Q 12-04-186JEE MainJEE Main 2020 (6 Sep, Shift 2)Medium
A charged particle going around in a circle can be considered to be a current loop. A particle of mass $m$ carrying charge $q$ is moving in a plane with speed $v$ under the influence of a magnetic field $\vec B$. The magnetic moment of this moving particle is:
Answer: (D) $-\dfrac{mv^2\vec B}{2B^2}$
Radius $r = \dfrac{mv}{qB}$, current $I = \dfrac{qv}{2\pi r}$.
$$\mu = I\pi r^2 = \frac{qvr}{2} = \frac{qv}{2}\cdot\frac{mv}{qB} = \frac{mv^2}{2B}$$
For either sign of charge, the circulating particle's magnetic moment points opposite to $\vec B$ (the motion is diamagnetic), so
$$\vec\mu = -\frac{mv^2}{2B}\hat B = -\frac{mv^2\vec B}{2B^2}$$
Solution by Sreeraj P, M.Sc Physics