Timeline for Electromagnetism problem: where does the magnetic field come from?
Current License: CC BY-SA 3.0
20 events
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| Jan 11, 2018 at 23:02 | comment | added | Voulkos | @Michael Seifert I think that the last plus "+" in the rhs of your equation must be minus "-" $$ \frac{d}{dt}\left[ \int_{S(t)} {\bf E}\cdot dS\right]=\int_{S(t)}{\frac{\partial}{\partial t}\bf E}\cdot dS+\int_{S(t)}{(\nabla \cdot \bf E)} {\bf v} \cdot dS \boldsymbol{-} \int_{\partial S(t)} ({\bf v} \times {\bf E}) \cdot dl $$ if we keep the right hand rule of the orentiation of $\:\partial S\:$ with respect to the normal of the surface $\:S$. | |
| S Jan 11, 2018 at 22:02 | history | edited | Kyle Kanos | CC BY-SA 3.0 |
fix typo and add word to meet 6-character requirement
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| S Jan 11, 2018 at 22:02 | history | suggested | Ben51 | CC BY-SA 3.0 |
fix typo and add word to meet 6-character requirement
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| Jan 11, 2018 at 21:41 | review | Suggested edits | |||
| S Jan 11, 2018 at 22:02 | |||||
| Jan 11, 2018 at 21:03 | comment | added | Adrian | @freecharly Thank you so much for your addenda! Really helpful. | |
| Jan 11, 2018 at 20:49 | history | edited | freecharly | CC BY-SA 3.0 |
deleted 4 characters in body
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| Jan 11, 2018 at 20:27 | history | edited | freecharly | CC BY-SA 3.0 |
minor correction
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| Jan 11, 2018 at 20:21 | comment | added | freecharly | I added in my answer below a simple mathematical derivation of the surface integral on the right hand side of the correct 4th Maxwell equation for this problem. In contrast to Anton Fetisov, I don't believe that there is an electrical current flowing in wire circuit, which would lead to an additional magnetic field. | |
| Jan 11, 2018 at 20:15 | history | edited | freecharly | CC BY-SA 3.0 |
Added simple mathematical derivation
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| Jan 11, 2018 at 14:33 | history | edited | freecharly | CC BY-SA 3.0 |
Removed redundend equation and minor text addition
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| Jan 11, 2018 at 14:27 | comment | added | freecharly | I have edited and extended the text of my answer due to the surprising analysis of Anton Fetisov who showed that in this special case due to the induced charges in the wire, the correct right hand side of 4th Maxwell equation (1) for time varying integration surfaces/boundaries gives the same results as the incorrect right hand side, equation (2). | |
| Jan 11, 2018 at 4:41 | history | edited | freecharly | CC BY-SA 3.0 |
Edited and expanded answer following the answer of Anton Fetisow
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| Jan 9, 2018 at 23:37 | comment | added | freecharly | I do no see any reason why this answer should be incorrect. In the inertial system of the charged plane you have a vertical electric field constant in space and time and in the loop with the sliding wire there is no force (electric or magnetic) on the charge carriers in the direction of the wire and thus no current and no magnetic field. Therefore you have, indeed, in the considered coordinate frame a constant electric field in time and space and $\frac{\partial E}{\partial t}=0$. For time dependent surface/contour, the 4th Maxwell equation must have the time derivative inside the integral! | |
| Jan 9, 2018 at 19:33 | comment | added | Anton Fetisov | Actually no, he didn't do a mistake. You are assuming that the total electric field is unchanged, which is impossible given the changing magnetic field given a changing current. | |
| Jan 9, 2018 at 18:52 | comment | added | Voulkos | @Michael Seifert I meet the Liebniz's rule with the name "Helmholtz transport theorem". See my answer as user82794 (former 'diracpaul') therein : Why the induced field is ignored in Faraday's law?. | |
| Jan 9, 2018 at 18:39 | comment | added | hyportnex | @Michael_Seifert please write this out as a full answer for posterity! | |
| Jan 9, 2018 at 16:42 | comment | added | freecharly | @Michael Seifert - You describe it very clearly! A similar error is often made with the integral form of the Faraday-Maxwell equation and time varying surface (Induction Law). | |
| Jan 9, 2018 at 16:38 | comment | added | Michael Seifert | If I've translated the differential-forms version of Liebniz's rule correctly, then the correct version of the above statement is$$\frac{d}{dt}\left[ \int_{S(t)} {\bf E}\cdot dS\right]=\int_{S(t)}{\frac{\partial}{\partial t}\bf E}\cdot dS+\int_{S(t)}{(\nabla \cdot \bf E)} {\bf v} \cdot dS + \int_{\partial S(t)} ({\bf v} \times {\bf E}) \cdot dl.$$The second integral vanishes, but the third one is non-vanishing, and cancels out the error. | |
| Jan 9, 2018 at 16:16 | comment | added | Michael Seifert | Exactly. Just to clarify, the error is that it is only true that$$\int_{S} {\frac{\partial}{\partial t}\bf E}\cdot dS = \frac{d}{dt} \left[ \int_{S} {\bf E}\cdot dS\right]$$ if the surface $S$ does not change with respect to time. It's possible to use the Liebniz integral rule for differential forms to relate these two quantities, but there's a couple of terms that your professor forgot. I suspect including them would resolve the paradox. | |
| Jan 9, 2018 at 16:03 | history | answered | freecharly | CC BY-SA 3.0 |