GREPhysics.NET: GR8677 #25

GR8677 #25

Problem

GREPhysics.NET Official Solution

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.

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Comments

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! Reply to this comment
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) Reply to this comment

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$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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