GREPhysics.NET: GR9677 #89

GR9677 #89

Problem

GREPhysics.NET Official Solution

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Electromagnetism $\Rightarrow$ }Trajectory

The only physics involved in this problem is equating the centripetal force with the Lorentz Force, $mv^2/R=qvB$ . The rest is math manipulation and throwing out terms of ignorable order.

The radius of curvature used in the centripetal force equation is given by $R^2=l^2+(R-s)^2$ , and ETS is nice enough to make this geometry fairly obvious in the diagram enclosed with the original question.

Now, note that since $s<<l$ , after expanding the expression for $r^2$ , one can drop out terms of higher order. Thus, $R^2=l^2+(R-s)^2=l^2+R^2+s^2-2Rs\approx l^2+R^2-2Rs + O(s^2)$ . Canceling the R's on both side, one finds, $l^2=2Rs \Rightarrow R = l^2/(2s)$ . Plug this into the force equation above to find,

$mv/R = qB \Rightarrow 2smv/l^2=qB \Rightarrow p=mv=qbl^2/2s, <br />$

which is choice (D).

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Comments

kroner 2009-10-12 07:23:56	An equivalent way to find R is to see that $s = R(1-cos\theta)$ and $l = Rsin\theta$ for some small angle $\theta$ . First order approximations for small $\theta$ are $s = R\theta^2/2$ and $l = R\theta$ , so then $R = l^2/(2s)$ . Reply to this comment
iostream 2009-08-14 21:32:19	easier way: if the particle's momentum is very large, then the magnetic field's influence will be negligible - that is, it will move as if it were in a straight line. thus, the momentum should diverge as s approaches 0, from which we eliminate all choices except D and E. then, you realize the binomial approximation is never going to introduce a factor of 8, so the answer is (D). Reply to this comment
askewchan 2008-11-06 19:41:29	Also, we can eliminate (A) and (B) because we know that a larger $s$ implies greater acceleration which means lower momentum. The same goes for a smaller $l$ . In other words, if the particle has great momentum the radius of this arc will be larger, which gives a greater $l$ and lesser $s$ . Reply to this comment
cordercom 2008-10-05 18:23:40	At the very least we can eliminate choice (C) by units. Reply to this comment

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