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Magnetic Flux Through a Tilted Medium

  1. Jun 1, 2015 #1
    Imagine a ferromagnetic medium shaped as a cylinder (a ferromagnetic fiber) with a magnetic relative permeability of μr, tilted with an angle a, as shown in the picture.

    220px-Cylinder_geometry.svg.png
    I would like to prove analytically that the sum of the inductances measured along the x axis (angle is a) and y axis (angle is π/2 - a) is independent of the angle a. (Lsum = Lx + Ly independent of a).

    Each case, Lx and Ly is also the sum of the inductance of the projections of the fiber in the x and y axis.
    My approach consists in using Hopkinson's Law and first determining the Reluctance for each case, and then calculating the Inductance using L= N2/Reluctance, where N is a constant number of turns of a supposed magnetomotive force.

    Do you think it's possible?

    Thank you very much.
     
  2. jcsd
  3. Jun 1, 2015 #2

    marcusl

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    Don't understand your post. What is the inductance of a cylinder?
     
  4. Jun 1, 2015 #3
    It's the Inductance of a ferromagnetic medium shaped as a cylinder, that is crossed by a flux produced by a magnetomotive force.
    At least was this that I meant to say. Beg your pardon if I wasn't clear.
     
  5. Jun 1, 2015 #4

    marcusl

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    Inductance is a quantity that relates energy stored in a magnetic field to the currents producung the field. It also relates an induced emf to a changing current. You have no currents, hence no inductance.
     
  6. Jun 1, 2015 #5
    My mistake, forgot to mention that the magnetomotive force is produced by a current I (constant). That's my current.
     
  7. Jun 1, 2015 #6

    marcusl

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    Your problem doesn't make sense as stated and suggests that you don't understand inductance. Have you had an undergrad E&M course?
     
  8. Jun 1, 2015 #7
    Yes I have. I'll try to explain this model.
    Imagine that a magnetomotive force is produced by a number of turns N and a current I, being Fmm=N*I.
    That magnetomotive force creates a magnetic flux ∅ that only crosses the ferromagnetic cylinder.
    The inductance of the ferromagnetic medium/fiber can be determined using the following expressions:

    L = N× ∅ / I

    ∅ = Fmm/R

    L = N2/R

    where R is the reluctance of the ferromagnetic medium, given by:

    R= l/(μ0r*A)

    where l is the lenght of the path of the magnetic flux, μ0 is the magnetic constant (vacuum's permeability), μr is the relative magnetic permeabilty of the ferromagnetic cylinder and A is the area that is crossed by the flux.
     
  9. Jun 1, 2015 #8

    marcusl

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    These formulas are usually written in terms of the number of turns per unit length n, in which case [itex]L=\frac{n^2l}{R}[/itex]. Furthermore, this applies to long solenoids. Since your core is short, this will be an approximation at best.

    To your original question, however, inductance is a scalar quantity so you can't break it into x and y components.
     
  10. Jun 1, 2015 #9
    Yes, inductance is a scalar quantity but I can make a projection of the ferromagnetic fiber along x and y, with a length and cross area also projected, calculate both Inductances Lx and Ly, and then the "final" inductance would be L = sqrt(Lx^2 + Ly^2).
     
  11. Jun 1, 2015 #10

    marcusl

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    Well yes, you could do that, but I don't see the value since L is already independent of your angle, by definition.
     
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