- #1

Divergenxe

- 2

- 1

Suppose now we have a current density in cylindrical coordinates$$\vec{J}=J_{0}\hat{z}, \hspace{1 cm} 0 \leqslant \rho \leqslant r,$$

And the two relevant Maxwell's equations are:

$$\nabla \times \vec{B}=\mu_{0}\vec{J}\hspace{1 cm}[1]$$

$$\oint\limits_{\partial_\Sigma} \vec{B} \cdot d\vec{l}=\mu_{0}\iint\limits_{\Sigma}\vec{J}\cdot d\vec{S} \hspace{1 cm} [2]$$

Remarks: $$\frac{\partial \vec{E}}{\partial t}=0$$

From [1], I try to express the curl in cylindrical form:

$$(\frac{1}{\rho}\frac{\partial B_{z}}{\partial \phi}-\frac{\partial B_{\phi}}{\partial z})\hat{\rho}+(\frac{\partial B_{\rho}}{\partial z}-\frac{\partial B_{z}}{\partial \rho})\hat{\phi}+\frac{1}{\rho}(\frac{\partial (\rho B_{\phi})}{\partial \rho}-\frac{\partial B_{\rho}}{\partial \phi})\hat{z}=\mu_{0}J_{0}\hat{z}$$

How can I deduce the B Field only has a phi component from the above expression?

From [2], I try to interpret in this way:

suppose I set up a circular Amperian loop on the rho-phi plane, only the phi component of B Field contributes to the path integral on the left hand side, and the right hand size is the current, so I conclude the current induces B-Field with phi component only. But I hesitate immediately, what if the loop I set is inclined? I will immediately conclude the B-Field only has the inclined component using the previous argument, but this is not true! I must think something wrong. What do you think?