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GR8677 #25
Problem
 GREPhysics.NET Official Solution Alternate Solutions

Electromagnetism$\Rightarrow$}Lorentz Force

Recall the Lorentz Force, $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$.

$\vec{E}$ and $\vec{B}$ are parallel. The particle is released from rest, so the Electric force would propel it. The resulting velocity would be parallel to the electric field, but since the magnetic field is also parallel to that, there would be no magnetic force contribution. The particle thus goes in a straight line.

Alternate Solutions
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harmonxjim33
2019-10-01 14:06:03
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fredluis
2019-08-09 04:26:58
Yosun has the answer as 1 ohm, that 1/2 you\'re seeing is just a 1 with capital omega. pressure cleaning
joshuaprice153
2019-08-09 02:27:41
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UNKNOWNUMBER
2018-10-14 22:55:11
This is how I got the correct answer, can someone confirm if it\'s viable method or if I just got lucky? \r\n\r\nWe\'re given a constant magnetic field, this means that according at Faraday\'s law, the curl of the electric field = 0. Since the electric field is what propels the charge, it must move in a straight path because there is no curl; there is no rotational aspect so it can\'t be any of the other answers. \r\n
FutureDrSteve
2011-10-29 15:02:56
I did this one qualitatively using the right-hand rule. To find the force the B-field exerts on the particle, I first reasoned that the E-field would propel it on a vector parallel to the E-field. Pointing my fingers in that direction, it's impossible to curl them towards the B-field, since they're parallel. Therefore, the B-field doesn't act on the particle, and it should continue on its straight-line path.
ME
2009-10-11 16:16:13
Could it be that on GR9677: #86 you meant cycloid instead of helical as in choice B instead of C? Thank you so much for your site. I love it!
 LAStew2011-09-12 17:11:19 You mean answer (D)
a19grey2
2008-11-02 22:23:12
Note that if the fields were perpendicular to each other the electron would drift in a helical pattern as in choice (C). It seems ETS likes questions of this type (see: GR9677: #86)

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