A sphere of radius 10 mm that carries a charge of 8nC is attached at the end of an insulating string and is whirled in a circle. The rotation frequency is 100? rad/s. What average current does this rotating charge represent ?

Surface area of the sphere is $4{\mathrm{\pi r}}^2=4\times \mathrm{\pi }\times {\left(10\times {10}^{-3}\right)}^2=0.125 {\mathrm{m}}^2$. For a conducting sphere all charges will be on surface area, so surface charge density is $\delta =\frac{8\times {10}^{-9}}{0.125}=6.4\times {10}^{-8}{\text{c/m}}^2$. Current for any spinning sphere with constant charge density is given by $I=\omega r\sin \theta \delta$ for a particular point on the surface. Now to get the total current we have to integrate the above equation $\text{from} \theta =0 \text{to} \theta =\pi$. Now integrating we have
$I_{tot}={\int }_0^{\mathrm{\pi }}\omega r\sin \theta \delta =\omega r\delta {\int }_0^{\mathrm{\pi }}\sin \theta =2\omega r\delta$
Now we have r=10mm=.01m and $\omega =100\text{rad/s}$, so we have
$I=2\times 0.01\times 100\times 6.4\times {10}^{-8}=1.28\times {10}^{-7}$A