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Given the scenario that follows, what will be $y$-coordinate of the particle when it collides with the screen? A particle having charge $q = 1C$ and mass = $1Kg$ is released from rest at the origin. The region has Electric and Magnetic fields given by: $E = 10 \hat I N/C$ for $x \leq 1.8m$ and $B = -5 \hat k T$ for $1.8m \leq x \leq 2.4m$. A screen is placed at $x = 3.0m$ parallel to the $y-z$ plane. Gravity is to be neglected.

$\begin {array} {1 1} (a)\;\large\frac{0.6 ( \sqrt 3-1)}{\sqrt 3} & \quad (b)\;0.6 ( \sqrt 3-1) \\ (c)\;1.2 ( \sqrt 3+1) & \quad (d)\;\large\frac{1.2 ( \sqrt 3-1)}{\sqrt 3} \end {array}$


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A is the point $x = 1.8m$
Speed at $x = 1.8m = 6m/s$ (from energy conservation)
$R = \large\frac{mv}{qB}$$ = 1.2 m$
$B$ is the point $2.4\: m$
Angle $AOD = 30^{\circ}$
Arc $OD$ is a part of a circle with radius $1.2m$
And $CD$ makes an angle of $30O$ with the horizontal
The $y$-coordinate of $D$ is $ \large\frac{1.2 ( \sqrt 3 -1.5)}{\sqrt 3}$
Simple geometric argument will get you the coordinates of $C$.
Ans : (d)
answered Feb 19, 2014 by thanvigandhi_1
edited Mar 14, 2014 by balaji.thirumalai

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