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Magnetic force Particle trajectory

  1. Jan 4, 2013 #1
    1. The problem statement, all variables and given/known data

    A particle of charge −q is moving with a velocity [itex]\vec{v}[/itex]. It then enters midway between two plates where there exists a uniform magnetic field pointing into the page, as shown in attached image.

    1.png


    (b) Compute the distance between the left end of the plate and where the particle strikes

    2. Relevant equations

    [itex]\vec{F_{B}} = q\vec{v}\times\vec{B}[/itex]



    3. The attempt at a solution

    I discovered that the charged particle is going to be deflected downward. Then it is going to strike the bottom plate before leaving the region between two plates. So the distance between the end of the bottom plate and where the particle strikes is what this problem is asking for.

    See the following attached image.
    2.png


    [itex]y = yo -\frac{1}{2}at^{2}[/itex]

    [itex]y = 0[/itex] because it strikes the end of the plate.
    [itex]y_{0} = \frac{d}{2}[/itex]

    so [itex]t = \sqrt{da}[/itex]

    [itex]\vec{F_{B}} = q\vec{v}\times\vec{B}[/itex]

    to find the magnitude of acceleration:


    [itex]ma = qvB[/itex]

    [itex]a = \frac{qvB}{m}[/itex]

    then [itex]t = \sqrt{d\frac{qvB}{m}}[/itex]

    [itex]x = x_{0} + v_{0}t[/itex]

    [itex]x_{0} = 0[/itex]

    [itex]x= v_{0}t[/itex]

    because [itex]v_{0} = v[/itex]

    [itex]x = v\sqrt{d\frac{qvB}{m}}[/itex]


    Then the distance is [itex]l- v\sqrt{d\frac{qvB}{m}}[/itex]


    Is that correct? cause i have feeling that it is wrong.
     
    Last edited: Jan 4, 2013
  2. jcsd
  3. Jan 4, 2013 #2

    Simon Bridge

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    I only skimmed, but I think you have assumed that the deflecting force always acts in the -y direction. Recall - a charge moving in a uniform/constant magnetic-field follows a circular path. So ... what kind of acceleration would that be and what is it's relationship to the radius of the circle?
     
  4. Jan 4, 2013 #3
    i could do it by the radius too.

    2.png

    [itex] m \frac{v^{2}}{R} = qvB[/itex]

    so [itex] R = \frac{mv^{2}}{qvB}[/itex]


    [itex] R^2 = (R- \frac{d}{2} )^{2} + x^{2}[/itex]


    [itex] x^{2} = Rd - (\frac{d}{2})^{2}[/itex]

    [itex] x^{2} = d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2} [/itex]

    [itex] x = \sqrt{d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2}} [/itex]

    Then the distance is [itex] l - x = \sqrt{d\frac{mv^{2}}{qvB} - (\frac{d}{2})^{2}} [/itex]
     
    Last edited: Jan 4, 2013
  5. Jan 4, 2013 #4

    Simon Bridge

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    Compute the distance between the left end of the plate and where the particle strikes
     
  6. Jan 4, 2013 #5
    That's the problem. I am not understating what the problem is asking for.
     
  7. Jan 4, 2013 #6

    Simon Bridge

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    The bottom plate is the one that is going to be hit.
    The bottom plate has two ends.
    One end is on the left of the picture, the other is on the right.
    Which end did you measure distance from (per your diagram)?
    What is the length ##x## on your diagram?

    Also - the last line in your calculations (post #3) does not follow from the previous line.
     
  8. Jan 5, 2013 #7

    oh right. I was calculating for the right end, not the left end.

    so my [itex]x[/itex] is the desired distance (distance between the end of the left [itex]x_{0} = 0[/itex] and where the particle hits the plate [itex]x[/itex].
     
  9. Jan 5, 2013 #8

    Simon Bridge

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    Well done.
    I used to get caught out by that sort of thing all the time :)
     
  10. Jan 5, 2013 #9
    Thanks for your time. :)

    just to make sure. i can not find the distance doing the calculation of the first post. I have to use the radius (post 3).
     
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