\nabla \times \vec{B}=\left(\frac{1}{r} \frac{\partial B_{z}}{\partial \phi}-\frac{\partial B_{\phi}}{\partial z}\right) \mathbf{i}_{r}+\left(\frac{\partial B_{r}}{\partial z}-\frac{\partial B_{z}}{\partial r}\right) \mathbf{i}_{\phi}+\frac{1}{r}\left(\frac{\partial\left(r B_{\phi}\right)}{\partial r}-\frac{\partial B_{r}}{\partial \phi}\right) \mathbf{i}_{z} \nabla \times \vec{B}=\frac{1}{r} \frac{\partial\left(r B_{\phi}\right)}{\partial r} \mathbf{i}_{z} O Use the above result and Ampere's Law, deltax B = µoJ, to show that the current density is given by \vec{J}=\left(\frac{1}{r}-1\right) e^{-r}{\mathbf{i}_{z}}

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