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GR8677 #25
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Alternate Solutions |
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Comments |
harmonxjim33
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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 | I feel that may be a captivating point, it made me suppose a bit. Thanks for sparking my pondering cap. Now and again I am getting such a lot in a rut that I just really feel like a record. makeup artist Tampa | | 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!
LAStew
2011-09-12 17:11:19 |
You mean answer (D)
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| | 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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