Sol. 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 are lengths R d θ each in cross-section at angles θ and (– θ ) from plane of symmetry as shown in Fig. . In fact each arc represents a long straight current carrying wire.

Fig.

Cross sectional area of each of these wires is A = t (R. d θ ) as shown in Fig. .

Fig.

Since, electrical conductivity of wire material is σ = σ 0 cos θ , therefore, resistance of each wire

= =

∴ Current through each wire, di = =

Since, wires are long, therefore, magnetic induction at O, due to each wire,

dB' = = cos θ d θ

Let direction of current through the conductor be inward. Then direction of magnetic induction due to these two wires will be as shown in Fig. .

Fig.

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

dB = dB´. 2 cos θ = cos 2 θ d θ

∴ B = d θ =