Question

A non conducting disc of radius $R$, charge $q$ is rotating about an axis passing through its centre and perpendicular to its plane with an angular velocity $\omega$, charge $q$ is uniformly distributed over its surface. The magnetic moment of the disc is :

• A. $\frac{1}{4}q\omega R^2$
• B. $\frac{1}{2}q\omega R$
• C. $q\omega R$
• D. $\frac{1}{2}q\omega R^2$

Answer 1

Answer: (A)

$\frac{1}{4}q\omega R^2$

Charge per unit area$\left(\sigma \right)=\frac{q}{\pi R^2}$
Charge on ring of width $dr$$=\sigma \cdot 2\pi rdr$
Magnetic moment of this ring of width dr
$d\mu =i\cdot A$
$\int d\mu ={\int }_0^{R}T\cdot 2\pi rdr\frac{w}{2\pi }\cdot \pi r^2$$=\mu =\frac{qw{R}^2}{4}$