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A charged particle is projected with velocity $ \upsilon_0$ along the positive $x$ - axis. The magnetic field $B$ is directed along negative $z$ - axis for the region $0<x<L$. The particle emerges out at $x = L$ at an angle of $60^{\circ}$ with the direction of projection. Find the velocity with which the same particle is projected at $x=0$ along the $+ve\: x$ - axis so that when it emerges out at $x = L$, the angle made by it is $30^{\circ}$ direction of projection with the

$\begin {array} {1 1} (a)\;2\upsilon_0 & \quad (b)\;\sqrt{2}\upsilon_0 \\ (c)\;3\upsilon_0 & \quad (d)\;\sqrt 3\upsilon_0 \end {array}$


1 Answer

The situation becomes clear if we sketch a qualitative diagram however
only a basic understanding of the situation is required to solve the problem
$\sin \theta = \large\frac{L}{r} $$ = \large\frac{L}{\bigg(\Large\frac{mv}{qB} \bigg)}$
$\upsilon \: \alpha\: \large\frac{ 1}{\sin \theta}$
$ \therefore \upsilon_1 = √3 \upsilon_0$
Ans : (d)


answered Feb 22, 2014 by thanvigandhi_1
edited Sep 17, 2014 by thagee.vedartham

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