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B Vector potential of an infinitely long cylinder

  1. May 12, 2018 #1
    I have seen the other threads on an infinitely long wires vector potential.Its obvious that really small wires are just infinitely long cylinders:

    ∇xA=B
    ∫∇xA.da=∫B.da
    ∫A.dl = ∫B.da = φ(flux)
    For an infinite cylinder
    A.2πri=B.2πrih
    A=Bh
    A=μ0*I*h/(2π*r)

    Now for a cylinder of radius limr->0 => μ0I/2π which im sure is wrong (interesting multiply this by r and u get B field of a wire)

    Obviously for infinitely long wire:
    A = (μ0/I2π)(log(Λρ+√1+Λ2ρ2))

    Any help?
     
  2. jcsd
  3. May 13, 2018 #2

    Charles Link

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    Your equation using ## B ## to try to generate ## A ## is somewhat clever, but your path of integration is inconsistent with the area over which you took the flux. By Stokes theorem, the path of ## \oint A \cdot dl ## needs to be around the region of the flux of ## B ##. ## \\ ## The equation you need is of the form ## \vec{A}(x)=\int \frac{\mu_o}{4 \pi} \frac{\vec{J}(x')}{|x-x'| } \, d^3 x' ##. For ## \vec{J}=J_z \hat{z} ##, ## \vec{A} ## will only have a z-component.
     
  4. May 15, 2018 #3
    Hmm , so which path of integration would i take? or is that not possible?
     
  5. May 18, 2018 #4

    vanhees71

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    [corrected typos indicated in #5]

    This is a very tricky issue, because the infinite wire is a physical idealization, and you have to use a more clever formula to evaluate the potential (there's no such problem for ##\vec{B}##, which can be calculated with the standard Biot-Savart formulat). The reason is that the naive integral for ##\vec{A}## of course diverges for and infinitely long wire. In this case, it's much easier to directly integrate the differential equation for ##\vec{A}## with the ansatz
    $$\vec{A}=\vec{e}_z A(R),$$
    where I use standard cylinder coordinates ##(R,\varphi,z)##. Using the standard formulae for curl in cylinder coordinates you get
    $$\vec{B}=\vec{\nabla} \times \vec{A} = -A'(R) \vec{e}_{\varphi}, \quad \vec{\nabla} \times \vec{B}=-\frac{1}{R} \left (R A'(R) \right )' \stackrel{!}{=} \mu_0 \vec{j}.$$
    With
    $$\vec{j}=\frac{I}{\pi a^2} \Theta(a-R) \vec{e}_z=j_0 \Theta(a-R) \vec{e}_z$$
    you get by simple integration (for ##R<a##)
    $$A_{<}=-\frac{\mu_0 j_0}{4} R^2,$$
    where I used the fact that the potential is regular at ##R=0##. For ##R>a## you get
    $$A_{>}=C_1-C_2 \ln \left (\frac{R}{a} \right).$$
    Continuity demands ##C_1=-\mu_0 j_0 a^2/4##. The magnetic field is
    $$\vec{B}_<=\frac{\mu_0 j_0 R}{2} \vec{e}_{\varphi}, \quad \vec{B}_>=\frac{C_2}{R} \vec{e}_{\varphi}.$$
    Since there are no surface currents ##\vec{B}## must be continuous at ##R=a##, i.e.,
    $$\frac{C_2}{a}=\frac{\mu_0 j_0 a}{2} \; \Rightarrow \; C_2=\frac{\mu_0 I}{2 \pi},$$
    and thus
    $$\vec{B}_{<}=\frac{\mu_0 I R}{2 \pi a^2} \vec{e}_{\varphi}, \quad \vec{B}_{>}=\frac{\mu_0 I}{2 \pi R} \vec{e}_{\varphi}.$$
     
    Last edited: May 19, 2018
  6. May 18, 2018 #5

    Charles Link

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    One correction, if my algebra/calculus is correct when I looked up the curl in cylindrical coordinates: The curl of ## A ## generates a ## \phi ## component of ## B ##, (##B_{\phi}=- A'(R) ##), and when taking the curl of that, the term of interest, (the z component of the curl of ## B ##), has ## \frac{1}{R} ( \frac{\partial{( R \, B_{\phi})} }{\partial{R}}) ## so that it should be ##( \nabla \times \vec{B})_z=-\frac{1}{R}(R \, A'(R))' =\mu_o j_o ##. ## \\ ## (This is apparently a "typo" because the subsequent solutions for ## A(R) ## are correct, and they are the solutions that are obtained with this correction). ## \\ ## And one additional minor "typo": ## C_1=-\mu_o j_o a^2/4 ##, (where it should be ## j_o ## rather than ## r_u ##). ## \\ ## A very interesting solution. Thank you @vanhees71 !! :)
     
    Last edited: May 18, 2018
  7. May 19, 2018 #6

    vanhees71

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    Indeed! Thanks. I corrected the typos in the posting.
     
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