Amedical cyclotron used in the production of medical isotopes accelerates protons to 6.5 mev. the magnetic field in the cyclotron is 1.2 t what is the diameter of the largest orbit, just before the protons exit the cyclotron?

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × = 1/2 × 1.672 × ×v²

v = 3.5 × m/s

so

radius will be

radius =   ........2

radius =    

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

The diameter of the largest orbit is 6.52m

Explanation:

First, we will use only magnitudes, as the problem is not asking for directions or any vectorial result.

Second, besides the given data (K=6.5 MeV, B=1.2 T), we have to know that the mass of a proton is 938.3 MeV/C^2, or 1.6726 x 10-27 kg, and that the charge of an electron in magnitude is q=1.6 x 10-19 C.

With this in mind, we use the Lorent'z force:

where q is the electron's charge, v is the velocity of the particles and B is the magnetic field. We also use the definition of force in a circular movement:

where m is the mass of the particle and r is the radius. Then, by equalizing both expressions we can clear the radius r

and as we can see, we have that data except for the velocity, wich we can calculate using that

then

which is the velocity of the protons, and which we replace in our previous result

Finally, to calculate the diameter, we multiply r times 2.