Posted by cr on .
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.
(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!

Physics ~ work shown, pls help! 
bobpursley,
Bqv=mv^2/r
v=Bqr/m
v= .232*1.6E19*.118/1.67E27
Check that. Now compare it with whatever you did, I cant figure out what you did.
for information, if you posted this as your answer: .118/2.83x10^7
I don't understand what you did, and why did you not simplify this. I get about 4.17E5 m/s from that fraction.
Check my work and see what I got from my calcs. 
Physics ~ work shown, pls help! 
cr,
Calculating your v gave me 2.62285030e48
so I then divide .118/ans and get 4.4989e46 would this be right?