Question

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

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

Answer 1

Answer: (D)

Period of revolution of charged particle $q$ ,

$T=\frac{\text{circumference}}{\text{velocity}}$

or $T=2\pi R/v$

Circulating current , $I=q/T=q/\left(2\pi R/v\right)=qv/2\pi R$

Therefore magnetic moment of charged particle ,

$M=IA$ (where $A=$area of circular path)

$M=\left(qv/2\pi R\right)\left(\pi R^2\right)=\left(qvR/2\right)$ ..........eq1

Now , angular momentum of particle of mass $m$ ,

$L=mvR$ .............eq2

Dividivg eq1 by eq2 ,

$\frac{M}{L}=\frac{qvR/2}{mvR}=q/2m$