Angles Transversed with Centripetal Acceleration

  • Thread starter Durin
  • Start date
  • #1
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Homework Statement


Okay, so the problem is I need to find the radius of curvature for a proton moving through a uniform magnetic field with initial velocity v for some distance d and exiting at angle θ. I was able to get the correct answer; however, I think the apparent assumption made by the person making this problem is wrong.


Homework Equations



sin θ = opposite/hypotenuse.

The Attempt at a Solution



[PLAIN]http://img715.imageshack.us/img715/4710/diagram2.png [Broken]

This is what I think is wrong.

[PLAIN]http://img194.imageshack.us/img194/6984/diagram1l.png [Broken]

This is why I think it is wrong.

If my diagrams need to be further explicated, don't hesitate asking.
 
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Answers and Replies

  • #2
557
1
What exactly do you think it's wrong ? State it with words...
What is "h" ? The total lenght of the path ?
 
  • #3
11
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h is the height it exits at. It's solved for using the radius. d is the distance it travels along the x-direction.

What I think is wrong is the angle at the top left of the first diagram. According to the correct answer, it appears that the angle the two radii make at the enter and exit points is equal to the angle the exit velocity makes relative to the ground. I used geometry to show why I don't believe that is actually the angle.
 

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