Timeline for Why is curl of current density $\nabla \times \vec{J}$ equal zero?
Current License: CC BY-SA 3.0
11 events
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| Oct 29, 2017 at 19:22 | comment | added | Kyle Kanos | @jvriesem: If you think it wrong or missing more details, feel free to write your own answer. As it stands, I'm quite satisfied with the post in terms of what it says and that it answers OPs question (albeit in an indirect manner). | |
| Oct 29, 2017 at 19:19 | comment | added | jvriesem | It is NOT true to say that $\hat{r}/r^2$ may be taken out of the integral in general. The reason for this is that the actual quantity is $(\vec{r}-\vec{r}')/\|\vec{r}-\vec{r}'\|$, which is most certainly a function of $\vec{r}'$. The case in which you could take it out of the integral is if you were integrating in a circular loop (the typical example application). In this case, you could treat the vector integral as a scalar integral, because $\|\vec{r}-\vec{r}'\|$ and the cross product are fixed. Wikipedia fixed its eqn. since your post. Jackson's E&M has it right and provides discussion. | |
| Oct 29, 2017 at 19:10 | comment | added | jvriesem | This is either a question of either notation or assumption vs. generality. @GDumphart's version is more correct in general. The introductory texts that introduce the Biot-Savart law probably write it that way because they consider it implicit that $\vec{r}$ is actually $\vec{r}-\vec{r}'$, where $r$ is the point at which $\vec{B}(\vec{r})$ is sampled, and $\vec{r}'$ is a point in the integral being integrated over. In my opinion, this inconsistent use of $\vec{r}$ and $\vec{r}'$ muddies their notation and causes confusion. Check out Jackson's E&M. Wikipedia fixed it since your post. | |
| Apr 29, 2016 at 18:33 | history | edited | user36790 | CC BY-SA 3.0 |
added 50 characters in body
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| Oct 14, 2014 at 15:38 | comment | added | Kyle Kanos | @GDumphart: The wikipedia link (it was there originally, but not active somehow until I just fixed it) shows this is a valid Biot-Savart law, as do all 3 of my E&M texts. You can take $\hat r/r^2$ out of the integral (taking care of minus signs), there is nothing wrong with doing that. | |
| Oct 14, 2014 at 15:36 | history | edited | Kyle Kanos | CC BY-SA 3.0 |
corrected link
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| Oct 14, 2014 at 15:31 | comment | added | GDumphart | I am still suspicious about your initial formula being a valid Biot-Savart without using $(\boldsymbol r-\boldsymbol{r}') / \|\boldsymbol r-\boldsymbol{r}'\|^3$ in the integrand. Your way, one could even pull the $\hat{r}/\boldsymbol r^2$ out of the integral, rendering the distance of a current $\boldsymbol J( \boldsymbol r')$ to the magnetic field at $\boldsymbol r$ irrelevant. That most likely doesn't affect the correctness of all the other steps though. | |
| Oct 14, 2014 at 15:16 | comment | added | Kyle Kanos | @GDumphart: $\hat r$ is the unit vector, some authors choose to bold this, $\hat{\mathbf r}$, but I often neglect that as the hat signifies the unit vector. Since $\hat r=\mathbf r/\vert\vert\mathbf r\vert\vert$, then $\hat r/r^2\equiv\mathbf r/\vert\mathbf r\vert^3$. | |
| Oct 14, 2014 at 14:55 | review | Suggested edits | |||
| Oct 14, 2014 at 15:08 | |||||
| Oct 14, 2014 at 14:37 | history | edited | Kyle Kanos | CC BY-SA 3.0 |
forgot the parenthesis.
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| Oct 14, 2014 at 14:32 | history | answered | Kyle Kanos | CC BY-SA 3.0 |