I know that electron has spin and momentum. Momentum is $vm$ and $v $ is $\dfrac {dx,y,z}{dt} $. So how it could be possible to affect the particle the way it will change only direction, instead of speed? Constant magnetic field does not does work, but when particle moves relative to constant magnetic field it becomes variable, and variable induces an electric field, that should change energy of the particle
1 Answer
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How can magnetic field change direction, do not changing the velocity of particle?
Basically: by acting perpendicularly to the velocity. The force exerted by a magnetic field $\mathbf B$ on a particle of charge $q$ moving at velocity $\mathbf v$ is given by the Lorentz force, $$ \mathbf F = q\mathbf v \times \mathbf B, $$ and the work that this delivers to the particle over a displacement $\mathrm d \mathbf s = \mathbf v \, \mathrm dt$ is therefore $$ \mathrm dW = \mathbf F \cdot \mathrm d\mathbf s = (q\mathbf v \times \mathbf B) \cdot \mathbf v \, \mathrm dt = 0, $$ because the cross product of $\mathbf v$ with any other vector is always orthogonal to $\mathbf v$ itself.
On the other hand, some of your other comments require a more careful look:
when particle moves relative to constant magnetic field it becomes variable
This isn't true. If the magnetic field is constant then it will be constant regardless of what speed you're travelling at. Nevertheless, this
and [a] variable [magnetic field] induces an electric field
does get at a correct fact: even a constant magnetic field, when seen from a moving frame of reference, transforms into an electric field (plus a magnetic field, of course). However, when you do that frame transformation to the rest frame of the particle, you get an electric field but the particle is at rest, so that the work $\mathrm dW = \mathbf F \cdot \mathrm d\mathbf s = q\mathbf E \cdot \mathbf v \, \mathrm dt$ is still zero, because in the rest frame $\mathbf v =0$. (And of course if you do a partial transformation, then the explanation involves both factors.)
answered Sep 4, 2018 at 17:01
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$\begingroup$ Don't understand the last sentence: when particle is in the rest relative to magnet field, magnetic field becomes an electric field? Even if we talk about an electric field (without a magnetic field) You are wrong, because then $F=k\dfrac {q1*q2}{r^{2}} $ (edited for formatting -DZ) $\endgroup$– user205628Commented Sep 4, 2018 at 19:12
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$\begingroup$ The Coulomb field is irrelevant - the field picture is local, i.e. we only care what the local field is, not how it was made. The transformation of magnetic fields to electric fields is a relativistic effect; see e.g. Wikipedia or an advanced-enough EM textbook for details. If you start with a homogeneous magnetic field and you Lorentz-transform in a direction orthogonal to the field, you get a homogeneous electric field (which doesn't have the Coulomb form). This is standard and uncontroversial. $\endgroup$ Commented Sep 4, 2018 at 19:20
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$\begingroup$ I thought contracted electric field (in constantly moving particle frame) actually is that called constant magnetic field. No? $\endgroup$– user205628Commented Sep 4, 2018 at 20:10
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$\begingroup$ That last comment makes absolutely no sense to me - I have no idea what you're saying. $\endgroup$ Commented Sep 4, 2018 at 20:14
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$\begingroup$ You've linked me to page, there is an illustration of moving charge field. As You can see when charge moves the field lines contract. But there is a magnetic field lines to, that confusing, because I thought that the contracted electric field is actually what we call a magnetic field. You understand me? $\endgroup$– user205628Commented Sep 4, 2018 at 20:31