If I have an infinite cylinder with radius $R$ and a surface density of current that varies with time, I can use ampere's law to calculate the magnetic field. My question is why is possible to ignore the displacement current when calculating the magnetic field, in this case?
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It's been a long time since my electromagnetism course but could it be by the symmetry of the problem? In an infinte cylinder with surface density $\vec{\sigma}=\sigma(t) \vec{e}_{\rho}$, the electric field $\vec{E}$ should be perpendicular to the outer surface of the cylinder i.e $\vec{E}\sim \vec{e}_{\rho}$ with $E_{\varphi}=E_{z}=0$ and the magnetic field goes parallel to the cylinder axis i.e $\vec{B}\sim \vec{e}_z$ with $B_{\varphi}=B_{\rho}=0$. Then if we suppose that $\vec{B}$ is homogeneous in the azimutal coordinate $\partial_\varphi B_z=0$ which makes totally sense you get $(\nabla \times \vec{B})_{\rho}=0$.
vin92
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