Cyclotron magnetic field questions

[meaghan]

Homework Statement

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H− (the hydrideion, which has one proton and two electrons) to an energy of 5MeV to 20MeV. This ion has a mass very close to that of a proton because the electron mass is negligible−about 1/2000 of the proton’s mass. A typical magnetic field in such cyclotrons is 2.3 T

A)What is the speed of a 5.0-MeV H−?

B)If the H− has energy 5.0MeV and B= 2.3 T , what is the radius of this ion’s circular orbit?

Homework Equations

F = qVxB
F = mv^2/r
W= qxV
F = ma
K=1/2mv^2

The Attempt at a Solution

so for part a, i was thinking to find the kinetic energy using 5 MeV = 1/2mv^2 but i'd use the conversion first to go from eV to J
for part b, i was thinking of setting qVxB = mv^2/r

 

[drvrm]

so for part a, i was thinking to find the kinetic energy using 5 MeV = 1/2mv^2 but i'd use the conversion first to go from eV to J
for part b, i was thinking of setting qVxB = mv^2/r

where lies your problem?

 

[meaghan]

Not sure sure how to solve it or if I'm approaching it correctly

 

[berkeman]

Homework Statement

Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H− (the hydrideion, which has one proton and two electrons) to an energy of 5MeV to 20MeV. This ion has a mass very close to that of a proton because the electron mass is negligible−about 1/2000 of the proton’s mass. A typical magnetic field in such cyclotrons is 2.3 T

A)What is the speed of a 5.0-MeV H−?

B)If the H− has energy 5.0MeV and B= 2.3 T , what is the radius of this ion’s circular orbit?

Homework Equations

F = qVxB
F = mv^2/r
W= qxV
F = ma
K=1/2mv^2

The Attempt at a Solution

so for part a, i was thinking to find the kinetic energy using 5 MeV = 1/2mv^2 but i'd use the conversion first to go from eV to J
for part b, i was thinking of setting qVxB = mv^2/r

Your approach is correct. The only thing to watch out for is to be sure to check the velocity that you get against the speed of light. As long as it is less than a percent or two of c, then you don't have to make any corrections. If it is a significant fraction of c, you will need to use a relativistic correction to get the actual speed. Makes sense?

 

[meaghan]

yes it does! Relativity didn't factor into this at all since we haven't learned about it yet :)