Question 254 of 276

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 :

A$\dfrac{m{v}^{2}\vec{B}}{2{B}^{2}}$B$-\dfrac{m{v}^{2}\vec{B}}{2{\pi B}^{2}}$C$-\dfrac{m{v}^{2}\vec{B}}{{B}^{2}}$D$-\dfrac{m{v}^{2}\vec{B}}{2{B}^{2}}$

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Answer: (D) $-\dfrac{m{v}^{2}\vec{B}}{2{B}^{2}}$

$i=\dfrac{q}{T}=\dfrac{qv}{2\pi r}$
$M=i\times \pi {r}^{2}$
$=\dfrac{2v}{2\pi r}\times \pi {r}^{2}$
$=\dfrac{1}{2}qv\times \dfrac{mv}{qB}$
Direction of $\vec{M}$ is opposite of $\vec{B}$ therefore
$\vec{M}=\dfrac{-{mv}^{2}\vec{B}}{2{B}^{2}}$