A thin sheet of width L and thickness t has a long length . It is bent into the shape of a long thin walled pipe having a semicircular cross section. It carries a current I along its length. The current density is uniform over the entire cross section of area Lt. O is the centre of curvature of the middle section (perpendicular to length) of the conductor. Field at O is $B_a$. Now, the same sheet is bent so as to have a cross section in the shape of a quarter of a circle. It is given a current I/2 which distributes uniformly on its cross section. O is the centre of curvature of the middle cross section of the conductor. Field at O has magnitude $B_{b.}$.

 

Consider long thin wires 1 and 2, at symmetrical locations with respect to O. Field due to 1 and 2 cancel out in y direction but add in x direction.

$\therefore {\mathrm{B}}_a\text{ is along the extreme radius OA. }$

${\mathrm{B}}_a=2{\int }_0^{\mathrm{\pi }/2} {\mathrm{dB}}_1\sin \mathrm{\theta }$ (i)

When the sheet is bent to produce a quarter circle section, radius will be double. Since current has been halved, the value of current through an element indentical to the one
considered above will be same. Since the thin wire is now at 2R distance from O, value of dB will be half of that in equation.

(i) 

${\mathrm{B}}_{\mathrm{x}}={\int }_0^{\mathrm{\pi }/2} dB \sin \mathrm{\theta }{\text{ =∫}}_0^{\frac{\mathrm{\pi }}{2}}\frac{d{B}_1 }{2}\text{sin θ⇒ }{\mathrm{B}}_{\mathrm{x}}=\frac{{\mathrm{B}}_a}{4}$from symmetry of the situation, B_x and B_y cannot be different

$\therefore {\mathrm{B}}_{\mathrm{y}}={\mathrm{B}}_{\mathrm{x}}=\frac{{\mathrm{B}}_a}{4};\therefore {\mathrm{B}}_{\mathrm{b}}=\frac{\sqrt{2}{\mathrm{B}}_a}{4}=\frac{{\mathrm{B}}_a}{2\sqrt{2}}$

Obviously, B makes 45º with the x direction.