Answer:
[tex]M = \frac{4\pi R^4 \sigma \omega}{3}[/tex]
Explanation:
As we know that the magnetic moment of a system of rotating charge and its angular momentum is related by an equation given as
[tex]M = \frac{Q}{2m} L[/tex]
so we can write here angular momentum of the sphere about its diametrical axis as
[tex]L = I\omega[/tex]
here we have
[tex]I = \frac{2}{3} mR^2[/tex]
so we have
[tex]M = \frac{Q}{2m} (\frac{2}{3} mR^2)\omega[/tex]
[tex]M = \frac{QR^2\omega}{3}[/tex]
here we know that
[tex]Q = \sigma (4\pi R^2)[/tex]
so we have
[tex]M = \frac{4\pi R^4 \sigma \omega}{3}[/tex]