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(A) The current arrangement consists of two bent wires. The point $O$ is at a perpendicular distance $a$ from the segments $BC$ and $ET$.For a semi-in

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$\frac{\mu_0 I}{2 \pi a}$

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(A) The current arrangement consists of two bent wires. The point $O$ is at a perpendicular distance $a$ from the segments $BC$ and $ET$.For a semi-infinite wire,the magnetic field at a perpendicular distance $r$ from the end is given by $B = \frac{\mu_0 I}{4 \pi r}$.Segments $AB$ and $ED$ are directed towards the corners $B$ and $E$ respectively,and their contribution to the magnetic field at $O$ is zero because the point $O$ lies on the line of these segments.Segments $BC$ and $ET$ contribute to the magnetic field at $O$. Using the right-hand thumb rule,the magnetic field due to current in $BC$ at point $O$ is directed outwards (perpendicular to the plane).The magnetic field due to current in $ET$ at point $O$ is also directed outwards.Since both fields are in the same direction,the total magnetic field $B_0$ is:$B_0 = B_{BC} + B_{ET} = \frac{\mu_0 I}{4 \pi a} + \frac{\mu_0 I}{4 \pi a} = \frac{2 \mu_0 I}{4 \pi a} = \frac{\mu_0 I}{2 \pi a}$.

The magnitude of magnetic induction at the mid-point $O$ due to the current arrangement as shown in the figure will be:

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