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What is the radius of curvature for a proton moving in a magnetic field?
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[QUOTE="collinsmark, post: 4534636, member: 114325"] I mean the center of the circle will be above the x-axis, [I]on[/I] the y-axis somewhere. You don't know exactly how far up the y-axis it is yet, but don't worry, you can find that out later. On a figure, draw a point where you think the center of the circle might be (it will be somewhere up on the y-axis). The length of the that line is the radius of the circle. Now draw another line, with the same length that starts at the center-point of the circle and ends at the location where the proton exits the magnetic field. Connecting the bottom of the two, equidistant lines is the path of the proton, which makes an arc. All together, the two lines and the arc form a "wedge." It's in the shape of a piece of pie or a piece of pizza. How does the geometry of this wedge relate to [I]θ[/I]? By the way, on a different note, there is yet another triangle you can form. How does [I]θ[/I] relate to [I]h[/I] and [I]D[/I] [and the radius of the circle]? [Edit: On second thought, you probably don't need to calculate this if you don't want to. I don't think it's necessary to solve the problem.] [/QUOTE]
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What is the radius of curvature for a proton moving in a magnetic field?
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