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Jan 28, 2018 at 19:03 comment added Philip Wood @freecharly Great minds...
Jan 28, 2018 at 17:05 comment added freecharly @PhilipWood - I just saw that your counter-example (charges moving on a circle and their electric and magnetic field) is practically the same as I have given in an addition to my answer to fulfill the OP's condition of a "single source".
Jan 28, 2018 at 16:03 history edited Philip Wood CC BY-SA 3.0
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Jan 28, 2018 at 15:37 comment added freecharly @MahlomolaDanielCwele - A single counterexample disproves your assertion that "$\vec{E}\perp \vec{B}$ at all times"!
Jan 28, 2018 at 15:07 comment added Philip Wood I focused on what I could understand – the clear and unambiguous question in your first paragraph. [I note that the other answers do the same.]
Jan 28, 2018 at 14:52 comment added Mahlomola Daniel Cwele The key in your quote is "at all times". But, please do not focus on a single part of the question, as I went to some length to make sure that the question is as clear as possible.
Jan 28, 2018 at 14:38 comment added Philip Wood With respect, your question (as worded at the time I answered it) asked "can one argue that E⃗ ⊥B⃗ at all times?" A single exception to E⃗ ⊥B⃗ is enough to show that one $can't$ uphold the claim.
Jan 28, 2018 at 14:33 comment added Mahlomola Daniel Cwele It seems that you are interested in finding an isolated example. As the question clearly shows, the interest is in whether this kind of conclusion would hold IN GENERAL, and for ALL SUCH SOURCES, not just for an isolated example. in fact, I have already provided an example under which the magnetic and electric field are known to be perpendicular, thereby stressing that this not the problem with which I need help.
Jan 28, 2018 at 14:19 history edited Philip Wood CC BY-SA 3.0
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Jan 28, 2018 at 14:06 history answered Philip Wood CC BY-SA 3.0