Interpret the geometry

Two identical infinitely long wires are bent at right angles.

• Wire 1 has horizontal part $LP$ along the $x$-axis and vertical part $PS$ parallel to $y$-axis.
• Wire 2 has horizontal part $QM$ along the $x$-axis and vertical part $QN$ parallel to $y$-axis.
• The point $O$ lies on the $x$-axis such that $OP = OQ = 4\text{ cm} = 0.04\text{ m}.$

From the standard bent-wire arrangement, the two bends are symmetrically placed about $O$, so the horizontal segments lie along the same line as $O$.

Field due to horizontal segments at $O$

For any small current element on the horizontal parts $LP$ or $QM$, the vector from the element to $O$ is along the same line as $d\vec l$.

Hence, $d\vec B \propto d\vec l \times \hat r = 0.$

So, horizontal segments contribute zero magnetic field at $O$.

Field due to each vertical semi-infinite segment

Each vertical part ($PS$ and $QN$) is a semi-infinite straight wire at perpendicular distance $r = 0.04\text{ m}$ from $O$.

Magnetic field at distance $r$ due to a semi-infinite straight wire is $B_{\text{semi-inf}} = \frac{\mu_0 I}{4\pi r}.$

Therefore, field due to one vertical segment at $O$ is $B_1 = \frac{\mu_0 I}{4\pi (0.04)}.$

Net magnetic field at $O$

By right-hand rule, the fields due to the two vertical segments are in the same perpendicular direction at $O$, so they add.

Thus,

Solve for $I$

Simplify: $10^{-4} = \frac{4\times 10^{-7} I}{0.08}\times \frac{\pi}{2\pi}$ or directly, $10^{-4} = \frac{4\pi\times 10^{-7} I}{0.08\pi imes 2}?$

Better simplifying carefully:

Direction of magnetic field

Using the right-hand rule for the current directions shown in the figure, the magnetic fields due to both vertical segments at $O$ are directed into the page. Hence the resultant field is also into the page.

Match with options

• Current in each wire $= 20\text{ A}$
• Direction of magnetic field at $O$: perpendicular into the page

Therefore, the correct option is: $\boxed{\text{C}}$