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Ampere's Law on Current Carrying Loop

  1. Oct 6, 2013 #1

    I haven't seen anyone derive the magnetic field density (B) using ampere's law, only using Biot-Savart Law
    any reason why?

    if we cut the loop and loop at one end (of the new cut) and treat it as if it was a current carrying wire, then by ampere's law we'd get:

    B = u*I / 2*pi*r

    but however by the Biot-Savart Law we actually get

    B = u*I / 2*r

    anyone know why?
  2. jcsd
  3. Oct 6, 2013 #2


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    Ampere's law tells you that (line integral of B.dl along the loop) = μ0 * (flux of I through the loop). If you know from the geometry of the situation that B is constant along the loop, then calculating the line integral of B along the loop is easy. This is the case with a long straight wire. Here we draw a loop at a distance R from the wire. We know from symmetry that the value of B is everywhere constant along the loop, so the line integral of B along the loop is just 2*pi*R*B. In the case of a current carrying ring, no matter how you draw your Amperian loop, there is no way to draw it so that B is constant. So, while Ampere's law still holds, it is not very useful, since you don't know how to calculate the line integral. So you use the Biot-Savart law, which is more amenable to a general situation.
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