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I Uniform current in cylinder and straight wire: same field?

  1. Apr 28, 2016 #1
    Hello, friends! The tridimensional version of the Biot-Savart law says that the magnetic field generated at the point ##\boldsymbol{r}\in\mathbb{R}^3## by a tridimensional distribution of current defined by the current density ##\boldsymbol{J}## is$$\boldsymbol{B}(\boldsymbol{r})=\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x}) \times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x$$where ##V\subset\mathbb{R}^3## is the tridimensional region where the current is distributed.

    I intuitively suspect that, if ##V## is a cylinder of height ##h=b-a## and radius ##R##, whose symmetry axis is parallel to the unit vector ##\mathbf{k}##, flown through by a current having constant density ##\boldsymbol{J}\equiv J\mathbf{k}## on ##V## (and ##\boldsymbol{J}\equiv\mathbf{0}## elsewhere), this last expression of ##\boldsymbol{B}(\boldsymbol{r})##, with ##\boldsymbol{r}\in\mathbb{R}^3\setminus V##, equates that of a magnetic field generated, according to the linear version of the Biot-Savart law, by the current flowing through a straight wire, carrying the same current ##\pi R^2 J##, placed where the cylinder's axis is, i.e. $$\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x}) \times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d^3x=\frac{\mu_0}{4\pi}\int_a^b\frac{\pi R^2 J \mathbf{k}\times(\boldsymbol{r}-z\mathbf{k})}{\|\boldsymbol{r}-z\mathbf{k}\|^3}dz.$$Is that so, either for a cylinder of finite (##a,b\in\mathbb{R}##) or infinite (##a=-\infty##, ##b=+\infty##) length? If it is, how can it be proved?

    I have tried to explicitly use cylindrical coordinates, so that the field can be expressed as$$\boldsymbol{B}(\boldsymbol{r}) =\frac{\mu_0}{4\pi}\int_a^b\int_0^{2\pi}\int_0^R\frac{J\mathbf{k} \times(\boldsymbol{r}-(\rho\cos\theta\mathbf{i}+\rho\sin\theta\mathbf{j}+z\mathbf{k}))}{\|\boldsymbol{r}-(\rho\cos\theta\mathbf{i}+\rho\sin\theta\mathbf{j}+z\mathbf{k})\|^3}\rho\, d\rho d\theta dz$$ but I am not able to handle the denominator. I do not want to use Ampère's law because all the proofs I have seen (like this) use ##\boldsymbol{J}\in C^2(\mathbb{R}^3)## while here we have a discontinuous density.

    I ##\infty##-ly thank any answerer!
     
  2. jcsd
  3. Apr 28, 2016 #2

    Charles Link

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    With an infinite length, Ampere's law says the two are identical, but only outside the radius of the cylinder. I think I can come up with a counter example that shows the result is not valid in general. Consider a thin cylinder so that it is a disc. For the wire, it becomes a moving point charge and the cylinder is a moving disc of charge of the same total charge. Any point outside the cylinder should work if your equation is valid. I think you could easily show that points just outside the disc near the center (near the point charge, in front of the disc) clearly have a stronger magnetic field from the point charge than from the distribution of charge.
     
    Last edited: Apr 28, 2016
  4. Apr 28, 2016 #3

    Simon Bridge

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    The question is how to go about proving that the linear form of the Biot Savart law gives the same result as the general form for r>R for a current density J confined within radius R.

    That amounts to answering: does the magnetic field outside a wire behave as if all the current were concentrated at the center?
    Personally I'd demonstrate the result from the differential form of Maxwell's equations.
     
  5. Apr 28, 2016 #4

    Charles Link

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    When the observation point is outside the current distribution, the same differential equations ## \nabla \cdot B=0 ## and ## \nabla \times B=0 ## are obeyed for the ## B ## field everywhere outside the distribution. I think the solutions for the wire and the cylinder need to match everywhere outside the distribution or in all likelihood they don't. If you can find a point outside the distribution where the solutions do not match up, in general, the solutions simply will not match up. Outside of the cylinder, (e.g. in front of it), there is nothing that readily distinguishes between points r>R and points r<R. If it doesn't hold for one of these points, I'd be rather skeptical that the result holds for all r>R.
     
    Last edited: Apr 28, 2016
  6. Apr 28, 2016 #5

    Simon Bridge

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    Except you cannot have a finite cylinder of current.
     
  7. Apr 29, 2016 #6
    I thank you very much for your answers!
    I hoped that it can be proved just by using the tools of integration (and symmetry considerations), without using Ampère's law.
    That is because all the proofs I have seen of the law use differentiable current densities ##\boldsymbol{J}:\mathbb{R}^3\to\mathbb{R}^3##, in fact (although all the proofs of the entailment of Ampère's law by the Biot-Savart law that I have found in books and on line texts nonchalantly commute derivatives and integrals without explaining why that is allowed -I know that commutations between derivative/integral signs are not always permissible, in general- and without even explaining whether the integrals are intended to be Riemann or Lebesgue integrals or notations for linear forms) I think I have been able to prove (here) the entailment to myself, but under the assumption that ##\boldsymbol{J}\in C_c^2(\mathbb{R}^3)##, while here ##\boldsymbol{J}##, constantly ##J\mathbf{k}## on ##V## and constantly null elsewhere, is not differentiable.
     
  8. Apr 29, 2016 #7

    Charles Link

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    I have to believe that what you are proving is likely to require complete cylindrical symmetry including uniformity in the z-direction. The electrical field problem with a point charge requires spherical symmetry (although there is no requirement on the distribution as a function of "r").
     
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