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A charged particle enters a magnetic field with the velocity vector making a $45^{\circ}$ magnetic field. If the pitch of the helical path followed by the particle is $\wp$, its radius will be angle with the

$\begin {array} {1 1} (a)\;\large\frac{\wp}{\sqrt 2 \pi} & \quad (b)\;\large\frac{\wp}{ 2 \pi} \\ (c)\;\large\frac{\wp}{ \pi} & \quad (d)\;\large\frac{\sqrt 2\wp}{ \pi} \end {array}$


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$\wp = \large\frac{2 \pi m}{bq}$ $ v \: \cos\: 45^{\circ}$
 radius of the helix = $ \large\frac{mv\: \sin ( 45)}{Bq}$=$\large\frac{\wp}{2 \pi} $
Ans : (b)


answered Feb 19, 2014 by thanvigandhi_1
edited Sep 15, 2014 by thagee.vedartham

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