We know that the magnitude of the magnetic field due to a current if given by:
[tex]\begin{gathered} B=\frac{\mu_0I}{2\pi r} \\ \text{ where:} \\ \mu_0=4\pi\times10^{-7}\text{ }\frac{H}{m} \\ I\text{ is the current} \\ r\text{ is the distance to the point of interest } \end{gathered}[/tex]To help us calculate the distance we can draw a square:
To find the distance r we need to notice that we have a right triangle, applying the pythagorean theore, we have:
[tex]\begin{gathered} r=\sqrt{(\frac{0.14913}{2})^2+(\frac{0.14913}{2})^2} \\ r=\sqrt{(0.074565)^2+(0.074565)^2} \\ r=0.074565\sqrt{2} \end{gathered}[/tex]This means that the magnitude of the field due to each wire is:
[tex]\begin{gathered} B=\frac{(4\pi\times10^{-7})(2.338)}{(2\pi)(0.074565)} \\ B=4.43\times10^{-6} \end{gathered}[/tex]Now, we need to remember that the magnetic field is a vector, assuming the currents point out of the page and using the right hand rule we have that the direction of each field is given by:
From the diagram we notice that the line fields cancel with each other; let's prove it:
[tex]\begin{gathered} \vec{B}=<4.43\times10^{-6}\cos45,4.43\times10^{-6}\sin45>+<-4.43\times10^{-6}\cos45,4.43\times10^{-6}\sin45> \\ +<-4.43\times10^{-6}\cos45,-4.43\times10^{-6}\sin45>+<4.43\times10^{-6}\cos45,-4.43\times10^{-6}\sin45> \\ =<0,0> \end{gathered}[/tex]Therefore, the field at the center of the square is zero.
Note: We could have conclude the same by symmetry considerations but we prove it here so it is clearer.
