A charged particle going around in a circle can be considered to be a current loop. A particle of a mass $$m$$ carrying charge $$q$$ is moving in a plane with speed $$v$$ under the influence of magnetic field $$\vec{B}$$. The magnetic moment of this moving particle is:

We start with the definition of magnetic moment for a planar current loop. A charge $$q$$ completing one revolution in time period $$T$$ constitutes a current

$$I \;=\;\frac{q}{T}.$$

The particle moves in a circle of radius $$r$$ with speed $$v$$, so its period is obtained from the well-known relation $$T=\frac{\text{circumference}}{\text{speed}}$$ :

$$T \;=\;\frac{2\pi r}{v}.$$

Substituting this value of $$T$$ in the formula for current, we obtain

$$I \;=\;\frac{q}{\dfrac{2\pi r}{v}} \;=\;\frac{qv}{2\pi r}.$$

The area of the circular path is

$$A \;=\;\pi r^{2}.$$

By definition the magnitude of the magnetic moment is

$$\mu \;=\; I\,A.$$

Inserting the expressions for $$I$$ and $$A$$ gives

$$\mu \;=\;\left(\frac{qv}{2\pi r}\right)\!(\pi r^{2}) \;=\;\frac{qvr}{2}.$$

Its direction is perpendicular to the plane of the circle, along the unit vector $$\hat n$$ obtained from the right-hand rule for a positive charge. Hence in vector form

$$\vec\mu \;=\;\frac{qvr}{2}\,\hat n.$$

Next, we express $$r$$ in terms of the given magnetic field $$\vec B$$. Because the velocity is perpendicular to $$\vec B$$, the magnetic Lorentz force provides the necessary centripetal force. Writing this equality explicitly,

$$qvB \;=\;\frac{mv^{2}}{r}.$$

Solving for the radius,

$$r \;=\;\frac{mv}{qB}.$$

Substituting this value of $$r$$ back into $$\vec\mu$$, we obtain

$$\vec\mu \;=\;\frac{qv}{2}\left(\frac{mv}{qB}\right)\hat n \;=\;\frac{mv^{2}}{2B}\,\hat n.$$

The angular momentum $$\vec L$$ of the particle is $$\vec L = mvr\,\hat n$$, and for a positive charge $$\vec\mu$$ is parallel to $$\vec L$$, whereas for a negative charge it is antiparallel. The options given contain a negative sign, so the particle is to be regarded as negatively charged. Thus we write

$$\vec\mu \;=\;-\frac{mv^{2}}{2B}\,\hat B,$$

because $$\hat n$$ is opposite to $$\hat B$$ for a negative charge in this standard configuration. Since $$\hat B = \dfrac{\vec B}{B}$$, we finally have

$$\vec\mu \;=\;-\frac{mv^{2}}{2B^{2}}\;\vec B.$$

This matches Option D.

Hence, the correct answer is Option D.