Need help to calculate this physics about kinetic energy

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A proton with a kinetic energy of 4.9*10^-16J movesw perpendicularto a magnetic field of .26T. What is the radius of its circularpath?

this answer is on the physics homework help as #20 physics jameswalker chap 22, but i just don't get it. Could somebody leave me avery detailed explanation but make it easy for me to understand,thank you.
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Nov 20th, 2015

Thank you for the opportunity to help you with your question!

kinetic energy K = mv2/2
mass of a proton m = 1.67*10-27 kg
so 4.9*10-16 J = 1.67*10-27 kg *v2/2
v = √(2*4.9*10-16/1.67*10-27) =7.66*105 m/s
magnetic force F = qvB
Newton 2nd law: F = ma = mv2/r
so qvB = mv2/r
qB = mv/r
qBr = mv
∴r = mv/(qB) =1.67*10-27*7.66*105/(1.6*10-19*0.26)= 3.08*10-2 m = 3.08 cm

Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 20th, 2015

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