoid, right? And if we apply Ampere’s law on both cases, aren’t we supposed to get the same result? If not, how come? There must be some point that I haven’t figured out.
     
  5. Feb 13, 2018 #4

    Charles Link

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    For the first case, ## a>> L ##. The first case does not work once ## L ## starts to get large enough to make a short solenoid. ## \\ ## Meanwhile, Ampere's law only works for the long solenoid geometry. Biot-Savart works for any geometry. Biot-Savart can readily be computed on-axis for the solenoid of medium length. Let me see if I can find the result in a google and give you a "link": https://notes.tyrocity.com/magnetic-field-along-axis-of-solenoid/ This "link" really needs a figure to show what the angles ## \Phi_1 ## and ## \Phi_2 ## are, but perhaps it is somewhat apparent. Here is a "link" with a diagram. See p.2. The angles are called ## \theta_1 ## and ## \theta_2 ## in this diagram. http://www.pas.rochester.edu/~dmw/phy217/Lectures/Lect_27b.pdf And see the formula at the bottom of p.6. This second "link" is using cgs units, so a couple conversion factors are necessary to get to the MKS result. ## \\ ## Editing: You can even use the formula ##B=\frac{\mu_o nI}{2}( \cos(\Phi_1)-\cos(\Phi_2)) ## to work the case with ## a>>L ##, and you do get the formula ## B=\frac{\mu_o NI}{2a} ## that you presented above. (You let ##n=\frac{N}{\Delta} ##, (with ## L=\Delta ##), and ## \Phi_1=\frac{\pi }{2}-\frac{\Delta}{2a} ##, and ## \Phi_2=\frac{\pi}{2} +\frac{\Delta}{2a} ##. In the limit ## \Delta \rightarrow 0 ##, you get the first formula above).
     
    Last edited: Feb 13, 2018
  6. Feb 13, 2018 #5
    Wow, thanks for the link and the editing part. I should and sould not have posted this thread. I posted so that I could see all of these great derivations. I should not because I am afraid that I have wasted some of your time. I made a mistake interpreting the result for the N loops. That’s for the geometry when you have N loops in the same plane but not for the case by placing loops side by side. Of course the first case cannot use Ampere due to the bad symmetry.

    Btw, I couldn’t agree more with your motto~
     
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