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.

(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!

Answered by bobpursley
Bqv=mv^2/r
v=Bqr/m

v= .232*1.6E-19*.118/1.67E-27

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.
Answered by cr
Calculating your v gave me 2.62285030e-48

so I then divide .118/ans and get 4.4989e46 would this be right?
There are no AI answers yet. The ability to request AI answers is coming soon!