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Doubt with Ampere and Biot-Savart

  1. Aug 30, 2012 #1
    Hello,

    I think I'm terribly wrong by supossing Ampère-Maxwell and Biot-Savart are referred to the same concept of magnetic field B. For example, for calculating B near an infinite line, I used both, as I understood them, obtaining different expressions (see image). What is that that I don't get?

    Thanks
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2012 #2
    You didn't do the Biot-Savart integral right. Let the wire go up the z-axis, and the integral is

    [tex]B(\rho) = \int_{-\infty}^\infty \frac{\mu_0}{4\pi} \frac{I e_z \; dz' \times (\rho e_\rho - z' e_z)}{(\rho^2 + {z'}^2)^{3/2}}[/tex]

    You treated the denominator like it doesn't depend on [itex]z'[/itex], but it does. I write it [itex]z'[/itex] to emphasize that it is the integration variable (not the position we want to find the magnetic field at). Consult a table of integrals to easily find the antiderivative.

    In general, for some current density [itex]j[/itex], the Biot-Savart Law is

    [tex]B(r) = \int_{V'} \mu_0 j(r') \; dV' \times \frac{r - r'}{4 \pi |r - r'|^3}[/tex]

    Actual line currents (not densities) just reduce this integral from a volume to a line. Compare with the electric field from some charge density:

    [tex]E(r) = \int_{V'} \frac{\rho(r')}{\epsilon_0} \; dV' \frac{r - r'}{4 \pi |r - r'|^3}[/tex]

    for vectors [itex]r, r'[/itex]. You can see these are both really the "same" law. The function [itex](r-r')/4\pi|r-r'|^3[/itex] has special significance in 3D space. Wiki "Green's functions" if you're interested in learning about it.
     
  4. Aug 30, 2012 #3
    Thank you very much, that was really helpful, I appreciate it!

    Carlos
     
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