Question

A charge $Q$ is uniformly distributed over the 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 $\omega$. The associated magnetic dipole moment will be:

• A. $\frac{Q\omega }{2}\frac{h^2{\tan }^2\theta }{\sec \theta }$
• B. $\frac{Q\omega }{4}\frac{h^2{\tan }^2\theta }{\sec \theta }$
• C. $\frac{Q\omega }{2}{h}^2{\cos }^2\theta$
• D. $\frac{Q\omega }{4}{h}^2{\sin }^2\theta$

Answer 1

The correct option is B $\frac{Q\omega }{4}\frac{h^2{\tan }^2\theta }{\sec \theta }$

The surface charge density of cone,

$\sigma =\frac{Q}{\text{surface area}}=\frac{Q}{\pi Rl}$

$\sigma =\frac{Q}{\pi R\sqrt{R^2+h^2}}$


$\tan \theta =\frac{R}{h}$

or, $R=h\tan \theta$

$\Rightarrow \sigma =\frac{Q}{\pi R\sqrt{h^2+h^2{\tan }^2\theta }}$

$\sigma =\frac{Q}{\pi R\sqrt{h^2({\tan }^2\theta +1)}}$

$\left[\right(\because {\sec }^2\theta -{\tan }^2\theta =1)$ & ${\tan }^2\theta +1={\sec }^2\theta ]$

$\sigma =\frac{Q}{\pi Rh\sec \theta }$

$\sigma =\frac{Q}{\pi \left(h\tan \theta \right)h\sec \theta }=\frac{Q}{\pi h^2\tan \theta \sec \theta }$

Now let us consider a small element of width $dx$ at a distance $x$ from vertex as shown in the figure.

$\tan \theta =\frac{r}{x}$

$r=x\tan \theta$

$dq=\left[\right(2\pi r\left)dx\right]\sigma$

$dq=2\pi \left(x\tan \theta \right)dx\times \frac{Q}{\pi h^2\tan \theta \sec \theta }$

$dq=\frac{2xdxQ}{h^2\sec \theta }$

Now current due to rotation of this charge is,

$di=dqf=\left(dq\right)\frac{\omega }{2\pi }$

The magentic moment due to this element is,

$d\mu =di\left(A\right)$

$d\mu =i\pi x^2{\tan }^2\theta$

$d\mu =\frac{\pi \left(dq\omega \right)}{2\pi }{x}^2{\tan }^2\theta$

$d\mu =\frac{\omega }{2}\left[\frac{2xdxQ}{h^2\sec \theta }\right]x^2{\tan }^2\theta$

$\mu =\int d\mu =\frac{\omega Q{\tan }^2\theta }{h^2\sec \theta }{\int }_0^h{x}^{3}dx$

Putting the limits from $x=0$ to $x=h$ we get ;

$\mu =\frac{\omega Q{\tan }^2\theta }{h^2\sec \theta }{\left[\frac{x^4}{4}\right]}_0^h$

$\mu =\frac{\omega Q{\tan }^2\theta h^4}{4h^2\sec \theta }$

$\therefore \mu =\frac{\omega Q{h}^2{\tan }^2\theta }{4\sec \theta }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option (b) is the correct answer.