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Homework Statement
A proton moves in a circular path perpendicular to a 1.15T magnetic field. The radius of its path is 9.3mm. Calculate the energy of the proton in eV.
Homework Equations
F=ma=mv^2/r, F=qvB
The Attempt at a Solution
F=ma=mv^2/r=qvB
mv/r=qB
v=qBr/m=1.6*10^-19*1.15*0.0093=1022833
F=qvB=1.6*10^-19*1022833*1.15=1.882*10^-13
I don't know how to convert F to the energy...
Well, first of all, am I correct so far?