No.
velcity=qBR/m
KE=1/2 m v^2= 1/2 (qBR)^2 m
You plan to build a large, cheap cyclotron using the Earth's magnetic fields(1x10^-4 T) and orbiting just above the Earth's atmosphere (radius 5.7 x10 ^6 m) What energy, in volts, should protons be given to just circle the earth? q= 1.6 x10^-19 C, m= 1.67x10^-27
1 have used the formula,
Kinetic energy of orbiting proton
= B^2q^2r^2/2m joules and then converted it into MeV. Am i right?
3 answers
I have simplified it as (B^2q^2r^2) / 2m
Well, lets do the algebra, I can make errors.
v=qBR/m\
v^2 = (qBR)^2 /m^2
1/2 mv^2=1/2.... you are correct.
v=qBR/m\
v^2 = (qBR)^2 /m^2
1/2 mv^2=1/2.... you are correct.