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Homework Help: Two particles in a magnetic field

  1. Feb 25, 2005 #1
    Illustration:
    http://home.no.net/erfr1/images/2.jpg
    "A particle, X with mass m and charge q starts with an inital velocity zero, and follows the red path indicated on the drawing. It follows a half circle when entering the magnetic field.

    A new particle, Y with mass m(same) and charge 2q starts from the same place as X. Is the following true og untrue:

    1. The force on Y is double the force on X in the electric field.
    2. The force on Y is double the force on X in the magnetic field.
    3. The time Y uses on the half circle is half the time X uses.

    1. Fex=qE and Fey=2qE. So that's true.
    2. Fmx=qvB = q*sqrt(qU/m) * B.
    And Fmy = qvb = 2q * sqrt(2qU/m) * B. So this statement is false.
    3. I don't get this.

    It's half a circle, so v=Pi*r/T
    And I have tried the formula: v^2/r = 4Pi^2*r / T^2

    But everything I use involves either v, or r og both. And the radius of Y has to be smaller right? Since the mag. force on Y is the biggest. But I don't know the radius! And the speed is such an ugly expression, and I always end up with the square root of an expression, so the difference can't be 1/2.


    The book says that statement 3 is true...

    Please help.

    Thanks
     
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Feb 25, 2005 #2

    dextercioby

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    [tex] T_{X}=\frac{\pi r_{X}}{v_{x}} [/tex] (1)
    [tex] T_{Y}=\frac{\pi r_{Y}}{v_{Y}} [/tex] (2)
    [tex] r=\frac{mv}{qB} [/tex] (3)

    Use (3) for the 2 particles and then combine with (1) & (2) to reach the result.

    Daniel.
     
  4. Feb 25, 2005 #3
    Aaaah! Thank you. So the speed is ruled out and I get:
    Tx= Pi*m/qB
    Ty=1/2 * Pi*m/qB
    Ty=1/2Tx
    I can't believe I spent so much time on this :). Stupid me.

    Thanks!
     
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