Vector potential of an infinitely long cylinder

[Somali_Physicist]

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.2πr_i=B.2πr_ih
A=Bh
A=μ₀*I*h/(2π*r)

Now for a cylinder of radius lim_r->0 => μ₀I/2π which I am 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?

 

[Charles Link]

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.

 

[Somali_Physicist]

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.

Hmm , so which path of integration would i take? or is that not possible?

 

[vanhees71]

[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
$$\vec{B}_{<}=\frac{\mu_0 I R}{2 \pi a^2} \vec{e}_{\varphi}, \quad \vec{B}_{>}=\frac{\mu_0 I}{2 \pi R} \vec{e}_{\varphi}.$$

 

[Charles Link]

∇×→B=−1R(1RA′(r))′!=μ0→j.B→=∇→×A→=−A′(R)e→φ,∇→×B→=−1R(1RA′(r))′=!μ0j→.​

\vec{B}=\vec{\nabla} \times \vec{A} = -A'(R) \vec{e}_{\varphi}, \quad \vec{\nabla} \times \vec{B}=-\frac{1}{R} \left (\frac{1}{R} A'(r) \right )' \stackrel{!}{=} \mu_0 \vec{j}.

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 ! :)

 

[vanhees71]

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