If the magnetic force doesn't change the speed of a moving charge can it be thought of as an elastic scattering? If magnetic force is a  force only on a moving charge that doesn't change its speed even though it changes its velocity can it be thought of as an elastic scattering from a circular wall created by the field? I am asking this because for me is confusing how a perpendicular force cannot change the magnitude of the particle velocity if the Pithagorean theorem states that it should be like so.(The angles of this scattering should be very small to mimic a circle.)

 A: A magnetic field aligns the magnetic dipole of the charge. The charge emits a photon with two reactions:

*

*the photon momentum deflects the charge a little from its straight trajectory.

*the spin gets de-aligned again.

This repeats many times until the kinetic energy of the charge is exhausted by the photon emissions and the charge comes to rest in the center of a spiral.
So you are right about the polygonal motion (being in a spiral path).
A: I believe the question is much simpler that what current answers try to answer. There seems to be a simple confusion regarding the Lorentz force. As such the force has an effect on the acceleration not the velocity.
$$m\, \vec{a}(t) = \vec{F}_{Magnetic} = q \vec{v}(t) \times \vec{B}$$
This being said if you follow the vector algebra the acceleration will always be point towards the center of the circle and will be always perpendicular to the velocity. Since vector components can be thought of as independent you can see that having a perpendicular acceleration cannot change magnitude of the velocity vector, only its direction. The pythagorean theorem has nothing to say in this regard, since you are not adding vectors here, you could perhaps look at the problem with the time component being discrete and you will see how at every step in time the velocity direction is corrected but not it's length
$$\vec{v}(t+\Delta t) = \vec{v}(t) + \Delta t\, \vec{a}  $$
$$|\vec{v}(t+\Delta t)|^2 = |\vec{v}(t)|^2 + \Delta t^2 |\vec{a}|^2$$
where I have used the fact that acceleration is perpendicular to velocity to get the last expression. Now as you can see, once you take the limit of $\Delta t\rightarrow 0$ the acceleration part falls off extremely quickly and your vector indeed keeps its norm.
