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. will rate high

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kinetic energy K = mv^{2}/2 mass of a proton m = 1.67*10^{-27} kg so 4.9*10^{-16} J = 1.67*10^{-27} kg *v^{2}/2 v = √(2*4.9*10^{-16}/1.67*10^{-27}) =7.66*10^{5} m/s magnetic force F = qvB Newton 2nd law: F = ma = mv^{2}/r so qvB = mv^{2}/r qB = mv/r qBr = mv ∴r = mv/(qB) =1.67*10^{-27}*7.66*10^{5}/(1.6*10^{-19}*0.26)= 3.08*10^{-2} m = 3.08 cm

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