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Toroidal coil, magnetic field

  1. Nov 15, 2013 #1
    1. The problem statement, all variables and given/known data
    A toroidal coil,with N turns of wire carrying a current I, is uniformly wound around a
    toroidal core that has a mean radius a, a cross-sectional area A= ∏(d/2)^2, and is made of a linear, isotropic homogeneous material with relative permeability μr. You may assume that the cross-sectional diameter, d, is very much smaller than a, d < a.

    a, Use Ampère’s law to derive a formula for the magnetic field B inside the torus, and hence the total magnetic flux through the toroidal solenoid

    b, A narrow air-gap of width w is now made, by removing a small sector of the toroidal core, so that the gap is a fiftieth of the circumference of the toroid. Show that the magnetic field in this gap is given by;

    B = μrμ0NI/ 2∏a - w + wμr



    2. Relevant equations

    ∫B.ds= B∫ds

    flux= B*A

    3. The attempt at a solution

    For part a I got B=μ0μrNI/2∏a

    and The flux ,phi = B*A = μ0μrNId^2/8

    b, μ0μrNI = Bc.dsc + Bg.dsg

    where Bc is the magnetic field in the toroidal core and Bg in the gap. Dsc is the area of the core and Dsg the area of the gap

    Any tips on where to begin with part b would be much appreciated,

    Thanks
     
  2. jcsd
  3. Nov 15, 2013 #2

    mfb

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    Staff: Mentor

    Okay.
    The right side is not B*A.

    Good. What is the relation between B- and H-field in the material / in the air gap? You can assume that the field is perpendicular to the surface of the material.
     
  4. Nov 16, 2013 #3
    Hello, thanks for helping, so now

    Flux , phi = ∫B.dA = μ0μrNId^2/8

    B=μ0(H + M) for the toroidal material but we have magnetic linearity so:

    Bc= μ0μrHc and Bg=μ0Hg

    where Bc and Hc represent the B-field and H-field in the toroid and Bg and Hg in the air gap.

    but the total flux trough each area of the circuit must be the same

    so Acμ0μrHc = Agμ0Hg

    where Ac and Ag represent the cross sectional area of the toroid and air gap respectively.

    So Hg = Acμ0μrHc / Agμ0

    Is this the right direction? it gets quite messy after this point
     
  5. Nov 16, 2013 #4

    mfb

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    Staff: Mentor

    Again, the left side and the right side are different things.

    At some point, it gets interesting to consider a line integral along the toroid core...
     
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