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 $B$. The magnetic moment of this moving particle:

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

Answer 1

The correct option is C

$-\frac{m{v}^{2}B}{2B^2}$

Step 1 : Given data

Mass of the particle $=m$

Charge of the particle $=q$

Magnetic field $=B$

Step 2 : To find the magnetic moment

Magnetic moment

$M=iA$

$$\begin{gathered} A={\mathrm{\pi r}}^2 \\ \mathrm{M}=\left(\frac{\mathrm{q}}{\mathrm{T}}\right)\times {\mathrm{\pi r}}^2=\frac{{\mathrm{q\pi r}}^2}{\left({\displaystyle \frac{2\mathrm{\pi r}}{\mathrm{v}}}\right)}=\frac{\mathrm{qvr}}{2} \\ \mathrm{M}=\frac{\mathrm{qv}}{2}\times \frac{\mathrm{vm}}{\mathrm{qB}} \\ \mathrm{M}=\frac{{\mathrm{mv}}^2}{2\mathrm{B}} \end{gathered}$$

The direction of magnetic moment $\left(M\right)$ is opposite to magnetic field

$$\begin{gathered} M=\frac{m{v}^2}{2B}\times \left(-B\right) \\ =-\frac{m{v}^2\overrightarrow{B}}{2B^2} \end{gathered}$$

Hence the magnetic moment $M=-\frac{m{v}^2\overrightarrow{B}}{2B^2}$

Therefore the correct answer is Option C