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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?