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Why does moving electron produce magnetic field? 
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#1
Sep1407, 03:14 PM

P: 16

Why does moving electron produce magnetic field around it? Thank you.



#2
Sep1407, 07:41 PM

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P: 21,871

A moving charge produces a current, or rather a time varying electric field in space.
The subject was also discussed here http://www.physicsforums.com/showthread.php?t=57712 


#3
May2011, 03:43 PM

P: 1

i think what he's asking in lamon's terms is how the photons/laybons bounce off eachother to create a gravitational field, and i was wondering the same thing. is it because moving through space/time gives the particle new posatively charged particles to react with when it looses them?
thanx 


#4
May2011, 04:20 PM

P: 842

Why does moving electron produce magnetic field?
The magnetic field is a relativistic correction to the electrostatic field. 


#5
May2011, 06:42 PM

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#6
May2011, 10:06 PM

P: 969

I like formula (1539): [tex]B=\frac{v \times E}{c^2} [/tex] as it shows how weak the magnetic field is compared to the electric field. They're using SI units, so for meaningful comparison, you'd have to multiply the magnetic field by c: [tex]B=\frac{v \times E}{c} [/tex] To detect the magnetic field of a single charge is even worse, since the Lorentz force has a u/c x B, so you'd get a proportionality to uv/c^{2}. I read somewhere that this asymmetry between electricity and magnetism is the result of symmetry breaking at low energy, but that makes no sense to me. Even if the particle is traveling at light speed, although the magnitudes of the two fields is about the same, that's just about all they have in common. What makes more sense to me is this: of course an electric charge/current will favor electric fields over magnetic fields. A magnetic charge/current, if they existed, would favor magnetic fields over electric fields. 


#7
May2111, 03:18 AM

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I had read that the magnetic field is causes by length contraction of the particle and it associated electric field. Is that true? 


#8
May2311, 11:09 AM

P: 842




#9
May2311, 01:28 PM

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#10
May2311, 01:48 PM

P: 842

The only relativistic derivation of the magnetic field I have ever seen was dependent on a positive charge as part of the system. 


#11
May2311, 08:52 PM

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#12
May2311, 09:04 PM

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#13
May2411, 04:05 PM

P: 262

As far as the Einstein OEMB paper is concerned I can only find a part where he Lorenz transforms existing magnetic fields in his chapter:
As for the site mentioned in post #5 my maths is not up to scratch but if it refers to equations 1512 – 1515 : then as usual it refers back to existing magnetic fields. However it would be nice for me and others if someone could simplify and see how this works for a straight conductor, ie can someone work out B parallel with a conductor and also B at a right angle with this conductor. (with his site in mind). I still would like to see the transformation of a electrical field into a magnetic one. I have never ever seen a satisfactory solution. 


#14
May2411, 04:22 PM

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I don't understand your underlined "existing" emphasis. If you have no EM field in one frame then you have no EM field in any frame. You have to have something existing in one frame in order to transform it into another frame. In this case you have a purely electric field in one frame and in all other frames you will have a mixed electric and magnetic field. But you cannot have no EM fields in one frame and some EM fields in another. 


#15
May2411, 04:46 PM

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#16
May2511, 01:05 PM

P: 842

I found the writing that I last read on the relative correction to the electrostatic field. It depends on the fact that both positive and negative charges be present. Perhaps this is now out of date information. Here it is though, very easy to understand explanation.
http://galileo.phys.virginia.edu/cla...el_el_mag.html 


#17
May2511, 05:27 PM

P: 262

@ RedX. Ok put B=0 into 1515 that’s exactly what I’m after! The copy/past doesn’t work here so I’ll write out: B’ perpendicular = gamma x V x E/c^2. Lets work out gamma x V/c^2. This is: (V/c^2+V^3/2c^4+…..) now it is customary to ignore higher powers of c^2 (but if you really want you can keep them in). The result is: B=  V x E/c^2 + …. And this to me is a bit of a revelation ie: a magnetic field can be caused by a travelling E field, where V is subject (in a first degree) to the every day Galilean relativity, that is relativity without Lorenz contraction. Also this formula shows me the absence of the 2 in the denominator which you would get when working out B, using Lorentz contraction. In short: this formula makes more sense to me then any other explanation. 


#18
May2511, 06:46 PM

P: 842

So I am lost. Did someone just determine that pure relative motion causes a charge to be accelerated perpendicular to the direction of travel?



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