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Velocity of electron shot from a solenoid.

  1. Feb 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A certain solenoid(50cm long with 2000 loops) carries a current of 0.70A and is in vacuum. An electron is shot at an angle of 10deg to the solenoid axis from a point on the axis.

    Find the speed of the electron if it is just to miss hitting the inside of the 1.6cm diameter solenoid?

    2. Relevant equations
    B=unI
    r=mv/qB
    v=qrB/m



    3. The attempt at a solution

    B=(12.5*10^-7)(2000/0.5m)(0.7)/sin10=0.020T

    v=0.20T * 0.008m*1.6*10^-19C=2.8*10^7m/s<<<the manual says the velocity is 1.4*10^7 don't understand why it's half of what I got. Please help.
     
  2. jcsd
  3. Feb 26, 2012 #2

    gneill

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    Staff: Mentor

    The loops that the electron will perform will not occupy the whole of the solenoid cross section. Remember that the electron is starting from the solenoid axis.
     
  4. Feb 26, 2012 #3
    does that make the angle double?
     
  5. Feb 26, 2012 #4

    gneill

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    No. Try drawing the situation as viewed looking into the solenoid (the circular cross section). The electron appears at the center of the circular cross section with some radial velocity (the portion of the initial velocity that is radially directed). How large a circle can it describe?
     
  6. Feb 26, 2012 #5
    Only half a circle? Because the entire circle isnt in the entire plane?
     
  7. Feb 26, 2012 #6

    gneill

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    No, a full circle but its size is dictated by its starting point! How large a circle can you draw within the solenoid's circular cross section if you start (and end) at its center?
     
  8. Feb 27, 2012 #7
    Hey you don't have to yell yknow. ;) I'm trying to get this.

    The circle's diameter appears to be .08m
     
  9. Feb 27, 2012 #8

    gneill

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    Right, so that makes its radius...
     
  10. Feb 27, 2012 #9
    0.04m so that's why.


    The little circle represents the electron's circular component of velocity while r helps us determine the linear component?

    Then we find the component perpendicular to r
     
  11. Feb 27, 2012 #10
    Ahhh the electron acquires helical motion due to being at an acute angle to the magnetic field. That's what the little circle represents.
     
  12. Feb 27, 2012 #11

    gneill

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    Bingo! :wink:
     
  13. Feb 27, 2012 #12
    Great!

    Hopefully im not undoing everything here but the circle is half as big b/c the field lines are at r and 2r(middle of solenoid cross section). respectively? I may be overthinking just wondering why and if the diameter is attributed to those velocity components. (In short what keeps me from drawing a circle the entire diameter of the solenoid? Or any other diameter?)
     
  14. Feb 27, 2012 #13

    gneill

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    The maximum diameter of the trajectory depends upon where the particle enters the field. A circle is a closed path and must return to the same x-y location repeatedly. The point of entry must be on the circumference of the circle -- there's no choice in the matter, the particle is where it is when it begins looping, and you're given that it enters on the solenoid axis which is the center of the cross section.

    attachment.php?attachmentid=44455&stc=1&d=1330357728.gif
     

    Attached Files:

  15. Feb 27, 2012 #14
    Okay that's the issue then I didn't know exactly where the axis fell. My textbook isn't that good it seems. So basically the electron starts from the center there and continues in it's path as discussed. For some reason I thought it shot out from the wire in the spring. Although the path is helical it doesn't go any further than that axis point as it moves.

    I was scouring the net to find a diagram on the axis! :D
     
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