Question

A semicircular ring of radius R uniformly charged with chage Q is rotated with angular velocity $\omega$ about point P as shown in figure. Its magnetic dipole moment will be.

• A. $\frac{Qw{R}^2}{4}$
• B. $\frac{Qw{R}^2}{3}$
• C. $\frac{Qw{R}^2}{2}$
• D. $Q\omega R^2$

Answer 1

The correct option is D $Q\omega R^2$

As $\frac{\mu }{L}=\frac{Q}{2m}\Rightarrow \mu =\frac{Q}{2m}L$
Where $L=I_p\omega =2m{R}^2\omega$
$\Rightarrow \mu =\frac{Q}{2m}\times 2m{R}^2\omega$
which gives $\mu =Q\omega R^2$