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Motion in a Magnetic Field Radius

  1. Feb 17, 2012 #1
    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) = (2.2 X 105 m/s, 1.4 X 105 m/s).

    http://www.smartphysics.com/images/content/EM/12/h12_bendtheta.png [Broken]

    I have the velocity, but I can't figure out the radius without the magnetic field. Help?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 17, 2012 #2

    berkeman

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    The radius will connect the input and output vectors. Can you show all of your work so far?
     
    Last edited by a moderator: May 5, 2017
  4. Feb 17, 2012 #3
    I know that the proton will move in a circle since the velocity is perpendicular to the field, which I figured to be in the negative z direction. I used the components of the velocity to determine that the angle the proton leaves the field is about 32.55 degrees, which is also the angle from input to output. I have the equation for the radius of the circle being R=mv/qB, where m is the mass of the proton, v is its velocity, q is its charge, and B is the field strength.
     
  5. Feb 17, 2012 #4

    berkeman

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    So both velocity vectors are tangent to that circular path. the input vector position at the origin gives you one point on the circle, and the other point is at the exit at (D,h). Use the fact that the output velocity vector has to be tangent to the circular path to give you more equations to solve the problem.
     
  6. Feb 18, 2012 #5
    I'm having serious problems with this one. I can't seem to get equations that will help me out. All I've gotten is that the vector equation is <2.61e5*cosθ, 2.61e5*sinθ>. There's the circle equation, x[itex]^{2}[/itex] + y[itex]^{2}[/itex] = R[itex]^{2}[/itex]. There's the force equation, F=[itex]\frac{mv^{2}}{R}[/itex]. There's also s=Rθ, but I can't figure out s to solve for R. However, none of these get me anywhere, and I'm positive they are not what you are referring to.
     
  7. Feb 18, 2012 #6
    Nevermind, I got it! I don't know if it was the correct method, but I used the equation sinθ=D/R.
     
  8. Nov 6, 2012 #7
    Can someone please explain how to arrive at this equation for R?
     
  9. Nov 6, 2012 #8

    berkeman

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  10. Nov 6, 2012 #9
    Yes r=m*v/q*B but in this case the value of the magnetic field is not given. I understand that the path is an arc from the entry to exit point but do not know how to arrive at an equation for the radius of curvature of this path.
     
  11. Nov 6, 2012 #10
    drawing lines perpendicular to the velocity vectors and noticing that the radius of this curvature lies on the yaxis, you can then connect the exit and entry points with both a straight line and an arc. You need to find the angle at the top of this isosceles triangle in order to be able to find the length of the arc??
     
  12. Nov 6, 2012 #11

    berkeman

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    Well it's been 9 months since any posts in this thread, and the diagram link in the OP no longer works. Do you have a diagram for your problem?

    The key was to remember that the velocity vector is always tangent to the circular path of the charged particle, IIRC. That gives you the extra equation that you need to solve it...
     
  13. Nov 6, 2012 #12
    Diagram is attached.
     

    Attached Files:

  14. Nov 6, 2012 #13

    berkeman

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    So you should be able to figure out the radius...
     
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