A charged particle $(c h a r g e = q; m a s s = m)$ is rotating in a circle of radius $^{'} R^{'}$ with uniform speed $^{'} V^{'}$ . Ratio of its magnetic moment $(μ)$ to the angular momentum $(L)$ is

\frac{q}{2 m}

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\frac{q}{m}

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\frac{q}{4 m}

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\frac{2 q}{m}

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Solution

The correct option is D $\frac{q}{2 m}$
Period of revolution of charged particle $q$ ,
$T = \frac{circumference}{velocity}$
or $T = 2 π R / v$
Circulating current , $I = q / T = q / (2 π R / v) = q v / 2 π R$
Therefore magnetic moment of charged particle ,
$M = I A$ (where $A =$ area of circular path)
$M = (q v / 2 π R) (π R^{2}) = (q v R / 2)$ ..........eq1
Now , angular momentum of particle of mass $m$ ,
$L = m v R$ .............eq2
Dividivg eq1 by eq2 ,
$\frac{M}{L} = \frac{q v R / 2}{m v R} = q / 2 m$

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