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Particles in a Magnetic Field Question

  1. Mar 14, 2017 #1
    1. The problem statement, all variables and given/known data
    A beam of doubly ionized particles (i.e., twice the elementary charge) is accelerated across a potential difference of 2000 V in a mass spectrometer. They are then passed perpendicularly through a magnetic field of 0.085 T resulting in a radius of curvature 12.5 cm. Calculate the mass of the unknown ion.

    2. Relevant equations
    Fc= mv2/r
    Fm= qvB
    ΔV=ΔE/q


    3. The attempt at a solution
    We are supposed to derive a formula for q/m without velocity. I think the solution probably involves the centripetal force being equal to something but I'm not sure what. Any help would be appreciated, thank you!
     
  2. jcsd
  3. Mar 14, 2017 #2

    gneill

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    Staff: Mentor

    In your third relevant equation, where does the ΔE end up?
     
  4. Mar 14, 2017 #3
    ΔE is the change in potential energy experienced by the particle. I could find that, but I'm not sure what I would do from there.
     
  5. Mar 14, 2017 #4

    gneill

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    Staff: Mentor

    How is the change in energy expressed by the particle?
     
  6. Mar 14, 2017 #5
    Sorry, I'm not quite sure what you mean! To be honest I'm not even sure if I need that equation.
     
  7. Mar 14, 2017 #6

    gneill

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    Oh, you definitely need it :smile:

    What happens to a charged particle when it "falls" through a potential difference? What about the particle changes?
     
  8. Mar 14, 2017 #7
    Haha ok good! I guess it gains kinetic energy and loses potential energy, would that ΔE be equal to 1/2mv2?
     
  9. Mar 14, 2017 #8

    gneill

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    Why, yes it would! :smile:
     
  10. Mar 14, 2017 #9
    Perfect! Ok, so I can say that ΔE= 6.4x10-16 J which would also equal 1/2mv2. After that I'm still not sure what to do!
     
  11. Mar 14, 2017 #10

    gneill

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    You now have an expression involving m and v. What's another relationship that involves m and v once the particle enters the magnetic field?
     
  12. Mar 15, 2017 #11
    Fc= mv2/r ?
     
  13. Mar 15, 2017 #12

    gneill

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    Try it.
     
  14. Mar 15, 2017 #13
    OH!! So I can say Vq= 1/2mv2 and mv2/r= qvB... So I solved for v on both sides and got v=qBr/m and v=√2Vq/m and then combine those, solve for q/m and so q/m=2V/B2r2! I plugged in the numbers and got the right answer, 9.0x10-27 kg. Thank you so much for your help, glad I finally got it :)
     
  15. Mar 15, 2017 #14

    gneill

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    Staff: Mentor

    Well done! Glad I could help.
     
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