When the shell rotates, current is induced due to motion of charge. To calculate magnetic induction at centre of the shell, rotating shell can be assumed to be composed of thin circular current carrying rings. Such a ring can be assumed as follows: Consider a radius of the shell inclined at angle '$\mathrm{\theta }$ ' with the axis of rotation. This radius is rotated about the axis keeping ,, constant. Thus a circle is traced as shown in fig.
When the shell rotates, current is induced due to motion of charge. To calculate magnetic induction at centre of the shell, rotating shell can be assumed to be composed of thin circular current carrying rings. Such a ring can be assumed as follows: Consider a radius of the shell inclined at angle '$\mathrm{\theta }$ ' with the axis of rotation. This radius is rotated about the axis keeping ,, constant. Thus a circle is traced as shown in fig.

Its radius, r = Rsin$\mathrm{\theta }$
Distance of its centre from centre of the shell, x = Rcos$\mathrm{\theta }$
Now consider another radius inclined at angle $(\mathrm{\theta }+\mathrm{d\theta })$. It is also rotated in the same way and another circle is traced. The portion between two circles forms a circular ring.
Area of this ring $=2\mathrm{\pi rRd\theta }=2{\mathrm{\pi R}}^2\sin \mathrm{\theta d\theta }$
Charge on this ring, $\mathrm{dq}=\mathrm{\sigma }.2{\mathrm{\pi R}}^2\sin \mathrm{\theta d\theta }$
Hence, current associated with the ring considered,  $\mathrm{i}=\frac{\mathrm{S}}{2\mathrm{\pi }}\mathrm{dq}={\mathrm{\sigma SR}}^2\sin \mathrm{\theta d\theta }$
Since, centre of the shell is a point lying on the axis of a circular soil of radius r, carrying current iat a distance x from centre of
the coil, therefore, magnetic induction at centre of the shell due to this coil is $\mathrm{dB}=\frac{{\mathrm{\mu }}_0{\mathrm{ir}}^2}{2{\left({\mathrm{r}}^2+{\mathrm{x}}^2\right)}^{3/2}}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{\sigma SRsin}}^3 \mathrm{d\theta }$
Hence, resultant magnetic induction at centre of the shell $\mathrm{B}=\int \mathrm{dB}=\frac{1}{2}{\mathrm{\mu }}_0\mathrm{\sigma SR}{\int }_0^{\mathrm{f}} {\sin }^3\mathrm{\theta d\theta }=\frac{2}{3}{\mathrm{\mu }}_0\mathrm{\sigma SR}$
Its radius, r = Rsin$\mathrm{\theta }$
Distance of its centre from centre of the shell, x = Rcos$\mathrm{\theta }$
Now consider another radius inclined at angle $(\mathrm{\theta }+\mathrm{d\theta })$. It is also rotated in the same way and another circle is traced. The portion between two circles forms a circular ring.
Area of this ring $=2\mathrm{\pi rRd\theta }=2{\mathrm{\pi R}}^2\sin \mathrm{\theta d\theta }$
Charge on this ring, $\mathrm{dq}=\mathrm{\sigma }.2{\mathrm{\pi R}}^2\sin \mathrm{\theta d\theta }$
Hence, current associated with the ring considered,  $\mathrm{i}=\frac{\mathrm{S}}{2\mathrm{\pi }}\mathrm{dq}={\mathrm{\sigma SR}}^2\sin \mathrm{\theta d\theta }$
Since, centre of the shell is a point lying on the axis of a circular soil of radius r, carrying current iat a distance x from centre of
the coil, therefore, magnetic induction at centre of the shell due to this coil is $\mathrm{dB}=\frac{{\mathrm{\mu }}_0{\mathrm{ir}}^2}{2{\left({\mathrm{r}}^2+{\mathrm{x}}^2\right)}^{3/2}}=\frac{1}{2}{\mathrm{\mu }}_0{\mathrm{\sigma SRsin}}^3 \mathrm{d\theta }$
Hence, resultant magnetic induction at centre of the shell $\mathrm{B}=\int \mathrm{dB}=\frac{1}{2}{\mathrm{\mu }}_0\mathrm{\sigma SR}{\int }_0^{\mathrm{f}} {\sin }^3\mathrm{\theta d\theta }=\frac{2}{3}{\mathrm{\mu }}_0\mathrm{\sigma SR}$