Is the Magnitude of Proton's Acceleration Constant?

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SUMMARY

The discussion centers on the acceleration of a proton, defined by the equation a = CvxB, where a is the acceleration vector, v is the velocity vector, B is the magnetic field vector, and C is a constant. It is established that while the direction of the velocity changes due to the magnetic field, the magnitude of both the velocity and acceleration remains constant. The key conclusion is that the scalar product of acceleration and velocity is zero, indicating that no work is done on the proton, thus preserving its kinetic energy.

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A proton has the acceleration a=CvxB, where a is the cross product of v and B (a,v,B are vectors). C is a constant and the magnetic field is constant. Prove that the magnitude of the v and a are constant.

I know I have to somehow prove the derivative of a or v to be zero, but i have no idea how to start. i know if v is constant, then a must be 0.

a=CvxB
(da/dt)=C(dv/dt)xB

i also wrote down |a|=|Cv||B|sintheta, but that doesn't help at all.
 
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If the acceleration is the vector product of the velocity and something else, the acceleration is at right angles to the velocity. The scalar product of acceleration and velocity must then be zero. (The scalar product of two vectors is the product of their magnitudes and the cosine of the angle between them. Cosine of 90 degrees is zero.) But this scalar product is the power per kilogram (work per kilogram per second) delivered by the force.

The derivative to time of the velocity is not zero; it is the acceleration. But it merely changes the direction of the velocity in this case, not its magnitude. So the kinetic energy does not change, and no work is done.
 
solved it, thanks!
 
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