Which one the following options represents the magnetic field $\overrightarrow{B}$ at O due to the current flowing in the given wire segments lying on the xy plane ?

The correct option is:
$\overrightarrow{B}=-\frac{{\mu }_{0}I}{L}\left(1+\frac{1}{4\sqrt{2}\pi }\right)\hat{k}$

Problem Diagram:

Step 1: Breakdown of the Wire Segments
To find the total magnetic field $\overrightarrow{B}$ at point O (the origin), we divide the wire carrying current I into distinct segments:

1. Straight segment along the line connecting to the horizontal axis: Wire segments directed along line-of-sight pointing directly toward or away from point O contribute zero magnetic field because the current element $d\overrightarrow{l}$ and position vector $\overrightarrow{r}$ are parallel or anti-parallel ($d\overrightarrow{l}\times \overrightarrow{r}=0$).
Specifically:
- The horizontal segment of length L/2 lying on the line extending to O has a field of zero at O.
- The horizontal straight segment pointing towards O has a field of zero at O.
- The vertical straight segment below point O of length 3L/4 lies along the line passing through O, so its magnetic field at O is also zero.

2. Finite vertical straight wire segment (left):
Consider the vertical straight wire of length L carrying current I upwards, located at a perpendicular distance d = L to the left of point O.
- One end of this segment is at the level of point O (angle $$₁ = 0^o).
- The lower end makes an angle $$₂ = 45^o with the perpendicular line to point O, since the length is L and the perpendicular distance is L (tan $$₂ = L/L = 1 ⇒ ₂ = 45^o).

Using the Biot-Savart Law formula for a finite straight wire:
$B_{\text{wire}}=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

Substituting $d=L$, ${\theta }_1=0^{\circ }$, and ${\theta }_2={45}^{\circ }$:
$B_{\text{wire}}=\frac{{\mu }_{0}I}{4\pi L}\left(\sin 0^{\circ }+\sin {45}^{\circ }\right)=\frac{{\mu }_{0}I}{4\pi L}\left(0+\frac{1}{\sqrt{2}}\right)=\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}$

By the right-hand thumb rule, current going upwards produces a magnetic field directed into the page at point O (along the $-\hat{k}$ direction):
${\overrightarrow{B}}_{\text{wire}}=-\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}\hat{k}$

3. Semicircular arc (top):
The upper circular segment is a semicircle of radius $R_1=L/2$ carrying current I counterclockwise.
The magnetic field at the center of a semicircle is given by:
$B_{\text{semi}}=\frac{{\mu }_{0}I}{4R_1}$

Substituting $R_1=L/2$:
$B_{\text{semi}}=\frac{{\mu }_{0}I}{4(L/2)}=\frac{{\mu }_{0}I}{2L}$

4. Quarter-circular arc (bottom right):
The lower circular segment is a quarter-circle of radius $R_2=L/4$ carrying current I clockwise.
The magnetic field at the center of a quarter-circle is given by:
$B_{\text{quarter}}=\frac{{\mu }_{0}I}{8R_2}$

Substituting $R_2=L/4$:
$B_{\text{quarter}}=\frac{{\mu }_{0}I}{8(L/4)}=\frac{{\mu }_{0}I}{2L}$

By the right-hand rule, both circular arcs produce magnetic fields directed into the page ($-\hat{k}$ direction):
${\overrightarrow{B}}_{\text{arcs}}=-\left(\frac{{\mu }_{0}I}{2L}+\frac{{\mu }_{0}I}{2L}\right)\hat{k}=-\frac{{\mu }_{0}I}{L}\hat{k}$

Step 2: Total Magnetic Field at Point O
Summing up the contributions from all effective segments:
${\overrightarrow{B}}_{\text{total}}={\overrightarrow{B}}_{\text{arcs}}+{\overrightarrow{B}}_{\text{wire}}$

${\overrightarrow{B}}_{\text{total}}=-\frac{{\mu }_{0}I}{L}\hat{k}-\frac{{\mu }_{0}I}{4\sqrt{2}\pi L}$

Factoring out $-\frac{{\mu }_{0}I}{L}\hat{k}$:
$\overrightarrow{B}=-\frac{{\mu }_{0}I}{L}\left(1+\frac{1}{4\sqrt{2}\pi }\right)\hat{k}$