A conductor of length $\ell$ has shape of a semi-cylinder of radius $R(\ll <\ell )$. Cross section of the conductor is shown in Fig. Thickness of the conductor is $t(\ll R)$ and conductivity of its material varies with angle $\theta$ only, according to the law $\sigma ={\sigma }_0\cos \theta$.

If a battery of emf $V$ and of negligible internal resistance is connected across its end faces, calculate magnetic induction at mid point $O$ of the axis of the semi-cylinder.

When battery is connected across end faces of the semi-cylinder, a current begins to flow along its length But conductivity of cylinder material is not uniform therefore, current density is also non-uniform.
To calculate magnetic induction at mid point $O$ of axis of the semi-cylinder, considering two equal elemental arc lengths $Rd\theta$ each in cross-section at angles $\theta$ and $(-\theta )$ from plane of symmetry as shown in Fig. (a). In fact each arc represents a long straight current carrying wire.

Cross sectional area of each of these wires is $A=t(R.d\theta )$ as shown in Fig. $\left(b\right)$.

Since, electrical conductivity of wire material is $\sigma ={\sigma }_0\cos \theta$, therefore, resistance of each wire

                                                                 $=\frac{\ell }{\sigma A}=\frac{\ell }{t{\sigma }_{0}R\cos \theta d\theta }$

Current through each wire, $di=\frac{\mathrm{V}}{\text{ resistance }}=\frac{t{\sigma }_{0}VR\cos \theta d\theta }{\ell }$

Since, wires are long, therefore, magnetic induction at $\mathrm{O}$, due to each wire,
$d{B}^{\text{'}}=\frac{{\mu }_{0}di}{2\pi R}=\frac{{\mu }_0{\sigma }_{0}Vt}{2\pi \ell }\cos \theta d\theta$

Let direction of current through the conductor be inward. Then direction of magnetic induction due to these two wires will be shown in Fig.$\left(c\right)$

Resultant of these magnetic induction is normal to plane of symmetry. Its magnitude

$dB=d{B}^{\text{'}}·2\cos \theta =\frac{{\mu }_0{\sigma }_{0}Vt}{\pi \ell }{\cos }^2\theta d\theta$
$B=\frac{{\mu }_0{\sigma }_{0}Vt}{\pi \ell }{\int }_{\theta =0}^{\theta =\pi /2}{\cos }^2\theta d\theta =\frac{{\mu }_0{\sigma }_{0}Vt}{4\ell }$