From 9d15d37bed3f2d96017ffa373b0d52df26f7903a Mon Sep 17 00:00:00 2001 From: Preston Pan Date: Thu, 25 Jan 2024 16:46:22 -0800 Subject: add code block in website --- mindmap/magnetostatics.org | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) (limited to 'mindmap/magnetostatics.org') diff --git a/mindmap/magnetostatics.org b/mindmap/magnetostatics.org index ea763e7..dd9d172 100644 --- a/mindmap/magnetostatics.org +++ b/mindmap/magnetostatics.org @@ -56,7 +56,7 @@ Due to the [[id:2a543b79-33a0-4bc8-bd1c-e4d693666aba][inverse square]] law, we k \begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} \end{align*} -The first directional derivative is zero because $\vec{J}$ does not depend on the same coordinates as $\vec{\nabla}}$ +The first directional derivative is zero because $\vec{J}$ does not depend on the same coordinates as $\vec{\nabla}$ with the same reasoning as for the divergence, so we have: \begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} -- cgit