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

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

Answer 1

The correct option is D $-\frac{m{v}^2\overrightarrow{B}}{2B^2}$

Length of the circular path, $l=2\pi r$

Current, $i=\frac{q}{T}=\frac{qv}{2\pi r}$

Magnetic moment, $M=\text{Current}\times \text{Area}$

$=i\times \pi r^2=\frac{qv}{2\pi r}\times \pi r^2$

$M=\frac{1}{2}q.v.r$

Radius of circular path in magnetic field, $r=\frac{mv}{qB}$

$\therefore M=\frac{1}{2}qv\times \frac{mv}{qB}\Rightarrow M=\frac{m{v}^2}{2B}$

Direction of $\overrightarrow{M}$ is opposite of $\overrightarrow{B}$, therefore
$\overrightarrow{M}=\frac{-m{v}^2\overrightarrow{B}}{2B^2}$

(By multiplying both numerator and denominator by $B$).