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(D) The magnetic moment $\mu$ of a current loop is given by $\mu = I A$. Here,the current $I$ produced by the revolving charge $q$ is $I = \frac{q}{T}

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$\frac{qvR}{2}$

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(D) The magnetic moment $\mu$ of a current loop is given by $\mu = I A$. Here,the current $I$ produced by the revolving charge $q$ is $I = \frac{q}{T}$,where $T$ is the time period of revolution. The time period $T$ for a particle moving in a circle of radius $R$ with speed $v$ is $T = \frac{2 \pi R}{v}$. Substituting $T$ into the expression for $I$,we get $I = \frac{q}{2 \pi R / v} = \frac{qv}{2 \pi R}$. The area $A$ of the circular path is $A = \pi R^2$. Therefore,the magnetic moment is $\mu = I A = \left( \frac{qv}{2 \pi R} \right) (\pi R^2) = \frac{qvR}{2}$.

$A$ charged particle (charge $q$) is moving in a circle of radius $R$ with uniform speed $v.$ The associated magnetic moment $\mu$ is given by

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