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Path of electrons motion

  1. Nov 28, 2015 #1
    1. The problem statement, all variables and given/known data

    A beam of electrons travels east at speed, v = 6.10x 10^6 m/s . It passes through a magnetic field , B = 0.450T , directed from top to bottom. Describe exactly the path of the electrons motion?
    2. Relevant equations


    3. The attempt at a solution
    The path of a charged particle moving in a plane perpendicular to a uniform magnetic field is a circle. The electron is moving east at v = 6.10x10^6m/s and the force on it deflects it downward. Because the force is always perpendicular to v, the magnitude of v does not change, and the electron moves at a constant speed. The electron moves clockwise in a circle.

    I think this is right but I was told after to answer it , quantitatively? Eg. How large is the circle.

    How can I show that? Thanks!
     
  2. jcsd
  3. Nov 28, 2015 #2

    gneill

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    Check the deflection direction that you've claimed. Won't it change continuously of the path is circular? What's the initial direction of deflection?

    For the calculation you'll want to look up the force experienced by a charge moving through a magnetic field. Then think circular motion...
     
  4. Nov 28, 2015 #3
    The force would be 4.39x10^-13 N, and I should equate that with the centripetal force right? Except that gives me r= m (6.10x10^6 m/s) / (1.60x10^-19C) ( 0.450T) ....the initial direction of deflection would be downward?
     
  5. Nov 28, 2015 #4

    haruspex

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    Yes, and it's a circle in that plane. So what is the geometric relationship between the magnetic field lines and the circular path?
    The magnetic field is vertical, right? It's a magnetic field, not an electrostatic field.

    Are you saying it gives the wrong answer? What answer do you get? How do you know it's wrong?
     
  6. Dec 1, 2015 #5
    I dont think its right because of my equation from what I showed above ^
     
  7. Dec 1, 2015 #6
    And why is deflection of is downward wrong?
     
  8. Dec 1, 2015 #7

    gneill

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    The magnetic field is oriented vertically. The electron's deflection must be in a direction perpendicular to both the field and the direction of motion of the electron.
     
  9. Dec 1, 2015 #8
    So if the magnetic field is vertically from top to bottom, and the path of the electron is moving east, then the only way it can be perpendicular to both is to move out of the page or into the page?
     
  10. Dec 1, 2015 #9

    gneill

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    Right. Or put another way, if the field is vertical then the deflection must be horizontal.
     
  11. Dec 1, 2015 #10
    So using the right hand rule, it moves clockwise in a circle, which is into the page?
     
  12. Dec 1, 2015 #11

    gneill

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    If the electron had a positive charge that would be true....

    The right hand rule assumes that it is a positive charge that's moving in the magnetic field. Either switch to the left hand rule for the electron or reverse the direction that the right hand rule gives you.
     
  13. Dec 1, 2015 #12
    Ive never used the left hand rule haha! So your saying flip what I just said, so it would be deflected towards me (out of the page) and would rotate counterclockwise
     
  14. Dec 1, 2015 #13

    gneill

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    The sense of the rotation depends upon your point of view. If viewed from "above" it will turn clockwise. If viewed from "below" it will turn counterclockwise.
     
  15. Dec 1, 2015 #14
    But the deflection is on the horizontal plane regardless....how do I get the size of the circle, I equated the force experienced by a charge moving through a magnetic field which I got 4.39 x10^-13 N. Then think circular motion should be r = (mv/qb) , but I cannot find m? I get r = m (6.10x 10^6m/s) / (1.50x10^-19 C)( 0.450 T)
     
  16. Dec 1, 2015 #15

    gneill

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    You should be able to look up the charge and mass of an electron. They are fundamental constants.
     
  17. Dec 1, 2015 #16
    Centripetal Force = 7.8 x10^5 m
    Magnetic Force = 4.39x10^-13 N

    if they have to equal eachother, would I use F= m (v^2) / r ?

    I used r= (mv) / (qB) from before and found 7.8 x10^5 m....im not sure how to equate them when I found both answers?
     
  18. Dec 1, 2015 #17
    Nevermind, I used Bqv = mv^2 / r

    so r = (9.1 x 10^-31kg)(6.10x10^6m/s) ^2 / (4.39x10^-13N)

    resulting in = 7.7 x 10^-5 m
     
  19. Dec 1, 2015 #18

    haruspex

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    The equation r= (mv) / (qB) is derived from F=qBv and F=mv2/r. There's nothing to be gained by going around that loop again.
    You know q, B and v, and you can look up m. So calculate r.
     
  20. Dec 1, 2015 #19
    I did that above ^^^, does that look correct? very small circle.
     
  21. Dec 1, 2015 #20

    haruspex

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    It's a very strong field.
     
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