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

Answer 1

Answer: (A)

The magnetic moment of charge, $\mu=n i A$

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, $A=\pi R^{2}$
$\therefore \mu=1 \times \frac{q v}{2\, \pi R} \times \pi R^{2}=\frac{q v R}{2}$
$\Rightarrow L=l \omega=m R^{2} \times \frac{V}{R}=m v R$
$I=$ Moment of inertia of loop
$\therefore \frac{\mu}{L}=\frac{n i A}{m v R}=\frac{q v R / 2}{m v R}=\frac{q}{2 m}$