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Electrons in magnetic field

  1. Jun 12, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=59503&stc=1&d=1371095175.png


    2. Relevant equations



    3. The attempt at a solution
    Let magnetic field be inside the plane of sheet (towards -ve z-axis)
    At any instant of time, let the components of velocity of electron be ##v_x## and ##v_y##.
    Then ##F_x=ev_yB_o## (i) and ##F_y=-ev_xB_o## (ii)
    Substituting ##v_x=dx/dt, v_y=dy/dt, F_x=md^2x/dt^2## and ##F_y=md^2y/dt^2## and dividing (i) and (ii), I end up with ##(d^2x)dx=-(d^2y)dy## and solving this with wolfram alpha gives ##y^2=k-x^2##. k=0 as this curve passes through origin. This gives ##y^2=-x^2## which is not possible. :confused:
     

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  3. Jun 13, 2013 #2

    ehild

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    What are the trajectories of the electrons in magnetic field? Can you find them without Wolframalpha? And what is k?

    ehild
     
  4. Jun 13, 2013 #3
    Because the curve passes through origin. :)
     
  5. Jun 13, 2013 #4

    ehild

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    You did nonsense with the second derivatives.
    d2y/dx2 is not (dx)2 divided by (dy)2, that is, square of the tangent. Think of a parabola y=x2. y'=dy/dx=2x, y"=d2y/dx2=2, (y')2=4x2...

    Remember your high-school studies. What are the trajectories in magnetic field?
     
    Last edited: Jun 13, 2013
  6. Jun 13, 2013 #5
    :confused:
    The trajectory depends on the velocity of electrons and direction of magnetic field. In this case, if I consider an electron which is projected along the y-axis, it will hit the x-axis at a distance of ##mv_F/{eB}## from the origin (the trajectory would be of a semicircle). If an electron is projected long the axis, it will disappear as it goes in ##y<0## region. But I am confused as to what happens when it is thrown at an angle (except 90) with x-axis. I am still clueless. :(
     
  7. Jun 13, 2013 #6

    ehild

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    The velocity, that is tangent to the trajectory makes and angle with the x axis. The radius is perpendicular to it. You get the radius of the circle from the speed and B. Can you find the position of centre of the circle?

    I edited my previous post, read it.
     
  8. Jun 13, 2013 #7
    And I never said that it is. I had the following:
    [tex]m\frac{d^2x}{dt^2}=e\frac{dy}{dt}B_o[/tex]
    [tex]m\frac{d^2y}{dt^2}=-e\frac{dx}{dt}B_o[/tex]
    I divided these two equations and ended up with
    [tex](d^2x)dx=-(d^2y)dy[/tex]
    Plugging this in wolfram alpha gave me ##y^2=k-x^2##.

    Let the electron be projected at an angle ##\theta##, then is the radius ##mv_F\sin\theta/(eB_o)##? :confused:
     
  9. Jun 13, 2013 #8

    ehild

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    Why? What angle does the velocity enclose with the magnetic field?

    ehild
     
  10. Jun 13, 2013 #9
    Sorry, the angle with magnetic field is ##\pi/2## but I still have no idea. :(
     
  11. Jun 13, 2013 #10

    ehild

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    So what is the radius of the trajectory with the given speed of the electrons?

    I have to leave. Hope, you can proceed. Make a drawing.

    ehild
     
  12. Jun 13, 2013 #11
    I still don't have any idea. Please give me a few hints. :(
     
  13. Jun 13, 2013 #12

    ehild

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    What do you mean with the d-s in [tex](d^2x)dx=-(d^2y)dy[/tex]?

    Wolframalpha interpreted d as constant. You can cancel them, and get xdx=-ydy, which is far away from the original problem.

    As for electron in magnetic field: what angle does the magnetic force enclose with the velocity? What is the trajectory? Maybe a circle? If it is circle, what is the centripetal force? Given the velocity, what is the radius that corresponds to that centripetal force?
     
    Last edited: Jun 13, 2013
  14. Jun 13, 2013 #13
    :confused:
    Looks here: Wolfram Alpha

    EDIT: Okay, I get, but what should I do now?

    It follows the path of circle only when it is projected along the y-axis. I don't see how can I find the centripetal force for the general case. :(
     
  15. Jun 13, 2013 #14

    ehild

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    You have a velocity vector in the x,y plane. Magnitude VF. You have a magnetic field B in the z direction.
    What is the direction of the Lorentz force with respect to the velocity and magnetic field?
     
  16. Jun 13, 2013 #15
    The direction of Lorentz force is perpendicular to both velocity and magnetic field. Do you ask me this:
    [tex]ev_FB_o=\frac{mv_F^2}{R}[/tex]
    [tex]R=\frac{mv_F}{eB_o}[/tex]
     
  17. Jun 13, 2013 #16

    ehild

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    exaxtly. So is the trajectory an arc of circle at arbitrary angle?
     
  18. Jun 13, 2013 #17
    Yes?
     
  19. Jun 13, 2013 #18

    ehild

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    Yes, of course. If a force of constant magnitude acts perpendicularly to the velocity the trajectory is a circle. Now it is only part of a circle as the electron disappears after reaching the x axis. And where is its centre? Make a drawing.
     
  20. Jun 13, 2013 #19
    If an electron is projected along y-axis, the centre of the circle is on the x-axis at a distance ##R## from the origin and if it is projected at an angle, the centre lies in ##y<0## region. Right?

    If an electron is projected at angle ##\theta## with the x-axis, it reaches the x-axis again at a distance of ##2R\sin \theta##.

    EDIT: Got it! Thanks! The question asked for the greatest value of x and it is greatest when projected along y-axis. Thank you ehild! :smile:
     
    Last edited: Jun 13, 2013
  21. Jun 13, 2013 #20

    ehild

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    The problem asked the position of the detector where it gets the greatest signal. How can be the magnitude of the signal depend on the angle the electron is projected at?
     
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