# Finding magnetic field of highly symmetric systems by Ampere's Law

Recall how we find the Electric field of spherically symmetric charge distribution, cylindrically symmetric charge distributions and also charge distributions with planar symmetry (infinite flat sheet), by Gauss's Law and arguements based on symmetry alone. This make me think that we can do the same in magnetism by using Ampere's law and symmetry alone. But I failed. So I would like to know if I have used the wrong method of arguments.

So take an infinite straight long current carrying wire as the simplest example. WLOG, let it be lying on the $z$ axis with current flowing in the positive $z$ direction. So using cylindrical coordinate system $$\vec{B}(\rho,\phi,z)=B_\rho(\rho,\phi,z)\hat{\rho}+B_{\phi}(\rho,\phi,z)\hat{\phi}+B_z(\rho,\phi,z)\hat{z}$$

Now by translational symmetry along the wire, we conclude that nothing can have $z$-dependence. And by rotational symmetry about $z$ axis, nothing can have $\phi$ dependence. Hence $$\vec{B}(\rho)=B_\rho(\rho)\hat{\rho}+B_{\phi}(\rho)\hat{\phi}+B_z(\rho)\hat{z}$$

Now if you flip the current upside down, $B_{\rho}(\rho)\hat{\rho}$ is unchanged. While it should have become $-B_{\rho}(\rho)\hat{\rho}$ because superimposing the flipped current and the original current gives zero current and therefore zero field.