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Electromagnetics density of Toroid

  1. May 24, 2014 #1
    Hello
    I have some problems to obtaining Electromagnetic density via Ampere's Law.

    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html

    my problem is why we don't calculate and sum Inflow and outflow of Ampere's closed loop path.to clearing my question please take a look at below picture.
    tor.gif
    in upper side of Toroid current direction is going into center of Ampere's closed loop path however down side current direction is going out subsequently input current and output current in Ampere's closed loop path should be zero.

    what is my misunderstand?

    Thanks
     
  2. jcsd
  3. May 24, 2014 #2
    You are misunderstanding the fact that the Amperean loop in the figure is constructed BETWEEN the innner and the outer radius of the toroid and not at a radius greater than or equal to b. Here, the direction of every current element is into the plane of the paper (or according to you, to the center of the toroid).

    Had the amperean loop been constructed at a radius of 'b' (outer radius of toroid) or more, the total current crossing the loop would be zero.

    Hope this helps! :)
    -Adithyan
     
  4. May 24, 2014 #3
    You're mistaken all the currents are going through the loop in the same direction. There is no upside and down side. Your misunderstanding seems to be a different (and incorrect) interpretation of what is meant by "to go through the loop".
     
  5. May 27, 2014 #4
    Thanks for your response.
    but my question is some thing else.(sorry if i explained so weak)
    as you see this picture
    7913629700_1401179559.jpg
    two different current direction see in Ampere's closed loop path.one of them wire is top of Toroid and we see the direction of current is going to center and the bottom current's wire that i specify with dot is going out.so sum of them should not be zero?
     
  6. May 27, 2014 #5

    rude man

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    The blue dashed line defines a round surface, area = pi r^2. The current always goes INTO that surface. The fact that the current goes OUT OF the page is irrelevant because that is not part of the surface.

    Ampere's law states that the integral of the B field around a defined surface = mu times the total current through that surface.
     
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