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I have some confusion regarding the magnetic force. I know that the magnetic field created by a moving charge or current EXERTS a force on any moving charge or current that is present in the field. But when trying to understand the motion of a charged particle in an uniform magnetic field, the youtube video I saw explained it like this: "The magnetic force on a charged particle ALWAYS points perpendicularly with respect to the velocity and magnetic field. Whenever a force acts on an object perpendicular to its motion, the object will undergo circular motion--this creates a centripetal acceleration)"

I am confused. Is the charged particle exerting a force on itself? Or what is the force that acts on the charged particle that is moving? If the charge particle creates a force due to the magnetic field, is the force it creates itself the force that makes it undergo a circular motion?

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As far as I understand your question you are asking whether a moving charge (creating a magnetic field because of its motion) can exert a force on itself. The answer is a clear no. Think about a conductor, which caries a current. You know that the magnetic field around the conductor is circular. That means the magnetic field is tangential to the surface of the conductor. Due to cylindrical symmetry no Lorentz force acts on the moving charges. Of course the created magnetic field may act on another moving particle but that is another issue.

As Andy have already said the force that acts on a moving charge in a magnetic field is Lorentz force, which is given by $\vec{F}=q(\vec{v}\times \vec{B})$, where $\times$ indicates a cross product. You see that the Lorentz force is always perpendicular to the velocity of the particle because a cross-product is an operator (call it a thing if you like), which “outputs” a vector, which is both perpendicular to $\vec{v}$ and $\vec{B}$ in this case. Recap: $$\vec{c}=\vec{a}\times \vec{b} \implies \vec{c} \perp \vec{a} \quad \& \quad \vec{c} \perp \vec{b}$$ (The direction of the vector is determined by the right-hand rule.) Now since the force is always perpendicular to the velocity vector, the charged and moving particle starts to move in circle$\mathrm{s}^\dagger$ (or in helices in more complex scenarios).

Assuming that the particle moves in circles you can equate the Lorentz force and the centripetal force, which arises from a circular motion and has nothing whatsoever to do with the magnetic field. For more information about centripetal force see Wikipedia.

If you want to see a more mathematical derivation and explanation, I can edit my answer accordingly.

$^\dagger$There are some additional conditions for this fact but I discarded them because they would clutter the argument.

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  • $\begingroup$ I'm finding it very hard to explain my confusion. I know of the equation F⃗ =q(v⃗ ×B⃗ ). What I'm confused about is how to interpret that. Does the charged particle CREATE/EXERT? a force equal to q(vxB), or would a force that is being exerted on a charged particle have a force equal to q(v x B)? For example, in my book it gives the example that if a charged particle is moving parallel to the magnetic field vector, the force is 0. So it almost seems like the force ARISES based on the fact that a field and a velocity are in some position with a certain angle between them $\endgroup$ – FrostyStraw Mar 15 '15 at 22:29
  • $\begingroup$ which doesn't sound like "the force exerted on the charged particle is 0" but sounds more like "the force created by the charged particle based on where its velocity and magnetic field vectors are in relation to each other is 0" @gonenc $\endgroup$ – FrostyStraw Mar 15 '15 at 22:29
  • $\begingroup$ so I guess I'm confused about who is exerting a force, and who the force is being exerted on $\endgroup$ – FrostyStraw Mar 15 '15 at 22:31
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    $\begingroup$ So when using or talking about that equation, the assumption is that there is alreay an existing magnetic field? So the B in q(v x B) refers to the magnetic field that charged particle is moving in? I feel like it would be helpful if my book had explicitly said this, unless this is some common sense assumption that doesn't need to be stated. @gonenc $\endgroup$ – FrostyStraw Mar 15 '15 at 22:47
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    $\begingroup$ @FrostyStraw Exactly as you've said it! The $B$ in the formula is an “external field“ ie the field in which the charged particle is moving. $\endgroup$ – Gonenc Mar 15 '15 at 22:53
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The charged particle is not exerting a force on itself. The force that acts on the charged particle is created by the magnetic field. By the Lorentz force law, ignoring the effects of the electric field, $F = q(v × B)$, where $v$ is the velocity and $B$ the magnetic field. The cross product of two vectors is perpendicular to both vectors, so the force produced by the magneto field is perpendicular to the velocity, which makes the particle undergo circular motion. In short, the magnetic field produces the force which makes the particle undergo circular motion.

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    $\begingroup$ But what produced the magnetic field? I thought it was charges that produced magnetic fields. So is this just...an invisible charge that already had created a magnetic field, and the charge in question is placed in that field? @Andy $\endgroup$ – FrostyStraw Mar 15 '15 at 19:16
  • $\begingroup$ You are right, charges produce magnetic fields. The charge in question is placed on a magnetic field produced by other charges. $\endgroup$ – user70720 Mar 15 '15 at 19:19
  • $\begingroup$ Okay. But does the charge placed in the magnetic field also produce its own magnetic field/exert a force? And does that affect the magnetic field it's in? @Andy $\endgroup$ – FrostyStraw Mar 15 '15 at 20:20
  • $\begingroup$ It creates a magnetic field, its own magnetic field, but the net force is from the magnetic field from other objects. If you put in a second charged object it would experience the magnetic field from the machine, as well as the magnetic field produced by the first particle. $\endgroup$ – Faraz Masroor Mar 15 '15 at 22:44
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Check this out:

http://iweb.tntech.edu/murdock/ph4610/MagForcesN3.pdf

This explains that Newton's 3rd Law does not hold for two infinite, straight wires exerting Biot-Savart forces on each other, but once they are considered closed wires -as they should be in real world-, Newton is right again.

A second side-effect -and not less important- is that no closed loop with stationary current can exert a force onto itself.

I hope it's useful.

Cheers

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Electrons, protons, neutrons, positrons, ... have magnetic moments. If such a particle is located inside a magnetic field this field will align the particles magnetic moment, by the way like every magnetic dipole will be aligned.

Now, if the electron is moving into a magnetic field his magnetic dipole will be aligned too. But this alignment is accompanied by a gyroscopic effect. Gyroscopic effect means that every rotating body act again the force which try to align it. But the electron has not only a magnetic moment but he spins too. That is the reason why the electron will be distracted. Since any distraction leads to a photon emission and this emission is directed against the gyroscopic effect, the direction of the electrons magnetic moment Falls back - more or less - in his previous orientation. Now the game starts again. Meanwhile the electron due to the photon emission loses energy, the velocity slows down and the path of the electron is a spiral. If to be precise it is a spiral made from "tangerine slices".

Please note that magnetic fields interact and electric fields interact but there isn't interaction between an electric and a magnetic fields. Note that a moving neutron in a magnetic field will be distracted too.

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    $\begingroup$ PSA(because this has been flagged): Please do not flag answers you think are factually incorrect, but downvote them instead. Flags are meant for link-only answers and stuff that doesn't try to answer the question. If you think something is factually incorrect, leave a comment and/or downvote. $\endgroup$ – ACuriousMind Mar 15 '15 at 22:49

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