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.
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?
3 answers
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
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]