Magnetism/electric fields/mass spectrum help

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The discussion revolves around the design of a mass spectrometer, specifically focusing on the use of a velocity selector to control the velocity of ionized particles before they enter the magnetic deflection chamber. Participants explore the required potential difference between charged plates in the velocity selector and the relationship between electric and magnetic forces acting on the ions. The charge of the singly ionized argon ions is confirmed to be 1.6 x 10^-19 C, and the equations governing the system are discussed, particularly how to derive the potential difference needed. Clarification is sought on the formulas used and the roles of the electric and magnetic fields in the setup. The conversation emphasizes the importance of understanding the forces at play to accurately determine the necessary parameters for the mass spectrometer's operation.
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Homework Statement



When the ionized particles entering a mass spectrometer are simply accelerated across the potential difference between two charged plates, there is always the chance of an uncontrolled variation in their velocities due to their random thermal motion before they undergo linear acceleration. An elaboration of the basic mass spectrometer design better controls the incoming velocity of the ions by passing them through a velocity selector before they enter the magnetic deflection chamber. Suppose that the same uniform magnetic field, having a strength 0.750 T, is used in both the velocity selector and the deflection chamber. In this chamber, singly ionized argon ions with a mass of 6.63×10−26 kg are to be deflected through semi-circular arcs of radius 12.0 cm. (a) If the oppositely charged plates of the velocity selector are separated by 2.00 cm, what is the required potential difference between them? (b) If you were to add a velocity selector to the diagram of the mass spectrometer, as shown in the illustration, which plate would be positively charged with respect to the other?

Homework Equations



B=.75
m=6.63e-26
r=.12m
d=.02m
q=1.6e-19 ? not sure on this

m=(qr^2B^2)/(2*deltaV)


deltaV= (gr^2B^2)/(2m)

for velocity selectors v=E/B
E=deltaV/deltax


The Attempt at a Solution



deltaV = 9773.76

I'm not sure if I'm using the right equations or if I have done the right thing so far. I would rearrange the equations to get v=deltaV/(deltax * B), but I that doesn't really get me anywhere since I'm solving for deltaV of the velocity selector and not the chamber. I'm confused on this one. Any help would be greatly appreciated.
 
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Can you show a picture of the set-up and trajectory of the ions?

ehild
 
Here's a link to the diagram. I hope it works. I wasn't able to copy/paste it out so I'm not sure how visible it is.


http://i42.tinypic.com/2i6pehg.jpg
 
The argon ions are singly ionized. They miss 1 electron. The charge is therefore 1.6 E-19 C.

I do not understand the formula you used: m=(qr^2B^2)/(2*deltaV)

What is deltaV?

In the chamber, the ion travels along a semicircle of radius R=12 cm with speed v. The centripetal force is equal to the magnetic force: mv^2/R = qvB. You know all data, find the speed v.

The velocity selector let's go a particle across without deflection if the electric and magnetic forces cancel. qE=qvB, that is E=vB. You know v, get the electric field, and the potential difference from that.

The electric force points from the positive plate to the negative one. You have to decide which plate is + and - so as the magnetic force is opposite to the electric force. For that you need to know the direction of the magnetic field (I can not see on the diagram, but you certainly do).

ehild
 
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