To find the magnetic field at the midpoint $O$, we use the magnetic field due to a finite straight current-carrying wire.

Since the figure is not visible here, this standard question usually corresponds to two perpendicular semi-infinite wires meeting at a corner, with point $O$ at equal perpendicular distance $a$ from both segments.

1. Field due to one semi-infinite straight wire

For a finite straight wire, the magnetic field at perpendicular distance $r$ is

For a semi-infinite wire, one angle is $90^\circ$ and the other is $0^\circ$:

Here, $r=a$, so field due to one segment is

2. Contribution of the second semi-infinite wire

The second wire is symmetrically placed at the same distance $a$ from point $O$, so it also produces

Using the right-hand rule, both fields at $O$ are in the same direction, so they add.

3. Net magnetic field at $O$

So the magnitude of magnetic induction is

4. Matching with options

This corresponds to:

• A: $\frac{\mu_0 I}{\pi a}$
• B: $\frac{\mu_0 I}{4\pi a}$
• C: $\frac{\mu_0 I}{2\pi a}$
• D: $0$

Hence, the correct option is

5. Comparison with stored answer

Stored correct answer: A

My derived answer: C

These do not match. The stored answer appears inconsistent with the standard result for two perpendicular semi-infinite wire segments at equal distance $a$ from the midpoint/corner region. Each contributes $\frac{\mu_0 I}{4\pi a}$, so total is $\frac{\mu_0 I}{2\pi a}$, not $\frac{\mu_0 I}{\pi a}$.

If the actual figure has a different geometry, the result could change, but for the usual arrangement corresponding to this question, option C is correct.