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Motion in a magnetic field [Magnetism]

  1. Nov 25, 2012 #1
    1. The problem statement, all variables and given/known data

    This is not hw, but practice for my upcoming exam. Any help will be appreciated :)

    A charged particle of mass m = 6.7X10-8 kg, moving with constant velocity in the y-direction enters a region containing a constant magnetic field B = 1.6T aligned with the positive z-axis as shown. The particle enters the region at (x,y) = (0.98 m, 0) and leaves the region at (x,y) = 0, 0.98 m a time t = 860.0 μs after it entered the region.

    https://www.smartphysics.com/Content/Media/Images/EM/12/h12_bend90.png [Broken]

    What is Fx, the x-component of the force on the particle at a time t1 = 286.7 μs after it entered the region containing the magnetic field.

    2. Relevant equations
    F = q(v x B)
    F = ma
    a = v^2/r
    d = vt
    d = pi/2 r

    3. The attempt at a solution
    So far I have found the correct speed that the particle is going. It was 1789.98 m/s. I got that by knowing the distance the particle will travel will be r* pi/2. Radius is simply the distance in either the x or y given. With that distance I know time, thus I used d=vt to solve for v.

    Now I am confused by the Fx and Fy components and what I need to do. I'm assuming I need to brush up on mechanics.

    I tried solving at first by keeping that velocity the same throughout the problem. So I did F=ma, where a is v^2/r. Can someone lead me to the right way and explain why this will not work? I'm having issues with pinpointing the correct values and what I need to solve in order to find the force at that time. Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Nov 25, 2012 #2

    Redbelly98

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    Welcome to Physics Forums.

    Your method will give you the correct magnitude of the force. But question doesn't ask for the magnitude, it asks for the x-component of that force. If you can figure out the direction in which the force acts, then you can get the x-component.
     
    Last edited by a moderator: May 6, 2017
  4. Nov 25, 2012 #3
    Thank you.

    I believe the force is acting in the negative x direction since it exits that way. I'm confused with not only the component but with time. Usually to get the component of x I multiply by cos to get the vector. However, since there is no angle given, I am confused. I cannot visually see the angle it will be at that time and I'm not sure how to get that.
     
  5. Nov 25, 2012 #4

    Redbelly98

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    Note quite. This is true only as the particle is first entering the magnetic field.
    Let's think about what this equation is saying. For a magnetic force, the direction of F depends on the direction of v and B, and is at a right angle to v. The direction of v changes, so the direction of F will be changing also.

    Since you know the particle travels in a circular path, see if you can use the time information to figure out how far along the circle it has traveled. If you can get that, you can figure out the direction of v and therefore F.
     
  6. Nov 26, 2012 #5
    Ok yes, I follow. v and B and F all have direction. B is constant in this field going out of the page. Like you said, v and F are changing because B is constant. I tried also using d =vt. I can find that distance with the given time. I am now stumped with what to do with that. Parametric? hmm.

    Also, thank you very much for your patience. I really appreciate the assistance.
     
  7. Nov 26, 2012 #6

    Redbelly98

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    You're welcome.
    Good.
    You know the path that the particle follows, and you know the distance along that path it travels in the given time of 286.7 μs. From that information, I would figure out where the particle is located after the given time of 286.7 μs.
     
  8. Nov 26, 2012 #7
    I apply d=vt, where the distance will now be 0.513 m (1789.98 m/s * 286.7E-6). I'm not too sure how to apply this distance to get force component.
     
  9. Nov 26, 2012 #8

    Redbelly98

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    Where is the particle located when it has traveled 0.513 m?
     
  10. Nov 26, 2012 #9
    Somewhere on the circle... hmm
     
  11. Nov 27, 2012 #10
    I do not see how I can apply this to find the component.
     
  12. Nov 27, 2012 #11

    Redbelly98

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    Here's a figure depicting how far the particle has traveled in 286.7 μs:

    ArcLength.png
    Note, figure is not to scale.

    You know r and s.
    From those you can find the angle through which the particle has traveled in its "orbit".
    Once you find the angle, figure out the direction in which the particle is moving at 286.7 μs.
    From the direction in which the particle is moving, find the direction of the force.
     
    Last edited by a moderator: Apr 15, 2017
  13. Nov 27, 2012 #12
    AH! I completely overlookd the arc length. I have calculated the right answer! Thank you! Just applied the correct sign for directions of the components and it was correct. Thank you very much, RedBelly98
     
    Last edited by a moderator: Apr 15, 2017
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