Question

A non conducting disc of radius R 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

Step 1: Given data and terms used

Charge per unit area(surface charge density)$=\sigma =\frac{q}{\pi r^2}$

Charge on the ring of width $dr=\sigma \cdot 2\pi rdr$

The magnetic moment of this ring of width dr is $\text{dμ=i⋅A}$

Step 2: Calculation of magnetic moment

$\int d\mu ={\int }_0^{R}T.2\pi rdr\frac{W}{2\pi }.\pi r^2=\mu =\frac{qw{R}^2}{\text{4}}$

Hence magnetic moment is

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

The correct option is A​