Question

Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV.

(a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Express your answer in Joules (the deuteron mass is 3.34×10−27kg)
b)Starting from a negligibly small velocity, how many full rotations does the deuteron need before it reaches this maximum energy?
c) What is the time it takes for the deuteron to make one complete rotation when its energy is about 500 keV and when it is about 5 MeV? Ignore possible relativistic effect

Answer 1

We need the equation of motion of a charged particle in a magnetic field which is

eq 1 Bev=mv2/r

Since the charged particle moves in a circular path in a magnetic field if its velocity is constant then the magnetic force is just equal to the centripetal force. We see the velocity is related to the radius so the maximum velocity and therefore energy is related to the maximum radius which is 2m. Knowing the velocity we find the kinetic energy (non-relativistic).

The number of revolutions is easy since we know the maximum energy and the increase in energy per rotation which is given (remember there are two impulses per full rotation).

The time for a revolution is also not difficult. For each revolution lets say the particle has a circular path of radius r and velocity v. Therefore the distance traveled for each revolution is

eq 2 distance per rev.=2πr

. and the time over a distance with a velocity v is

eq 3 time=distance/velocity

We know v from our first equation. Now if you can write the equation for time of revolution in the general case by combining equations 1,2,3 you will save some work on this third part.