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Magnetic Field in a rectangular toroid

  1. Apr 2, 2012 #1
    Consider a toroidal structure with a rectangular cross-section. If the toroid is defined by
    the surfaces r = 1cm and r = 4cm and the planes z = 0 cm and z = 2 cm, and the surface
    current density on the surface defined by r = 4cm is given by -60az A/m.
    (a)
    Specify the current densities on the surfaces at r = 1, z = 0, and z = 2.
    (b)
    Find the expression for H inside the toroid (i.e. in the region 1 < r < 4 cm and 0 <
    z < 2cm).

    I have the solution, I just don't really understand it. The current density, K, for r=1 was found to be 240 A/m.

    To find the field inside the toroid ampere's law is used [itex]\oint[/itex]H dl = 2∏r = ∫Kr=1dθ from θ=0 to 2∏. Giving the final answer of H=15/(2∏r)

    So in this question K is different at r=4 and r=1. Is it changing due to distance from the sources of the field, or is it uniform and I'm missing some simple math? Since K is different at different radii, how come we integrate only integrate with Kr=1 and not other any other K's? I mean it is changing and I'm assuming since the cross section of the toroid is a square loop, all four sides would contribute to the field inside the toroid, not just r=1 .

    Wouldn't I need to calculate for Kr=4 and Kz=0 =Kz=2 and sum them up? Hope that made sense, thanks in advance.
     
  2. jcsd
  3. Apr 2, 2012 #2

    collinsmark

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    Homework Helper
    Gold Member

    I think you might have left out an H and an r somewhere in various places.
    Okay, that sounds reasonable.
    You can add them in if you want. But the real trick is to realize what is the contribution from these other sides?

    Let's look at Ampere's law for magneto-statics.

    [tex] \oint _P \vec B \cdot \vec{dl} = \mu_0 I_{enc} [/tex]

    And realizing [itex] \vec B = \mu_0 \vec H [/itex] for vacuum/air (which is the case for this problem), we can rewrite that in a form more like you used,

    [tex] \oint _P \vec H \cdot \vec{dl} = I_{enc} [/tex]

    The left hand side of the equation describes a closed path integral. The right hand side of the equation is the current enclosed within the path. Currents outside the path don't count.

    For this problem, choose a path that circles along within the complete torus structure itself. How much current is within the path? In other words, how much total current flows through the looped path (on the inside)?

    [Edit: Or another way to ask the problem, are the currents on top, bottom, and at r = 4 outside the path or inside the path? If they don't actually pass through the path/loop itself, they don't contribute.]
     
    Last edited: Apr 2, 2012
  4. Apr 3, 2012 #3
    So to clarify, I can pick a circle as my path anywhere within the circle and make it so only the Kr=1 goes through cross section of the path. The Kz are parallel with the path so they don't go through and don't contribute. Also Kr=4 can be ignore since it's outside of my path,making only Kr=1 left.

    I'm surprised I didn't get that, since this is almost the same as gauss law. Thanks again for the help.
     
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