$\text{'}Q\text{'}$ charge is uniformly disturbed over the same surface of a right circular cone of semi-vertical angle $\theta$ and height $h$. The cone is uniformly rotated about its axis at angular velocity $\omega$. Calculate the associated magnetic diplole movement.

 

A ring radius $r$ with charge $q$ rotating with angular velocity $\omega$ is equivalent to a dipole of dipole moment

$\mu =NIA=qfA=q\frac{\omega }{2\pi }A$

where $f$ frequency of rotation $\left(Hz\right)$ , $\omega$ angle frequency of rotation.

Dipole moment of differential ring,

$d\mu =\frac{Q}{\pi R\mathcal{l}}2\pi r\frac{d\mathcal{x}}{\cos \left(\theta \right)}.\frac{\omega }{2\pi }\pi r^2 where l = slant height.$

From similar triangles (see Fig.) we have

$\frac{R}{r}=\frac{h}{h-\mathcal{x}}$

$d\mu =\frac{Q\cos \left(\theta \right)}{\pi Rh}\frac{d\mathcal{x}}{\cos \left(\theta \right)}\omega \pi {\left(\frac{h-\mathcal{x}}{h}R\right)}^3$

  $\mu =\frac{Q\omega R^2}{h^4}{\int }_0^h{(h-\mathcal{x})}^{3}d\mathcal{x}$

    $=\frac{Q\omega R^2}{h^4}.{\left|\frac{{(h-\mathcal{x})}^4}{4}\right|}_h^0$

     $=\frac{Q\omega R^2}{h^4}.\frac{h^4}{4}=\frac{Q\omega R^2}{4}=\frac{Q\omega }{4}{h}^2{\tan }^2\left(\theta \right)$