Answer to Question #343410 in Electricity and Magnetism for Cecille

Question #343410

a proton moving horizontally at 3.7 x 10 to the fifth power ms and thirst and a uniform magnetic field of 0.08 e60 directed upward at 35 degrees with the horizontal find the magnitude of the magnetic force acting on the proton

1
Expert's answer
2022-05-22T14:38:40-0400
"F=qvB\\sin\\theta\\\\\n=1.6*10^{-19}*3.7*10^5*0.08*\\sin 35^\\circ\\\\\n=2.7*10^{-15}\\:\\rm N"


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