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(D) The magnetic moment $\mu$ of a particle with charge $Q$ moving in a circular path of radius $R$ with speed $v$ is given by the formula:$\mu = I A$

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(D) The magnetic moment $\mu$ of a particle with charge $Q$ moving in a circular path of radius $R$ with speed $v$ is given by the formula:$\mu = I A$Where $I$ is the equivalent current, $I = \frac{Q}{T} = \frac{Q v}{2 \pi R}$, and $A$ is the area of the circular path, $A = \pi R^2$.Substituting these values, we get:$\mu = \left( \frac{Q v}{2 \pi R} \right) (\pi R^2) = \frac{Q v R}{2}$From this expression, it is clear that the magnetic moment $\mu$ depends only on the charge $Q$, speed $v$, and radius $R$.Since the mass $m$ of the particle does not appear in the formula for the magnetic moment, changing the mass while keeping $Q$, $v$, and $R$ constant will not affect the magnetic moment.Therefore, the magnetic moment remains unchanged.

$A$ particle of mass $m$ and charge $Q$ moving with a speed $v$ in a circular path of radius $R$ has a magnetic moment $\mu$. If the mass of the particle is doubled and it maintains the same speed while revolving in the same circular path, then the magnetic moment will be

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