Q. A non-conducting disc of radius $2 \,m$ is rotated about its axis of symmetry perpendicular to the plane of disc with a uniform angular speed $\omega$. The disc carries a charge whose density is given as $\sigma=A-B r$, where $A$ and $B$ are positive constants and $r$ is distance from the centre of disc. The value of $A / B$ for which magnetic field at the centre of disc will be zero is
Moving Charges and Magnetism
Solution:
$d i=\frac{d q}{\left(\frac{2 \pi}{\omega}\right)}=\frac{\omega}{2 \pi}(\sigma .2 \pi r d r)=\omega(A-B r) r d r$
$d B=\frac{\mu_{0} \omega(A-B r) r d r}{2 r} ; B=\frac{\mu_{0} \omega}{2} \int\limits_{0}^{2}(A-B r) d r$
$0=\left[A r-B \frac{r^{2}}{2}\right]_{0}^{2}$
$2 A-2 B=0, \frac{A}{B}=1$
