GREPhysics.NET: GR9677 #71

GR9677 #71

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

Electromagnetism $\Rightarrow$ }Particle Trajectory

One can get a reasonable approximation for the deflection angle as follows.

Assuming that there is no magnetic field, one has from the Lorentz force $F=ma=qE=qV/d$ , where one neglects gravitational acceleration. The acceleration is constant, and it is $a=qV/(dm)$ .

Recalling the baby physics kinematics equation, $y=0.5at^2 \Rightarrow dy=atdt$ and the fact that $x=L=vt \Rightarrow dx=vdt$ and $t=L/v$ , one can calculate the angle as $\tan \theta \approx dy/dx = \frac{atdt}{vdt}=at/v=\frac{qVL}{v^2dm}$ . Take the arctangent to get choice (A).

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

dstahlke
2009-10-09 10:57:43

The diagram threw me off here... The deflection of the final velocity is $arctan(dy/dx)$ but the dotted lines on the diagram form a triangle suggesting they want $arctan(\Delta y / \Delta x)$ which I think would be equal to $arctan(dy/2dx)$ . That wasn't one of the options so in this case I suppose guessing the nearest answer would work.

Reply to this comment

haro
2007-04-13 04:25:01

It is simplier to think of tan $\theta$ = $v_{y}/v_{x}$

Jeremy
2007-10-15 12:01:19

I agree. From $a=qV/dm$ use $a=v_y/t$ , $t=L/v_x$ , and finally, the equation you gave: $\tan{\theta}=v_y/v_x$ .

Reply to this comment

yosun
2005-11-09 14:42:33

tachyon: thanks for the typo-alert; it has been corrected.

Reply to this comment

tachyon
2005-11-09 13:36:53

typo-the correct answers is A

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

<a href="http://www6.shoutmix.com/?grephysics">View shoutbox</a>
ShoutMix chat widget

1GHz Smartphone for 1 cent - best deal on a Samsung Galaxy Phone

Free 3D Virtual World. Largest User-Created Fantasy World. Don't Play, Experience. Join Now!

Bored, got time to kill, and want to earn micro-cash by mindlessly clicking on links? Join CrownGPT. Click Offers. Click the Paid to Click button... then click to your heart's delight.

FREE 3D Virtual World Chat - Join Second Life Today!

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone