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Cathode Ray Tube with magnetic field

  1. Aug 31, 2015 #1
    1. The problem statement, all variables and given/known data

    A solenoid is placed beneath a CRT that produces a magnetic field of 1.10 x -2. The CRT is 1 meter away from a screen.
    The electrons that make up the beam were accelerated from rest through a potential difference(V) at the beginning of the CRT. Acceleration of the electrons in the magnetic field is 9.08 x 1014 m/s2.
    What is the potential difference required to accelerate the electrons.

    See image for diagram http://imgur.com/OLIRC3j


    2. Relevant equation
    E=Energy
    m=mass
    v=velocity
    q=charge
    V=potential difference
    d=distance
    a=acceleration


    V=E/q
    E=(1/2)mv2
    v(final)2=v(initial)2 + 2ad



    3. The attempt at a solution
    I figure that I can solve for v(final) because I have all the variables. I can then use v(final) in the E=(1/2)mv2 formula to solve for E. Once I have E, i should be able to plug it into V=E/q.

    Any help would be appreciated
     
  2. jcsd
  3. Aug 31, 2015 #2

    haruspex

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    Yes, that sounds like a plan, though you did not explain how you will find the velocity.
    My advice is to solve it entirely symbolically, not plugging in any numbers until you have a final formula. You may find there's some cancellation.
     
  4. Aug 31, 2015 #3

    v(final)2=v(initial)2 + 2ad

    Initial velocity is 0 because it is at rest, so final velocity will be equal to whatever "2ad" turns out to be.
    So E will look something like this
    E = (1/2)m(2ad)2
    Therefore, potential difference should be
    V=E/q
    V=((1/2)m(2ad)2)/q

    I am unsure if it is okay if I use final velocity in my final equation, or if i should use average velocity?
     
  5. Aug 31, 2015 #4

    haruspex

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    I don't think so. Where are you getting a and d from for this?
     
  6. Aug 31, 2015 #5
    a and d are given to us in the question. This formula is one of the kinematics formula provided to me.
    V final 2 = Velocity initial 2 + 2ad
     
  7. Aug 31, 2015 #6

    haruspex

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    Be specific. What are you using for a and d, out of the supplied information?
     
  8. Aug 31, 2015 #7
    For d I am using the distance of the screen from the CRT, which is 1 meter.
    For a, I am using the acceleration of the electrons in the magnetic field, which is 9.08 x 1014 m/s^2
     
  9. Aug 31, 2015 #8

    haruspex

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    In the equation v2=2ad, the acceleration must be acting along the distance d. Where is the magnetic field act in relation to the 1m, and in what direction?
     
    Last edited: Aug 31, 2015
  10. Aug 31, 2015 #9

    The magnetic field is acting in a clockwise direction, which is why it pushes the electron the right side of the screen.

    Hmmm, I know that E=Vq, therefore Vq=(1/2)mv2 so i need to find v.

    Since the problem tells me the acceleration in the magnetic field, would I be able to use Force=massxacceleration to find the magnetic force?
    Then with the magnetic force, use this formula:
    Magnetic Force =qvB
    (B=Magnetic Field)
    ,to solve for v?
     
  11. Aug 31, 2015 #10

    haruspex

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    Yes, that's how you need to find v here.
     
  12. Aug 31, 2015 #11
    Ok, thank you.
     
  13. Aug 31, 2015 #12
    Hi haruspex. Second part of the question here. Would appreciate any advice on this one. The only thing I can come up with is reversing the current of solenoid, but not sure how to prove this with the formulas.
     

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  14. Aug 31, 2015 #13

    rude man

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    Since haruspex may still be in bed (Australia), can I suggest (1) what effect is there on the mag. force when B is reversed, and (2) does the magnitude of the deflection depend on the magnitude of B?
     
  15. Aug 31, 2015 #14
    Thanks rude man, I should be able to figure it out from here...
     
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