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(B) The magnetic field at point $O$ due to the straight wire segment lying on the axis passing through $O$ is $B_1 = 0$ because the point $O$ lies on

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$\frac{\mu _0 I}{4\pi r}\left( \sqrt{2} - 1 \right)$

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(B) The magnetic field at point $O$ due to the straight wire segment lying on the axis passing through $O$ is $B_1 = 0$ because the point $O$ lies on the axis of the wire.For the second wire segment, the perpendicular distance $d$ from point $O$ to the wire is $d = r \sin 45^{\circ} = \frac{r}{\sqrt{2}}$.The magnetic field due to a finite wire segment is given by $B = \frac{\mu_0 I}{4 \pi d} (\sin 	heta_1 + \sin 	heta_2)$.Here, one end is at infinity $(	heta_1 = 90^{\circ})$ and the other end makes an angle $	heta_2 = 45^{\circ}$ with the perpendicular.Thus, $B_2 = \frac{\mu_0 I}{4 \pi (r/\sqrt{2})} (\sin 90^{\circ} - \sin 45^{\circ}) = \frac{\mu_0 I \sqrt{2}}{4 \pi r} (1 - \frac{1}{\sqrt{2}}) = \frac{\mu_0 I}{4 \pi r} (\sqrt{2} - 1)$.The total magnetic induction at $O$ is $B = B_1 + B_2 = \frac{\mu_0 I}{4 \pi r} (\sqrt{2} - 1)$.

Find the magnetic induction at point $O$ in the given figure.

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