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I Magnetic field by infinite wire: convergence of integral

  1. Aug 13, 2016 #1
    Let ##\boldsymbol{l}:\mathbb{R}\to\mathbb{R}^3## be the piecewise smooth parametrization of an infinitely long curve ##\gamma\subset\mathbb{R}^3##. Let us define $$\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0 I}{4\pi}\int_\gamma\frac{d\boldsymbol{l}\times(\boldsymbol{x}-\boldsymbol{l})}{\|\boldsymbol{x}-\boldsymbol{l}\|^3}=\frac{\mu_0 I}{4\pi}\int_{-\infty}^{+\infty}\frac{\boldsymbol{l}'(t)\times(\boldsymbol{x}-\boldsymbol{l}(t))}{\|\boldsymbol{x}-\boldsymbol{l}(t)\|^3}dt.$$A physical interpretation of the integral is that ##\boldsymbol{B}## represents the magnetic field generated by an infinitely long wire ##\gamma## carrying a current ##I##, whith ##\mu_0## representing vacuum permeabilty.

    Can we be sure that the integral converges in general and, if we can, how can it be proved?

    I am posting here rather than in calculus because I suppose that the best way to approach the problem is by using the techniques of Lebesgue integration.

    I notice that every component of the integral is the difference of two terms having the form ##l_i'(t)(x_j-l_i(t))\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}##, and I see that ##|l_i'(t)(x_j-l_i(t))|\|\boldsymbol{x}-\boldsymbol{l}(t)\|^{-3}\le |l_i'(t)||x_j-l_i(t)|^{-2}##, but the absolute value does not allow me to use the rule ##l_i'(t)dt=dl_i##...

    Thank you so much for any answer!
     
    Last edited: Aug 13, 2016
  2. jcsd
  3. Aug 13, 2016 #2

    Orodruin

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    No, it is relatively easy to construct a counter-example. For example, consider the case when your curve just is a circular loop going round and round. A single turn of the loop will result in a finite magnetic field, but you have an infinite number of loops and the integral therefore does not converge.
     
  4. Aug 19, 2016 #3
    Then, when we state laws like Ampère's ##\oint\boldsymbol{B}\cdot d\boldsymbol{x}=\mu_0 I_{\text{enclosed}}## with ##\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0 I}{4\pi}\int_\gamma\frac{d\boldsymbol{l}\times(\boldsymbol{x}-\boldsymbol{l})}{\|\boldsymbol{x}-\boldsymbol{l}\|^3}## I suppose we impose other, stricter, assumption on ##\gamma## than its piecewise smoothness. What other conditions?
    I heartily thank you and anybody wishing to add for your answer(s)!
     
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