Asked by David
In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 8.4% of the speed of light while moving in a circular path of radius 0.551 m. What magnitude of magnetic force is required to maintain the deuteron in a circular path?
Answers
Answered by
bobpursley
centripetal force= magnetic force
m v^2/r=Bqv
B= m v/qr
at .084 c, I would first work it ignoring relativistic changes in mass, then rework it considering them.
mass m is not in u units, but in kg.
m v^2/r=Bqv
B= m v/qr
at .084 c, I would first work it ignoring relativistic changes in mass, then rework it considering them.
mass m is not in u units, but in kg.
Answered by
LOLLLL
OK SO F=MA
M= 2.00U --> kg --> 3.32107773 × 10^-27 kg
A= v^2/r
v=8.4% speed light, => (299792458)*8.4/100...
Plug into F=MA
M= 2.00U --> kg --> 3.32107773 × 10^-27 kg
A= v^2/r
v=8.4% speed light, => (299792458)*8.4/100...
Plug into F=MA
Answered by
Kris
T=Mg[(v^2/rg)-1]
or
T=[(3.32107773×10^-27)*g][(((299792458)*8.4)^2)/(g*0.551 m))-1]
or
T=[(3.32107773×10^-27)*g][(((299792458)*8.4)^2)/(g*0.551 m))-1]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.