Question

A charged particle (charge $=q;$ mass $=m$ ) is rotating in a circle of radius $R$ with uniform speed $V$. Ratio of its magnetic moment $(\mu )$ to the angular momentum $(L)$ is

• A. $\frac{q}{2m}$
• B. $\frac{q}{m}$
• C. $\frac{q}{4m}$
• D. $\frac{2q}{m}$

Answer 1

Answer: (A)

The magnetic moment of charge, $\mu =niA$

and angular momentum of charge
$L=m\cdot v\cdot R$
Here, $n=1$
Current, $i=\frac{q}{T}$
Time-period, $T=\frac{2\pi R}{V}$
Equivalent current, $i=\frac{q}{2\pi R/v}=\frac{q\cdot v}{2\pi R}$
Area of loop or circle, $A=\pi R^2$
$\therefore \mu =1\times \frac{qv}{2\pi R}\times \pi R^2=\frac{qvR}{2}$
$\Rightarrow L=l\omega =m{R}^2\times \frac{V}{R}=mvR$
$I=$ Moment of inertia of loop
$\therefore \frac{\mu }{L}=\frac{niA}{mvR}=\frac{qvR/2}{mvR}=\frac{q}{2m}$