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We all know that when the velocity vector is perpendicular to the magnetic field the particle moves in a circular path. But since the particle is moving it is also creating its own magnetic field. This field is always opposite to the initial magnetic field that we applied. This should change the net magnetic field and hence, the flux through the circular path should also change. This means that the path of the particle should also change because the net magnetic field has changed. But it is taught to us that it doesn't.

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We all know that when the velocity vector is perpendicular to the magnetic field the particle moves in a circular path. But since the particle is moving it is also creating its own magnetic field.

That is true.

This field is always opposite to the initial magnetic field that we applied.

This is not generally true. It can't be, because direction of magnetic field of the moving particle depends on its charge; positive charge will create magnetic field opposite to negative charge.

This should change the net magnetic field and hence, the flux through the circular path should also change.

That is true.

This means that the path of the particle should also change because the net magnetic field has changed.

No, this does not follow from the previous statements. Just because total field changes, it does not mean that force on the particle has to change. However, in some cases the charged body indeed has different path than the ideal one predicted by the external field; the difference is often negligible, so textbooks do not expose this part of theory.

Particle's path is determined by its initial state and net force that acts on it during its motion. Net force acting on a charged particle in external electromagnetic field is usually given with very good accuracy by the formula:

$$ \mathbf F = q\mathbf E_{ext} + q\mathbf v\times \mathbf B_{ext}. $$ Notice the index $ext$: only external field is used when this equation is applied in practice. This simplification is successful in low energy experiments and devices, such as CRT tubes. The effect of the field of the particles on their motion is ignored.

There are cases where this equation is not accurate, though. For example, motion of electron bunches (little clouds made of billions of electrons) in cyclotrons shows that the bunches in circular motion are being damped down, so the above equation is not entirely accurate when applied to a single bunch.

The obvious reason is that charged constituents of the bunch produce their own EM fields that act on the other charged constituents and net result is that the bunch is heavier than sum of masses of the parts and it is being slowed down. So electron bunch would not circle in magnetic field indefinitely, but would grow in size and slow down. This effect is undesirable, hence the accelerators are built to periodically apply accelerating and volume shaping electric force on the bunches.

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  • $\begingroup$ How to calculate the magnetic field for these electrons in the accelerator near c? $\endgroup$ – HolgerFiedler Jun 30 '18 at 15:51
  • $\begingroup$ Calculate which magnetic field? That of the electrons or that of the accelerator? $\endgroup$ – Ján Lalinský Jun 30 '18 at 17:28
  • $\begingroup$ Of the electron $\endgroup$ – HolgerFiedler Jun 30 '18 at 18:52
  • $\begingroup$ In the answer I assumed that electron's field is the well-known retarded solution of the Maxwell equations with source term corresponding to single charged point particle. Different particles have different source terms, so they produce different fields. Field relevant for force acting on particle 1 due to particle 2 is determined by trajectory of the particle 2 in the past. $\endgroup$ – Ján Lalinský Jun 30 '18 at 19:14
  • $\begingroup$ I simply want to know what is the strength of the magnetic field of an electron on relativistic speed. $\endgroup$ – HolgerFiedler Jun 30 '18 at 22:28
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Everything you say is correct except the part about particle feeling its own field. It does feel its own field and this is called self interaction. It's effect is already incorporated into the particle's mass and charge. If there are many particles then the effect of their fields on each other is opposing the external field, so diamagnetic.

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... when the velocity vector is perpendicular to the magnetic field the particle moves in a circular path.

Not in a circular but in a spiral path. The particle emits periodically photons and this reduces its kinetic energy. So the best description is a spiral path made of tangerine slices.

But since the particle is moving it is also creating its own magnetic field. This field is always opposite to the initial magnetic field that we applied.

I take the liberty of having a dissenting opinion here. The magnetic dipole moment of an electron is its intrinsic property, the related magnetic field is existing independently from any surrounding circumstances. And this magnetic moment is aligned under the influence of the external magnetic field. In the two cases that the direction of movement is parallel to the external field or that the particle is not in relative motion to the field nothing happens after the alignment. In the opposite case the electron gets deflected due to the emission of photons and this disalign the electrons magnetic dipole again et voilà the game starts again until the exhausted particle comes to rest in the centre of the spiral.

This should change the net magnetic field and hence, the flux through the circular path should also change. This means that the path of the particle should also change because the net magnetic field has changed. But it is taught to us that it doesn't.

The external magnetic field works like a spring during the alignment (by this field) and the disalignment (from the photons momentum).

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