Question

A hollow cylinder has length $L$ and inner and outer radii $R_1$ and $R_2$ respectively. The cylinder carries a uniform charge density $\rho$. An expression for the magnetic moment as a function of the angular velocity $\omega$ of rotation of the cylinder about its axis is

• A. $\frac{1}{2}\pi \rho L\omega (R{{}_2}^4-R{{}_1}^4)$
• B. $\frac{1}{2}\pi \rho L\omega (R{{}_1}^4-R{{}_2}^4)$
• C. $\frac{1}{4}\pi \rho L\omega (R{{}_1}^4-R{{}_2}^4)$
• D. $\frac{1}{4}\pi \rho L\omega (R{{}_2}^4-R{{}_1}^4)$

Answer 1

The correct option is D $\frac{1}{4}\pi \rho L\omega (R{{}_2}^4-R{{}_1}^4)$

The cylinder can be visualized to consist of differential cylindrical shells of charge $dQ$. Let, one such cylindrical shell has radius $r$ and thickness $dr$.

Charge $dQ$ of differential cylindrical shell is

$dQ=\rho \left(2\pi rdr\right)L$

The current due to rotation of this charge is given by

$dI=\frac{dQ}{\left(2\pi /\omega \right)}=\rho \left(2\pi rdr\right)L\left(\frac{\omega }{2\pi }\right)$

The magnetic moment of this differential current loop

$d\mu =\left(dI\right)\left(\pi r^2\right)=\left[\rho \right(2\pi rdr\left)L\left(\frac{\omega }{2\pi }\right)\right]\pi r^2$

To find total magnetic moment, we integrate the above equation,

$\mu =\int d\mu ={\int }_{R_1}^{R_2}\left[\rho \right(2\pi rdr\left)L\left(\frac{\omega }{2\pi }\right)\right]\pi r^2$

$\Rightarrow \mu =\pi \rho L\omega {\int }_{R_1}^{R_2}{r}^{3}dr$

$\therefore \mu =\frac{1}{4}\pi \rho L\omega (R_2^4-R_1^4)$

Hence, option $\left(D\right)$ is the correct answer.