We consider a small ring making a polar angle θ with the equatorial plane.
\mathbf{di = \frac{\omega}{2\pi}\frac{\sigma (2\pi r^{2})\cos \theta d\theta}{4\pi r^{2}}=\frac{\omega \sigma}{4\pi}\cos \theta d\theta}
\mathbf{dB = \frac{\mu_{0}i\cos^{2}\theta}{2R}}
\mathbf{dB = \frac{\mu_{0}\omega\sigma \cos^{3}\theta d\theta}{8\pi R}}
B = \int^{\pi/2}_{-\pi/2}\mathbf{\frac{\mu_{0}\omega\sigma \cos^{3}\theta d\theta}{8\pi R}}
\mathbf{B = \frac{\mu_{0}\sigma \omega}{6\pi R}}
Shaswata Roy Sorry!I had mistaken sigma to be the net charge on the sphere.Multiply by 4pi R^2 to get the required result.
Upvote·0· Reply ·2013-08-20 11:15:05Soumyadeep Basu Thanks.