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Why does moving electron produce magnetic field around it? Thank you.
A moving electron alone actually does not produce a magnetic field. It requires electrons and protons to produce a magnetic field. As the electrons move relative to protons (ions) there is a relativistic charge per unit volume difference between the positive and the negative charges. This causes any external charges to feel a force we know as the magnetic field.Why does moving electron produce magnetic field around it? Thank you.
Actually, it does.A moving electron alone actually does not produce a magnetic field.
That was a pretty clever derivation. It's interesting that the electric field is strongest perpendicular to the moving charge, just like the case for acceleration where the radiated power goes as [tex]sin^2 \theta [/tex] (Larmor formula), and hence the field as [tex]E \; \alpha \; sin \theta [/tex].
Unfortunately, I can't make heads or tails of that.
It seems to me this article is based off the Maxwell equations, written as 4 vectors. I don't see how it gives a solution which shows an electron in an isolated system can give off a magnetic field.
Yes, a very good derivation. In words, a charge sitting at rest produces a Coulomb field. When you view it from a moving rest frame, the E field becomes partly a B field. That's why we call it Electromagnetism.http://farside.ph.utexas.edu/teachin...s/node125.html [Broken]
But then we'd have to call it... Magnetoelectrism!A magnetic charge/current, if they existed, would favor magnetic fields over electric fields.
"Becomes"? I don't see where this part is derived.When you view it from a moving rest frame, the E field becomes partly a B field.
There are at least two ways different from the website that you can use to derive the magnetic field of a constant-velocity point charge. One is to just use a Lorentz transform - just plug and chug. The second way is that most books already solve the Maxwell equations in terms of the sources, and there's an equation called the Feynman-Heaviside equation that tells you the field of a point particle - just set the acceleration to zero in those equations. The website does it yet a 3rd way, inserting the particle 4-velocity as way of introducing frame dependence. One possible point of confusion is the pre-factor [tex]\frac{1}{r'^3} [/tex] in front of the tensor. If r' is equal to say 5, then in the new frame that factor would still be numerically equal to 1/5^{3}, although written in terms of variables of the new frame: [tex]\frac{1}{r'(r,t)^3}[/tex] where [tex]r'=r'(r,t) [/tex] It's everything to the right of the pre-factor that produces the correct structure for tensor transformation. So if you tensor transformation that expression, don't replace [tex]\frac{1}{r'^3} [/tex] by [tex]\frac{1}{r^3} [/tex], which you could do only if the pre-factor were a scalar. But rather use [tex]r'=r'(r,t) [/tex]"Becomes"? I don't see where this part is derived.
The only relativistic derivation of the magnetic field I have ever seen was dependent on a positive charge as part of the system.
Einstein derived it in his original OEMB paper."Becomes"? I don't see where this part is derived.
I looked up the Feynman-Heaviside equation and as far as I can see it is also derived starting with existing electrical and magnetic fields.§ 6. Transformation of the Maxwell-Hertz
Equations for Empty Space. On the Nature
of the Electromotive Forces Occurring in a
Magnetic Field During Motion
Einstein derived the full expression for transforming arbitrary electromagnetic fields between reference frames. Then the link above took the known electric field (Coulomb's law - magnetic field = 0) in the frame where the charge is stationary and then simply calculated the magnetic field in a reference frame where the charge is moving.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:
I looked up the Feynman-Heaviside equation and as far as I can see it is also derived starting with existing electrical and magnetic fields.
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.
Just set B=0 in eqns 1512-1515. Then you can transform any purely electro-static problem (where B is equal to zero) into a electro- and magneto- static problem (where B is not equal to zero) .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...
I still would like to see the transformation of a electrical field into a magnetic one. I have never ever seen a satisfactory solution.
Hi DaleSpam can you give me the link to that paper? Yesterday I read one of his papers but either missed it or read the wrong paper.Einstein derived the full expression for transforming arbitrary electromagnetic fields between reference frames. Then the link above took the known electric field (Coulomb's law - magnetic field = 0) in the frame where the charge is stationary and then simply calculated the magnetic field in a reference frame where the charge is moving.
I did underline because I wanted to see how a purely electrical field transforms into a magnetic one.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
Here it is, see section 6.Hi DaleSpam can you give me the link to that paper? Yesterday I read one of his papers but either missed it or read the wrong paper.
[nitpick] I wouldn't call it a travelling E field. Fields don't have velocity. In one frame it is purely an E field, and in a frame which is travelling wrt that frame it is both an E and a B field. But there really isn't any sense in which the field itself has a velocity in either frame. [/nitpick]a magnetic field can be caused by a travelling E field
Well no not exactly. We have to work out E prime first. E’=gamma E, but that’s all the Lorentz contraction done. Then B= - (E’xV)/C^2.So I am lost. Did someone just determine that pure relative motion causes a charge to be accelerated perpendicular to the direction of travel?
All that does it provide a relativistic correction the magnetic field. It does not show that the magnetic field IS a relativistic correction to the electrostatic force.Well no not exactly. We have to work out E prime first. E’=gamma E, but that’s all the Lorentz contraction done. Then B= - (E’xV)/C^2.
Thanks @ Dale
Here it provides a Lorentz correction to the electric field. After this correction perhaps then we need to rethink this whole conversion from electric to magnetic field? No matter how hard you try its impossible to get a proper cross product just assuming further Lorentz corrections. But I’m open for suggestions.All that does it provide a relativistic correction the magnetic field. It does not show that the magnetic field IS a relativistic correction to the electrostatic force.
That is why I expected that if an electron in an isolated system it could not produce a magnetic field.Here it provides a Lorentz correction to the electric field. After this correction perhaps then we need to rethink this whole conversion from electric to magnetic field? No matter how hard you try its impossible to get a proper cross product just assuming further Lorentz corrections. But I’m open for suggestions.