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A particle of charge $q$ and mass $m$ moves in a circular orbit with radius $r$ and angular speed $ \omega$. The ratio of the magnitude of magnetic moment to its angular momentum depends on

$\begin {array} {1 1} (a)\;\omega\: and \: m & \quad (b)\;\omega \: and \: q \\ (c)\;m \: and \: q & \quad (d)\;\omega ,q\: and \; \: m \end {array}$


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To find the magnetic moment due to a single particle, it is advisable to consider the equivalent current due to it
$M = iA = (qf)( \pi r2 ) =\large\frac{ (q\: \omega\: r2 )}{2}$
$L = I \: \omega = mr^2 \omega$
$\large\frac{M}{L} = \large\frac{q}{2m}$
Ans : (c)
answered Feb 21, 2014 by thanvigandhi_1

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