I'm supposed to find the vector potential $A(x)$
- the field generated by an infinite conducting wire of section $\pi a^2$, in which flows a constant current density $j=j \cdot \theta(a-r)\hat{z}$ where $r=\sqrt{x_1^2+x_2^2}$
I got kinda stuck on the integral so i decided to ask for some assistance here on MSE instead of PSE, thanks for understnading
So since $j_x,j_y$ are both 0 we have $A_x, A_y=0$
Since $\nabla \times B=\nabla \times(\nabla \times A) = \nabla(\nabla \cdot A)-\nabla^2 A=\mu_0J$ and by definition $\nabla \cdot A = 0$ $\rightarrow$ $\nabla^2 A = -\mu_0J$ which is a Poission's eequation with known solution:
$$A(z)=\frac{\mu_0}{4\pi}\int\frac{J(r')}{\eta}d\tau'=\frac{\mu_0}{4\pi}\int\frac{j\cdot \theta(a-r)\hat{z}}{r}d\tau'=\hat{z} \ \frac{\mu_0j\theta}{4\pi}\int \frac{a-r}{r}d\tau'$$
So here is where i'm a bit unsure.
Can i just use cylindrical coordinates and say:
$$A(r)=\hat{z} \ \frac{\mu_0j\theta}{4\pi}\int \frac{a-r}{r}rdzdrd\psi$$
So can i say $a=\pi r^2$ and substitue in and get:
$$A(r)=\hat{z} \ \frac{\mu_0j\theta}{4\pi}\int \frac{\pi r^2-r}{r}rdzdrd\psi=\hat{z} \ \frac{\mu_0j\theta}{4\pi}\int (\pi r -1)rdzdrd\psi$$
and $r$ is a constant and if i just evaluate the integral: $$A(r)=\hat{z} \ \frac{\mu_0j\theta 2z(r^2-1)}{2}$$
But this just feels wrong...
Appreciate any help
PS: i realise i'm using different notations through out....hopefully it's still clear..
Let $\vec J(\vec r)=\hat z j$ for $\rho <a$ and $\vec J(\vec r)=0$. Applying the Biot-Savart Law, we find that
$$\begin{align} \vec B(\vec r)&=\frac{\mu_0}{4\pi}\int_{-\infty}^\infty\int_0^{2\pi} \int_0^a \frac{j \hat z \times (\vec \rho-\vec \rho')}{(\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi')+(z-z')^2)^{3/2}}\,\rho'\,d\rho'\,d\phi'\,dz'\\\\ &=\frac{\mu_0}{4\pi}\int_{-\infty}^\infty\int_0^{2\pi} \int_0^a \frac{j (\hat \phi \rho-\hat \phi' \rho')}{(\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi')+(z-z')^2)^{3/2}}\,\rho'\,d\rho'\,d\phi'\,dz'\\\\ &=\frac{\mu_0}{4\pi}\int_0^{2\pi} \int_0^a \frac{2j (\hat \phi \rho-\hat \phi' \rho')}{\rho^2+\rho'^2-2\rho\rho'\cos(\phi-\phi')}\,\rho'\,d\rho'\,d\phi'\\\\ &=\hat \phi \frac{\mu_0 j}{2\pi}\int_0^{2\pi} \int_0^a \frac{\rho-\cos(\phi') \rho'}{\rho^2+\rho'^2-2\rho\rho'\cos(\phi')}\,\rho'\,d\rho'\,d\phi'\\\\ &=\hat \phi \frac{\mu_0 j}{2\pi} \int_0^{\min(\rho,a)} \frac{2\pi}{\rho}\rho'\,d\rho'\\\\ &=\begin{cases} \hat \phi \frac{\mu_0 j a^2}{2\rho}&, \rho >a\\\\ \hat \phi \frac{\mu_0 j \rho}{2}&, \rho <a \end{cases} \end{align}$$
which could have been obtained more efficiently by exploiting symmetry and applying Amperes' Law.