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Ampere's Law using the Biot Savart Law

  1. Jul 23, 2008 #1
    Is it possible to derive Ampere's Law(circuital) using the Biot Savart Law and elementary calculus?

  2. jcsd
  3. Jul 23, 2008 #2
    Ampere's Law is equivalent to the conservation of charge i.e. the continuity equation.
  4. Jul 23, 2008 #3
    Can you provide the derivation please..

  5. Jul 24, 2008 #4
    Please help me! :D
  6. Jul 24, 2008 #5


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    You can derive it with vector calculus, but definitely not only with elementary calculus. But with elementary calculus it's possible to derive a simple special case of Ampere's circuital law, namely that for a closed path around a current carrying wire. There may be other (highly-symmetric) configurations for which Ampere's law can be derived from Biot-Savart, but I can't think of any at present.

    For the case of the current carrying wire, the magnetic field at a distance r from the wire can be found by Biot-Savart law and is given as [tex]\frac{I}{2\pi r}[/tex]. This itself can be derived by considering a particular setup, that of a current flowing upwards in a wire aligned along the z-axis towards z. By the Biot-Savart law:

    [tex]d\textbf{B} = \frac{\mu_0 Id\textbf{L} \times \textbf{r}}{4\pi r^3}[/tex].
    For this setup, [tex]d\textbf{L} = dz \textbf{k} \ \mbox{and} \ \textbf{r} = x\textbf{i} + y\textbf{j} + z\textbf{k}[/tex] and this makes the numerator of the dB expression [tex](x\textbf{j} - y\textbf{i})dz[/tex] by the cross product.

    So integrating the above from -infinity to +infinity along the z-axis of current flow gives:

    [tex]\frac{I\mu_0}{4\pi} \int_{-\infty}^{\infty} \frac{-y}{(x^2+y^2+z^2)^{3/2}} \textbf{i} + \int_{-\infty}^{\infty} \frac{x}{(x^2+y^2+z^2)^{3/2}} \textbf{j}[/tex]

    This reduces to the following after substituting for the limits of the integral expression:

    [tex]\frac{I\mu_0}{2\pi} \left( -\frac{y}{x^2+y^2} \textbf{i} + \frac{x}{x^2+y^2} \textbf{j}\right)[/tex].

    And the magnitude of the vector to be [tex]\frac{I\mu_0}{2\pi r}[/tex], where [tex]r=\sqrt{x^2+y^2}[/tex].

    So, with this one can prove Ampere's circuital law for the simple case of a current carrying wire. Starting with [tex]\oint \textbf{B} \cdot d\textbf{r}[/tex], note that the value of magnetic flux density does not change with the line integral around the wire at the same radius, so that means we can take B outside of the integral and the closed path integral reduces to the circumference of a circle about the wire. So this means that [tex]\textbf{B} \oint d\textbf{r} = B (2\pi r)[/tex]. And with the expression for B as derived above, [tex]\frac{I\mu_0}{2\pi r} 2 \pi r = I\mu_0[/tex].

    You can find a clearer and better derivation for this in any introductory physics textbook.
  7. Jul 25, 2008 #6


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    Take the curl of the integral in the B-S law.
    If done, carefully, this gives Maxwell's eqaution for the curl of B (for static fields).
    Applying Stokes' theorem to the curl of B gives Ampere's law.
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