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

lukenlow
2018-06-22 12:31:32

I\'m glad that I found this useful information in such a simple explanation, this knowledge will be very useful to me 192.168.l.l

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walczyk
2011-03-01 21:50:25

luckily i knew the equation of radius given chord length and chord height (2l and s); everyone's tactics are so much harder, but r = s/2 + l^2/2s, so you're missing a factor of s/2 but since l >> s, its dropped by ETS. then mv^2/r = qvb is all you neat to solve it: p = qbl^2/2s + qbs/2 wooo.

timtamm
2011-08-26 20:16:59

this solution is rather easy... I am a big big fan... of memorizing and plugging and chugging rather than trying to figure stuff out on a high stress/ timed test. THANKS

nontradish
2012-04-19 20:33:54

I agree with timtamm to a point. I have been using Dr. Brown's flash cards from Case Western Reserve and have been able to answer many questions off the top of my head. It's better if the physics makes sense too, so you don't have to rely solely memory under pressure;) Good luck to all those taking it on Saturday and in the future!!

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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 $R = l^2/(2s)$ .

aprilrussell
2018-05-30 03:48:38

At the very least we can eliminate choice (C) by units. \r\nhttp://fivenightsat-freddys.com

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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).

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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 $s$ .

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cordercom
2008-10-05 18:23:40

At the very least we can eliminate choice (C) by units.

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I\'m glad that I found this useful information in such a simple explanation, this knowledge will be very useful to me 192.168.l.l

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