Magnetic Field Mass Spectrometer Calculationsby AceInfinity Tags: intensity, magnetism, spectrometer 

#1
Jun2611, 02:49 AM

P: 26

Here's a practice diploma question I found on the internet, this isn't homework. I'm studying, I just need help on how to do this type of question.
http://www.paulway.com/physics30/p30dips/p30jan2002.pdf (Page 11 Numerical Response #5) Answer: 4.80x10^3 1. The problem statement, all variables and given/known data In the magnetic field mass spectrometer shown, the radius of curvature of a proton’s path is 3.00 m. What is the magnetic field intensity? 2. Relevant equations I'm guessing this one, but I don't see how radius would be added to solve it. 3. The attempt at a solution I couldn't even attempt this one, sorry, I tried, but maybe I have the wrong equation? I've posted other questions, but as I was reviewing I understand that I never really knew how to do this kind of problem. So for anyone to help me starting with the basics, would be helpful as any help at all is highly appreciated. 



#2
Jun2611, 03:00 AM

HW Helper
P: 6,210

Well the magnetic field is going to cause the proton to move in a circular path.
So the magnetic force provides the centripetal force mv^{2}/r. So equate that to BQv and solve for B. 



#3
Jun2611, 04:36 PM

P: 26

alright, so I have this:
mv^{2}/r = (1.67x10^{27}kg)(v^{2})/(3.00m) How do I solve for v in either equation when I don't have the Fm? 



#4
Jun2611, 04:41 PM

HW Helper
P: 6,210

Magnetic Field Mass Spectrometer Calculationsmv^{2}/r = Bqv so you can get v. 



#5
Jun2611, 04:55 PM

P: 26

Ahh, thanks, i'm going to try that...
.... > v = qBr/m v = (1.60x10^{19})(B)(3.00m)/(1.67x10^{27}kg) What exactly is B though in this equation? How do I get it 



#6
Jun2611, 08:29 PM

P: 26

Still requesting some help on this one. At this point i'm at a loss for how to solve this kind of problem at all. Sorry for being so impatient, but I'm trying to learn as much as I can for today about classical physics involving charge, momentum, mainly.
I know that the curvature of the proton's path is due to the inertia of the proton having a mass of 1.67x10^{27}kg I forgot to add in, that the voltage between the 2 first plates that the proton passes through is 1.00x10^{4} Volts. 



#7
Jun2611, 09:38 PM

HW Helper
P: 6,210

From Bq=mv/r you need to rearrange for B and you have m=1.67(10^{27}) kg. (If you did not do the part before to get v, you should consider conservation of energy for that.) 



#8
Jun2611, 09:48 PM

P: 26

Alright I got it:
B = [mv/r] ÷ q B = [(1.67x10^{27})(1.38x10^{6})/(3.00m)] ÷ 1.60x10^{19} B = 4.8012...x10^{3} However I just took the answer of the velocity from the answer key. I'm not quite sure how to solve for velocity in this case. If that could be the second (should have been the first) part of this thread? 



#9
Jun2611, 09:53 PM

HW Helper
P: 6,210

That should be correct now.



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