Four long, straight wires, each carrying a current of 5.0 A , are...

Let the side length of the square be $a = 5.0 \text{ cm}$ and the current in each wire be $I = 5.0 \text{ A}$.

The center $P$ of the square is at a perpendicular distance of $a/2$ from each of the four long, straight wires.

Using the right-hand grip rule, the direction of the magnetic field produced by each wire at point $P$ can be determined:

The top horizontal wire carries current to the right, producing a magnetic field at $P$ directed into the plane of the paper.

The bottom horizontal wire carries current to the right, producing a magnetic field at $P$ directed out of the plane of the paper.

The left vertical wire carries current upwards, producing a magnetic field at $P$ directed into the plane of the paper.

The right vertical wire carries current upwards, producing a magnetic field at $P$ directed out of the plane of the paper.

The magnitude of the magnetic field due to each long wire at a distance $a/2$ is $B = \dfrac{\mu_0 I}{2\pi (a/2)} = \dfrac{\mu_0 I}{\pi a}$.

Since the magnetic fields from the top and bottom wires are equal in magnitude but opposite in direction, their net contribution at $P$ is zero.

Similarly, the magnetic fields from the left and right wires are equal in magnitude but opposite in direction, so their net contribution is also zero.

Therefore, the total magnetic field at the center $P$ is zero.

Answer: zero