Magnetic Field Mass Spectrometer Calculations

[AceInfinity]

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

Homework Statement

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?

Homework Equations

I'm guessing this one, but I don't see how radius would be added to solve it.
[PLAIN]http://k.min.us/idLl6m.png

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.

 

[rock.freak667]

Well the magnetic field is going to cause the proton to move in a circular path.

So the magnetic force provides the centripetal force mv²/r. So equate that to BQv and solve for B.

 

[AceInfinity]

alright, so I have this:

mv²/r
= (1.67x10^-27kg)(v²)/(3.00m)

How do I solve for v in either equation when I don't have the Fm?

 

[rock.freak667]

alright, so I have this:

mv²/r
= (1.67x10^-27kg)(v²)/(3.00m)

How do I solve for v in either equation when I don't have the Fm?

Because you will have

mv²/r = Bqv

so you can get v.

 

[AceInfinity]

Ahh, thanks, I'm going to try that...

... > v = qBr/m

v = (1.60x10^-19)(B)(3.00m)/(1.67x10^-27kg)

What exactly is B though in this equation? How do I get it

 

[AceInfinity]

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^-27kg

I forgot to add in, that the voltage between the 2 first plates that the proton passes through is 1.00x10⁴ Volts.

 

[rock.freak667]

Ahh, thanks, I'm going to try that...

... > v = qBr/m

v = (1.60x10^-19)(B)(3.00m)/(1.67x10^-27kg)

What exactly is B though in this equation? How do I get it

B is the magnetic field strength, which is what you are looking for.

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^-27kg

I forgot to add in, that the voltage between the 2 first plates that the proton passes through is 1.00x10⁴ Volts.

You were told you to use the value of 'v' from the question before.

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.)

 

[AceInfinity]

Alright I got it:

B = [mv/r] ÷ q
B = [(1.67x10^-27)(1.38x10⁶)/(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?

 

[rock.freak667]

That should be correct now.