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 = 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
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Nov 20th, 2015
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