Question

A charge $q$ is spread uniformly over an isolated ring of radius $R$. The ring is rotated about its natural axis with an angular velocity $\omega$. Magnetic dipole moment of the ring is

• A. $\frac{q\omega R^2}{2}$
• B. $\frac{q\omega R}{2}$
• C. $q\omega R^2$
• D. $\frac{q\omega }{2R}$

Answer 1

Answer: (A)

Given, charge $=q$
Radius of isolated ring $=R$
Angular velocity of ring $=\omega$
Magnetic dipole moment of the ring,
$M=niA,n=1,M=iA$
The current in the ring, $i=q\times f$
where, $f=$ frequency of charge
$\omega =2\pi f$
$f=\frac{\omega }{2\pi }$
$\Rightarrow i=q\cdot \frac{\omega }{2\pi }$
Area, $A=\pi R^2$
$\Rightarrow M=q\cdot \frac{\omega }{2\pi }\cdot \pi R^2=\frac{1}{2}q\omega R^2$