Timeline for Line current density into a surface integral
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2020 at 7:02 | comment | added | Puk | I realize $\vec{K}$ is along $\hat{\varphi}$. The current across a small line element $d\vec{\ell}$ with a normal vector $\hat{n}$ within the surface is $dI=\vec{K}·\hat{n}d\ell$. With $d\vec{\ell}=\hat{z}d\ell$ and $\hat{n}=\hat{\varphi}$, this is $dI=\alpha d\ell$. | |
| Jun 14, 2020 at 6:39 | comment | added | Darkenin | First of all, thank you. I still don't understand from your answer when it is valid to multiply by $dz$ to get the current. The current flows in the $\hat{\varphi}$ direction in my case, not in the $\hat{z}$ direction. Can I still use it as $I =Kdz$ ? | |
| Jun 14, 2020 at 0:01 | history | edited | Puk | CC BY-SA 4.0 |
added 2 characters in body
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| Jun 13, 2020 at 22:48 | history | edited | Puk | CC BY-SA 4.0 |
added 334 characters in body
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| Jun 13, 2020 at 22:17 | history | answered | Puk | CC BY-SA 4.0 |