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Magnetic field motion different scenario

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with constant velocity enters a region containing a constant magnetic field that is directed along the z-axis at (x,y) = (0,0) as shown. The magnetic field extends for a distance D = 0.7 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (3.1 X 105 m/s, 1.4 X 105 m/s).


    2. Relevant equations



    3. The attempt at a solution
    i don't know how to even start this
     
  2. jcsd
  3. Feb 28, 2013 #2
    Can you post a picture?
     
  4. Feb 28, 2013 #3
    i can't even tell what direction the magnetic field is in
     

    Attached Files:

  5. Feb 28, 2013 #4
    You can discern the direction of the magnetic field the position of the particle before and after entering the magnetic field Remember, the force on a negative charge acts opposite in direction to the force on a positive charge.
     
  6. Feb 28, 2013 #5
    so, B will point out of the screen, and F in positive y direction (initially).
    i've tried using pytagoras' theorem to get a value for Vnet, but that would be for the exit velocity, not the velocity on entrance...?
     
  7. Feb 28, 2013 #6
    No, B will not be out of the screen. Look at the image.
     

    Attached Files:

  8. Feb 28, 2013 #7
    is there an alternate version to the right hand rule? i pointed my index finger towards velocity vector, and the force vector inwards using my middle finger, which gave me the magnetic field pointing outward. i understand that the positioning would be different each time, but is there any particular way to "start" the right hand rule interpretation?
     
  9. Feb 28, 2013 #8
    using this picture, the magnetic field is represented by the middle finger. but before, in other questions, the direction of the thumb indicated the magnetic field. why does the representation change?
     
  10. Feb 28, 2013 #9
    I have seen a couple variations to the rule. If you are not comfortable with it yet, I would stick to the one in your textbook. You want to ensure all three vectors are mutually perpendicular, and pointing a way from a common point (like an origin).
     
  11. Feb 28, 2013 #10
    It depends on the sign of the charge. This charge is positive, the other is negative. The right hand rule works for positive charges. If the charge is negative, you have to visualize the vector pointing in the opposite direction that the right hand rule tells you.
     
  12. Feb 28, 2013 #11
    OH!!!! FLASHBACK!!!! i just remembered an unconsciously learned moment from class. our prof actually DID say that negative and positive charges will change the RHR directions. thanks sandy.bridge for reminding me. won't forget it for my test tomorrow :D
     
  13. Feb 28, 2013 #12
    but getting back to the question, how do i get around solving the questions? so the velocity is towards positive x direction, force is towards positive y, and magnetic field is pointing into screen (negative z direction).
    if i use a consolidated equation v = (rQB)/m , i have two unknowns.
     
  14. Feb 28, 2013 #13
    Note: the components of the velocity may change, however, the magnitude of the velocity (speed) will remain constant!
     
  15. Feb 28, 2013 #14
    i realized that i didn't post the questions that needed answering, so here they are:
    1)What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?
    2)What is h, the y co-ordinate of the proton as it leaves the region conating the magnetic field?
    3) What is Bz, the z-component of the magnetic field? Note that Bz is a signed number.

    need a little guidance on how to solve them
     
  16. Feb 28, 2013 #15
    For starters, you should draw a diagram including any angles you know, along with the vectors for the velocity. You will have to use some geometry.
     
  17. Feb 28, 2013 #16
    i found θ=24 using arctan(vy/vx), and the trace of the radius with the center of the circle in the top left of the square, as the picture shows.
     

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  18. Mar 1, 2013 #17
    Your drawing is off. The velocity vector that you solved theta for will be tangent to the curve of the circle. The line connecting the center of the circle and the point both velocity components extend from will be perpendicular to the resultant velocity vector.
     
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