Question

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 magnetic field $\vec{B}$. The magnetic moment of this moving particle :

• A. $-\frac{m{v}^2\vec{B}}{B^2}$
• B. $-\frac{m{v}^2\vec{B}}{2\pi B^2}$
• C. $\frac{m{v}^2\vec{B}}{2B^2}$
• D. $-\frac{m{v}^2\vec{B}}{2B^2}$

Answer 1

Answer: (D)

Magnetic moment
$M=iA$
$M=\left(\frac{q}{T}\right)\times \pi r^2=\frac{q\pi r^2}{\left(\frac{2\pi r}{v}\right)}=\frac{qvr}{2}$
$M=\frac{qv}{2}\times \frac{vm}{qB}$
$M=\frac{m{v}^2}{2B}$
As we can see from the figure, direction of magnetic moment (M) is opposite to magnetic field
$\vec{M}=-\frac{m{v}^2}{2B}\hat{B}$
$=-\frac{m{v}^2}{2B^2}\vec{B}$