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Electron in magnetic field and center of circumference

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data

    We have an electron in 3D space in which there is also a magnetic uniform field B=Bz.
    At t=0, we have z=0 and $v_z=0$.
    We have to find the center of the circumference travelled by the electron using only x, y, v_x and v_y (coordinates and velocity of the electron at the general istant t) .

    2. Relevant equations
    $$L_F= qv×B$$ : this is the equation of Lorentz force.

    3. The attempt at a solution
    The first thing that I have done was finding the radius: $R=\frac{mv}{qB}$
    Then, I have thought that I could obtain the center considering:

    $$R_x= \frac{m \dot x}{qB} \rightarrow x-x_c=\frac{m \dot x}{qB} $$

    $$R_y= \frac{m \dot y}{qB} \rightarrow y-y_c=\frac{m \dot y}{qB} $$

    .. but probably this way is wrong.

    And so I have thought to use Lorentz equation

    $$F_{L,x}=q v_y B$$ $$ F_{L,y}=-q v_x B$$ $$F_{L,z}=0$$
    but I don't know how I can obtain the center using this equation... what can I do?

    Thank you so much!!
     
  2. jcsd
  3. Feb 9, 2013 #2

    mfb

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    If the electron moves in some direction at t=0, where is the center of the circle? In the direction of motion, behind it, left*, right*, something in between?
    *as seen from "above" (positive z), for example

    How can you find a point in that direction with a distance of R?
     
  4. Feb 9, 2013 #3
    on the left? (negative charge move in anticlockwise..)

    I have thought to use the euclidean distance between points, but I obtain only one relation and I don't obtain the coordinates of the center using the velocities.. can you help me? thanks a lot!
     
  5. Feb 9, 2013 #4

    mfb

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    Hmm, let's view that from below, along the magnetic field. Ok, to the left, so at a 90°-angle to the direction of motion. Do you know how you can find an orthogonal line in 2 dimensions? The velocity components will help. And the euclidian distance will be used, too.
     
  6. Feb 9, 2013 #5
    I'm not sure to have correctly understood your answer.

    If I consider the line that have the direction of motion, I obtain
    $$y-y_T=y'(x-x_T)$$

    and the radius belongs to the line

    $$y-y_T=-1/y'(x-x_T)$$

    (y' is v_y, isn't it?)

    and the euclidean distance between the position of the electron at generical istant t (x_T,y_T) and the center of the circumference (x_c,y_c) is

    $$sqrt ((x_T-x_c)^2+(y_t-y_c)^2))=R$$

    now I don't know what I have to do...
     
  7. Feb 9, 2013 #6

    mfb

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    What is yT?

    For simplicity, let's use x=y=0. You can add them afterwards.

    If the velocity vector is (x',y'), can you find an orthogonal vector?
     
  8. Feb 9, 2013 #7
    If the vector a=(a1,a2) si orthogonal to (x',y') then
    a1*x'+a2*y'=0..
     
  9. Feb 9, 2013 #8

    mfb

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    You can use this to generate some orthogonal vector. In 2 dimensions, there is a very nice way to chose a1 and a2.
     
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