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Equivilence between Biot-Savart's Law and Ampere's Law

  1. Apr 16, 2010 #1
    I am considering the equivilence between Biot-Savart's Law and Ampere's Law for a current loop. The form of the magnetic field from a current element in the Biot-Savart law becomes

    dB = [tex]\mu[/tex]oI dL sin[tex]\theta[/tex]/4[tex]\pi[/tex]r2

    which in this case simplifies greatly because the angle =90 ° for all points along the path and the distance to the field point is constant. The integral becomes

    B = [tex]\mu[/tex]oI/2r

    It would appear that there is sufficient symmetry to apply Ampere's Law, with the line integral

    [tex]\oint[/tex] B dL cos [tex]\theta[/tex] = [tex]\mu[/tex]oI

    of the enclosed value B dL cos [tex]\theta[/tex] reflecting the surface area of a torus.

    However, as I work this out I cannot get the same value as produced by Biot-Savart's Law.

    Is the problem that the B Field is not constant such that the line integral

    [tex]\oint[/tex] B dL cos [tex]\theta[/tex]

    cannot be easily calculated?
  2. jcsd
  3. Apr 16, 2010 #2
    First, we need to be clear that your result from the Biot Savart Law is only valid at one point in space: the center of the loop, to be specific. It is easy to calculate the field along the entire central axis of the loop as well. However, at other points, you probably need to do the integration numerically.

    Second, your last statement is essentially correct. There is not enough symmetry to find an integration path, which includes the center of the loop, and in which you know B is constant. This is why Biot Savart's Law is often more useful than Ampere's Law.

    One exception (as I'm sure you know) is the well-known case of a straight, infinitely long wire. Here symmetry allows Ampere's Law (static case) to be used for a quick derivation of the B field at all points in space.
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