Timeline for Condition for the magnetic field
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2014 at 19:42 | comment | added | Ant | @Gigi10012 thank you for the effort but I know maxwell's equation, my question was not about that. | |
| Apr 8, 2014 at 19:41 | comment | added | Ant | @Jim Thank you very much! Now it's more clear! :-D | |
| Apr 8, 2014 at 19:40 | vote | accept | Ant | ||
| Apr 8, 2014 at 19:29 | comment | added | user21420 | $\int E.dA = \frac{Q}{\epsilon_0}$. and $\int B.dA = \frac{Q_{magnet}}{k}$ but $Q_{magnet}$ doesn't exists so it's zero. but $\oint B.dl = \mu_0(I+ \epsilon_0 \frac{\partial\Phi_E}{\partial t})$ Watch this video for Maxwell's equations derivation youtube.com/watch?v=AWI70HXrbG0 (when current is flowing through wire then $B = \frac{\mu_0I}{2\pi r}$) and $\nabla . B = 0$ means that the sum of magnetic field inside of sphere is zero | |
| Apr 8, 2014 at 19:22 | comment | added | user21420 | $E = \frac{q}{4\pi\epsilon_0 r^2}$ and It's not zero when charge isn't zero. and $\nabla \times E$ is curl operator (cross product of $\nabla$ and $E$) example of curl operator: $\overrightarrow F = y^2\hat i + x^2\hat j$ then $\nabla\times \overrightarrow F= \begin{vmatrix} \hat i & \hat j & \hat k \\ {\partial \over \partial x} & {\partial \over \partial y} & {\partial \over \partial z} \\ y^2 & x^2 & 0 \\ \end{vmatrix} = \hat k \begin{vmatrix} {\partial \over \partial x} & {\partial \over \partial y}\\ y^2 & x^2 \end{vmatrix} = 2(x-y)\hat k$. | |
| Apr 8, 2014 at 17:22 | history | edited | Jim | CC BY-SA 3.0 |
added 1019 characters in body
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| Apr 8, 2014 at 17:10 | comment | added | Ant | I see. Could you please expand your answer and include why exactly we can simplify without concern and set $B = 0$ and why it has physical meaning? | |
| Apr 8, 2014 at 17:07 | comment | added | Jim | If the three conditions you mentioned are met, then $B$ is a constant in space and time. Because it cannot have an effect on any charges (which would change the $E$ field and thus, change$B$ making it not a constant), we can effectively re-scale the problem and set $B=0$ without any concern of consequences. It is a simplification tactic | |
| Apr 8, 2014 at 17:03 | comment | added | Ant | Okay but how can you get to that conclusion? $B = [3; \ 2; \ 1]$ satisfy all the conditions; why do you say $B = 0$? | |
| Apr 8, 2014 at 16:58 | comment | added | Jim | if all three of those are zero, then yes, you can say $B=0$. But it is a fairly uninteresting situation. | |
| Apr 8, 2014 at 16:57 | comment | added | Ant | But if $\nabla \times E = 0$? | |
| Apr 8, 2014 at 16:55 | history | answered | Jim | CC BY-SA 3.0 |