Magnetic field of a moving charged particle

AI Thread Summary
A moving charged particle generates a magnetic field that is observed by a stationary observer, who also perceives a changing electric field due to the particle's motion. The discussion emphasizes that the electric and magnetic fields are interrelated, with a changing magnetic field producing an electric field according to Faraday's law. It is clarified that there is not a separate electric field created by the moving magnetic field; rather, the observed fields are part of a single electromagnetic field. The distinction between the electric field of a stationary charged particle and that of a moving one is acknowledged, particularly in how they are perceived by an observer. Overall, the conversation highlights the complex interplay between electric and magnetic fields in the context of relative motion.
arul_k
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A moving charged particle produces a magnetic field around it, thus a stationary observer would notice a moving electric and magnetic field.

Does this moving magnetic field (as seen by the stationary observer) in turn produce another electric field?
 
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I think it's more accurate to look at it this way:

The particle has an EM field around it that is purely electrical as seen by an observer who is stationary with respect to the particle, and partly magnetic and partly electrical as seen by an observer in motion with respect to the particle. I don't think either of fields "create" a new field.

Someone correct me if I'm wrong.
 
You can't even create an electric field , it's already there , every charged particle has it as a fundamental entity.
All we do is play around with it and watch what happens , like two stationary particles one is charged and the other is not, it is the " observer" particle.
Now the observer particle sees an electric field from the charged particle , a stationary electric field with a fixed value at any given distance, which decreases in strength with distance.

Now if you start to move the charged particle , the observer particle sees the electric field lines from the charged particle to start to deflect so to say.
Now it sees a magnetic field.
Actually there is no magnetic field , there is only electric one , the magnetic field arises around the movement of charged particles.Or as distorted electric field lines due to different frames of reference which involves SR.
But if nothing would move and you could somehow keep the charged particles stationary there would be no magnetic field just electric one, as long as something moves doesn't matter if the observer or the charged particle , the one sees a magnetic field while the other sees an electric one.
 
Crazymechanic said:
Actually there is no magnetic field , there is only electric one , the magnetic field arises around the movement of charged particles.Or as distorted electric field lines due to different frames of reference which involves SR.

I'd be careful in saying there is no magnetic field. Think about the magnetic field of particles due to their intrinsic spin. There is no frame of reference where this magnetic field does not exist. At least not to my knowledge.
 
yes particles have this phenomenon due to spin but then again what is spin... movement.
Well ofcourse it's a fundamental phenomenon of matter , atleast from a classical point of view.Still without that movement you would just have an electric field possessing charged particle.
Even though a complicated matter , maybe better left out of the picture of this thread.
 
Crazymechanic said:
Actually there is no magnetic field , there is only electric one , the magnetic field arises around the movement of charged particles

This is incorrect. The electric field is not more fundamental than the magnetic field. In fact what is true is that there is a single (covariant) electromagnetic field.

Arul, if the magnetic field is time varying in the frame of the stationary observer then it will have assocaited with it a locally circulating electric field as per Faraday's law ##\nabla \times E = -\frac{1}{c}\partial_t B## in said frame.
 
Thanks Drakkith and Crazymechanic for your answers, but my question pertains to the magnetic field that is observed by the stationary observer.
For a moving charged particle a stationary observer will also observe a moving magnetic field, does this moving magnetic field give rise to an electric field apart from the particles original electric field.
 
WannabeNewton said:
This is incorrect. The electric field is not more fundamental than the magnetic field. In fact what is true is that there is a single (covariant) electromagnetic field.

Arul, if the magnetic field is time varying in the frame of the stationary observer then it will have assocaited with it a locally circulating electric field as per Faraday's law ##\nabla \times E = -\frac{1}{c}\partial_t B## in said frame.

Thanks WanabeNewton for your answer. If I understand correctly this locally circulating electric field is inependent of the charged particles electric field. If so, how would this locally circulating field interact with the charged particles original field?
 
It's not independent as the magnetic field the observer sees is caused by the traveling charged particle which has a certain charge and velocity as compared to the stationary observer , in other words charged particles are not independent of the respective fields nor are the fields free from the particles.

I'm sure wannabeNewton will elaborate on this whole thing.
 
  • #10
Crazymechanic said:
It's not independent as the magnetic field the observer sees is caused by the traveling charged particle which has a certain charge and velocity as compared to the stationary observer , in other words charged particles are not independent of the respective fields nor are the fields free from the particles.

I'm sure wannabeNewton will elaborate on this whole thing.


I know the field is not independent of the charged particle, what I asked is if the new circulating field is independent of the original electrical field
 
  • #11
A changing electric field produces a magnetic field and a changing magnetic field produces an electric field.

A charged particle moving at a constant speed produces a constant magnetic field. That constant magnetic field does NOT produce another electric field. If the charged particle were accelerating then it would produce a changing magnetic field which would produce another elecftric field. In particular, a periodically moving charged particle (as in harmonic motion) produces an electro-magnetic field.
 
  • #12
arul_k said:
A moving charged particle produces a magnetic field around it, thus a stationary observer would notice a moving electric and magnetic field.

Does this moving magnetic field (as seen by the stationary observer) in turn produce another electric field?
The complete field from an arbitrarily moving charged particle is given by the Lienard Wiechert potential:
http://en.wikipedia.org/wiki/Liénard–Wiechert_potential
 
  • #13
HallsofIvy said:
A changing electric field produces a magnetic field and a changing magnetic field produces an electric field.

A bar magnet moving with uniform velocity would produce an electric field around it, there is no accleration or changing magnetic field here, only relative motion of the magnetic field.

Similarly one one expect the moving magnetic field around the charged particle to produce an electric field.

DaleSpam, I read the wiki article, but could not find any direct reference to my question.
 
  • #14
The fields wouldn't interact. The charge itself creates a divergent part of the field and the time derivative of B will create a rotational part of the field. These two parts are generated separately, they should just add and produce an overall diverging and rotating field.
 
  • #15
mikeph already stated what I was going to state (sorry for the late reply) so at this point I just want to bring to note a phrase you keep using: " moving magnetic field". Perhaps you already know this but you keep using this phrase so let me just point out that the magnetic field does not move.
 
  • #16
arul_k said:
DaleSpam, I read the wiki article, but could not find any direct reference to my question.
You can use the formulas I provided to calculate the fields. Then you can directly see if the "moving fields" generate any electric fields.
 
  • #17
If I understand the question correctly, the OP seems to believe that the electric field produced by the varying magnetic field (as per Faraday law) is in addition to the electric field produced by the moving particle. That's incorrect. They are one and the same. The fields that appear in Faraday's equation are the total fields. There is nothing left out that must be added to it. How many different ways can I say that?
 
  • #18
WannabeNewton said:
mikeph already stated what I was going to state (sorry for the late reply) so at this point I just want to bring to note a phrase you keep using: " moving magnetic field". Perhaps you already know this but you keep using this phrase so let me just point out that the magnetic field does not move.

What I mean by "moving magnetic field" is the "magnetic field associated with a charged particle moving with uniform velocity wrt a stationary observer", it was just easier to use the term moving magnetic field.
 
  • #19
DaleSpam said:
You can use the formulas I provided to calculate the fields. Then you can directly see if the "moving fields" generate any electric fields.

Quite honestly, DaleSpam, the math is above me, I haven't studied advanced math. I would appreaciate it if you could tell me if the moving fields generate any electric fields.
 
  • #20
dauto said:
If I understand the question correctly, the OP seems to believe that the electric field produced by the varying magnetic field (as per Faraday law) is in addition to the electric field produced by the moving particle. That's incorrect. They are one and the same. The fields that appear in Faraday's equation are the total fields. There is nothing left out that must be added to it. How many different ways can I say that?

Thanks everyone for the answers.

I understand that there is only one electric field for a moving charged particle, but to put it differently, is there any difference between the static electric field of a charged particle and the electric field of a moving charged particle (wrt to a stationary observer) at non relativistic speeds?

Dauto, what do you mean b the term "total fields"
 
  • #21
if a stationary observer observes a stationary charged particle with respect to the observer then all he sees is an electric field.

If the stationary observer observes a charged particle that is in motion with respect to the observer (say passing by the observer) then the observer sees a magnetic field , while the charged particle still has an electric field in it's own frame of reference.
 
  • #22
arul_k said:
I would appreaciate it if you could tell me if the moving fields generate any electric fields.
Unfortunately, I don't understand what you mean well enough to answer directly. I don't know what you mean by a moving magnetic field, nor a moving electric field, nor another electric field. All I can say is that the equation I posted follows all of Maxwells equations and represents the total field from a moving charge.
 
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  • #23
arul_k said:
What I mean by "moving magnetic field" is the "magnetic field associated with a charged particle moving with uniform velocity wrt a stationary observer", it was just easier to use the term moving magnetic field.

I explained what I ment by moving magnetic field in post #18.
 
  • #24
Since the Lienard Wiechert fields are the fields associated with a charged particle with arbitrary motion, then according to your definition the "moving magnetic field" would be the Lienard Weichert magnetic field for a uniform velocity (i.e. set ##\dot{\beta}=0##).

However, given your definition the answer to your OP is tautologically "no" regardless of the actual fields.

A moving charged particle produces some magnetic field, B, and some electric field, E. You have defined the term "moving magnetic field" to refer to B, and the term "moving electric field" to refer to E. Regardless of what influence B may have on E, the total field produced by the moving charge is E. So, by definition only the moving electric field, E, is present, not another electric field (i.e. if any additional field were present then it would be part of E, by definition).
 
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  • #25
arul_k said:
I understand that there is only one electric field for a moving charged particle, but to put it differently, is there any difference between the static electric field of a charged particle and the electric field of a moving charged particle (wrt to a stationary observer) at non relativistic speeds?

Any answer to the question in post #20
 
  • #27
DaleSpam said:
Certainly.

OOps...sorry, allow me to rephrase...

As a kind request, if its not to much trouble, may I please have the answer to the question in post #20. :biggrin:
 
  • #28
yes arul there is a difference that is seen by a stationary observer when a charged particle flies right by or just stands still at a fixed distance.
The difference is that in a fixed point the observer sees a electric field and in the moving charged particle case it sees a magnetic field.
I and many others already said that I don't know why you didn't notice that.
 
  • #29
arul_k said:
OOps...sorry, allow me to rephrase...

As a kind request, if its not to much trouble, may I please have the answer to the question in post #20. :biggrin:
That was the answer: There certainly is a difference.
 
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  • #30
Crazymechanic said:
yes arul there is a difference that is seen by a stationary observer when a charged particle flies right by or just stands still at a fixed distance.
The difference is that in a fixed point the observer sees a electric field and in the moving charged particle case it sees a magnetic field.
I and many others already said that I don't know why you didn't notice that.

If you would be kind enough to read the question carefully you would notice that I have already mentioned in the very first post that a moving charged particle has a magnetic field associated with it in addition to its electric field. I never questioned the fact of the appearance of the magnetic field. The answer you have been giving me is already stated in the question.
 
  • #31
Or maybe your just not getting the answer provided to you.

There is no magnetic field of a moving particle in it's own frame of reference if I even can call it so.
The magnetic field is only to a observer which moves at a different speed than that of the moving particle or is stationary with respect to it.

there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.
 
  • #32
Crazymechanic said:
there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.
There is a difference between the electric field of a static charge and a moving charge. Again, I point to the Lienard Wiechert fields:
http://en.wikipedia.org/wiki/Liénar...onding_values_of_electric_and_magnetic_fields

In these equations the velocity as a fraction of the speed of light is given by ##\beta##. Thus, the case of a non-accelerating charge is given by ##\dot{\beta}=0## while the case of a stationary particle is given by ##\beta=0##. Even if you don't follow the meaning of the equation you can clearly see that it is a function of ##\beta## which means that the field depends on the velocity.
 
  • #33
Crazymechanic said:
there is no difference from a static electric field to that of a particle because the field of a single charged particle is fixed in value and is static if that is what you wanted to know.

Without even looking at the solution, this clearly cannot be true. Faraday's law and Gauss's law read ##\vec{\nabla} \cdot \vec{E} = 4\pi \rho## and ##\vec{\nabla} \times \vec{E} = -\frac{1}{c}\partial_t \vec{B}## so clearly the solution for an arbitrarily accelerating charge (time-varying delta function source) is in general different from the solution for the special case of a static charge, which simply has ##\vec{\nabla} \cdot \vec{E} = 4\pi Q\delta^3 (\vec{r} - \vec{r}')## and ##\vec{\nabla} \times \vec{E} = 0## where ##\vec{r}'## fixed.
 
  • #34
sorry wrong button .:)
 
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