Asked by cr
A proton is moving in a circular orbit of radius 11.8 cm in a uniform magnetic field of magnitude 0.232 T directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. Answer in units of m/s.
I came up with .118/2.83x10^-7 and it was incorrect.
I came up with .118/2.83x10^-7 and it was incorrect.
Answers
Answered by
bobpursley
Of course it is incorrect. Do you think an electron can possibly travel that slow speed?
Let me see your work.
Let me see your work.
Answered by
cr
(2pie(1.67x10^-27)/ (1.60x10^-19)x .232= 2.83x10^-7
.118/2.83x10^-7 wrong answer. I have entered three wrong answers and can not pass this if I can not get the right answer. Please help!
.118/2.83x10^-7 wrong answer. I have entered three wrong answers and can not pass this if I can not get the right answer. Please help!
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