# Integrating current density for cylindrical symmetry

Considering a constant current density, $$\vec{J}=\frac{I}{A}\hat{k}$$, and also assuming there is no displacement current in the situation, how would one apply Maxwell-Ampere's law? I keep running into infinite field whenever I try this approach:

$$\oint_c \vec{B} \cdot \vec{dl} = \mu_0(\iint_S(\vec{J_e}+\epsilon_0 \frac{\partial \vec{E}}{\partial t})\cdot \vec{dA})$$ $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\iint_S\vec{J_e} \cdot \vec{dA}$$ $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\iint_S\hat{J_e}J_e \cdot dA\hat{J_e}$$ $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\iint J_e dA$$ $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\iint \frac{I}{\pi r^2} rdrd\theta$$ $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\iint \frac{I}{\pi r} drd\theta, 0 $$\oint_c \vec{B} \cdot \vec{dl} = \mu_0\int \frac{1}{\pi}ln(\frac{a}{0}) d\theta = diverges$$

• Is your your current density $\vec{J}$ meant to be zero for $r>a$ ? Nov 13, 2023 at 22:11

If I understood you correctly, your current density is $$\vec{J}=\begin{cases} \frac{I}{\pi a^2}\hat{k} &, \text{ for } 0a \end{cases}$$

The mistake is in the step where you replace the current density $$J_e$$ by $$\frac{I}{\pi r^2}$$. It should rather be $$\frac{I}{\pi a^2}$$.

\begin{align} \oint_C \vec{B}\cdot d\vec{l} &= ... \\ &= \mu_0 \iint J_e\ dA \\ &= \mu_0 \int_0^{2\pi}\int_0^a \frac{I}{\pi a^2}r\ dr\ d\theta \\ &= ... \\ &= \mu_0I \end{align}

• Thank you so much! This was exactly the issue :) Nov 13, 2023 at 23:00