Timeline for Describing a circular current loop as delta functions
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 1, 2014 at 12:41 | comment | added | user22180 | yeah. see the current I is in the direction $\hat{\phi}$, and $r\sin\theta d\phi$ is the component of $d\vec{l}$ along the direction $\hat{\phi}$. Now already you have understood that when the ring is in xy plane that $\sin\theta$ may be omitted. But if the ring is in a plane parallel to xy plane then you require this formula.then $\delta(\cos\theta)$ is replaced by $\delta(\cos\theta-\cos{\theta_0})$. | |
| Aug 1, 2014 at 6:26 | comment | added | Ben | So when I did my line integral $\int I \cdot dl$ I should should have evaluated it as $\int I a\sin\theta d\phi$ right? | |
| Aug 1, 2014 at 5:43 | vote | accept | Ben | ||
| Aug 1, 2014 at 5:43 | comment | added | Ben | Yeah you're right, I wasn't following. I was thinking the $\sin \theta$ was included when the azimuthal angle was held constant (i.e. I thought $dS_\phi=r\sin\theta dr d\theta$ ). I see what you were saying now. thank you | |
| Aug 1, 2014 at 5:04 | comment | added | user22180 | I think you couldn't follow my answer properly. Take time to read it properly. If it doesn't answer your question completely, point out exactly where you have the problem in my answer. | |
| Jul 31, 2014 at 17:19 | comment | added | Ben | I would agree that a $\sin \theta$ belongs in the formula if he only integrated over $d\theta$ not $\sin\theta d\theta$ | |
| Jul 31, 2014 at 16:58 | comment | added | Ben | Thank you for the reply. Yes, a $\sin\theta$ shows from Jacobian $\sin\theta d\theta$ But, jackson integrates over the full solid angle $d\Omega =\sin\theta d\theta d\phi$ ... which would give a $\sin^2\theta$ I'll edit the question to show his next step | |
| Jul 31, 2014 at 7:54 | history | answered | user22180 | CC BY-SA 3.0 |