# B Vector potential of an infinitely long cylinder

1. May 12, 2018

### 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.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. May 13, 2018

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.

3. May 15, 2018

### Somali_Physicist

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

4. May 18, 2018

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

Last edited: May 19, 2018
5. May 18, 2018

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 !! :)