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(B) The wire segments lying on the same line as point $O$ (segments $1$ and $4$) produce zero magnetic field at $O$ because the point $O$ lies on thei

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$\frac{\mu_0 I}{2\pi r} \odot$

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(B) The wire segments lying on the same line as point $O$ (segments $1$ and $4$) produce zero magnetic field at $O$ because the point $O$ lies on their axes.For the two perpendicular segments (segments $2$ and $3$),each acts as a semi-infinite wire at distance $r$ from point $O$.The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_0 I}{4\pi r}$.Using the right-hand rule,both segments $2$ and $3$ produce a magnetic field directed outwards (represented by $\odot$) at point $O$.Therefore,the total magnetic field at $O$ is $B_{total} = B_2 + B_3 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4\pi r} = \frac{2\mu_0 I}{4\pi r} = \frac{\mu_0 I}{2\pi r} \odot$.

Two infinite wires bent in the form of an $L$ shape carry current $I$ as shown. What is the magnetic field at $O$?

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