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Electromagnetism and potential

  1. May 19, 2010 #1
    1. A particle of charge q=6E-6C and mass m=1E-10kg is fired out of a charge gun towards the uniform magnetic field pointing into the page, B=0.50T. The particle moves vertically up originally and then accelerated through potential difference of 200V. There is a diagram of the B field going "into" the page. a) How fast is the particle moving when it leaves the gun? b) Does the particle exit the magnetic field to the left or right? Why? c) How far to the side does the particle exit the field?

    Small attempt at diagram. The period is the particle. x is going "into" page
    x x x x
    x x x x
    x x .x x

    2. I'm not sure but these are the eqns I think might be relevant: F=ma, F=qvBsin(theta)It would help if even someone could just tell me all the eqns I need because I'm lost on this

    3. I think I'm supposed to use a=F/m and F=qvBsin(theta) but I don't know how to find v.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 19, 2010 #2
    What is E in your two equations?
    q=6E-6C and m=1E-10kg

    Also, I'm having trouble understanding the diagram you've described. The particle leaves the gun which is aimed vertically and is immediately accelerated vertically through a potential difference of 200V? Or does it travel for some distance before being accelerated? And does it enter the magnetic field after being accelerated or is the magnetic field present as soon as it leaves the gun? What is the scale of the magnetic field? It must be confined to a certain area if the question asks how the particle exits the field.
    Last edited: May 19, 2010
  4. May 19, 2010 #3
    sorry. E is a way to express x10^

    that was the only info given.

    The diagram is supposed to be the tail ends of arrows going away from you to signify direction of field. The particle is placed there as a period. And the professor labeled "P" on the board and wrote next to it "Vo" which I assume is V nought.
  5. May 19, 2010 #4
    Ok then I'm assuming your professor meant that at the particle's initial position the potential is equal to the constant V[tex]_{0}[/tex]. Think about the particle before and after it is accelerated, what does it gain in being accelerated through the potential difference of 200V?
  6. May 19, 2010 #5
    gaining more velocity? Sorry. I'm a bit slow on the uptake
  7. May 19, 2010 #6
    Which particle has a higher energy, a proton placed at a point where the potential is V[tex]_{0}[/tex] or a proton placed at a point where the potential is V[tex]_{1}[/tex]? Where V[tex]_{1}[/tex]>V[tex]_{0}[/tex].
  8. May 26, 2010 #7
    Usually since we are dealing with one particle, the place with the highest potential has the highest electrostatic potential energy, while in areas of lower potential, more work is done on them and therefore have more kinetic energy.

    Makes sense?
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