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Gee Wiz
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Homework Statement
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 = 0.64 m in the x-direction. The proton leaves the field having a velocity vector (vx, vy) = (2.2 X e5 m/s, 0.9 X e5 m/s).
What is R, the radius of curvature of the motion of the proton while it is in the region containing the magnetic field?
Homework Equations
F=qv*b
m*(v^2)/R=F
The Attempt at a Solution
Well, I first wanted to try to find theta. So i did arctan(.9e5/2.2e5). Then using that angle i did .64 (my dimension D) divided by sin of that angle. Which gives me the correct radius. Yet, when i look at the trig behind this is does not make sense to me. I would think i should use cos of the angle, but that is incorrect.
https://www.smartphysics.com/Content/Media/Images/EM/12/h12_bendtheta.png
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