Magnetic Field From Square Wire

Consider a very thin infinite long wire carries a current $I$ placed along z-axis (into the page). The magnetic field at $\vec{r}_0 = (x_0,y_0)$ is given by $\vec{B} = \frac{\mu_0 I}{2\pi r_{0}^2}(-\hat{k}\times\vec{r}_0)\:.$

Now, let consider the small partition of the wire with area $dA = dx dy$ placed at $\vec{r} = (x,y)$ . Since the current density is uniform along the cross section, then the partition carries current $dI = \frac{I}{A} dA = I dxdy$ . Using the result above, the magnetic field of the partition at $\vec{r}_0 = (x_0,y_0)$ is given by $\begin{aligned} d\vec{B} & = & \frac{\mu_0 dI}{2\pi |\vec{r}_{0} - \vec{r}|^2}\left[-\hat{k}\times(\vec{r}_{0} - \vec{r})\right] \\ & = & \frac{\mu_0 I}{2\pi} \:\frac{(y_{0}-y) \mathrm{i} - (x_{0}-x) \mathrm{j}}{(y_{0}-y)^2 + (x_{0}-x)^2} dxdy\:. \end{aligned}$

With $x_0 = 2 \:\mathrm{m}$ and $y_0 = 2\:\mathrm{m}$ , the x-component of magnetic field, $B_x = \frac{\mu_0 I}{2\pi}\int_{0}^{1}\int_{0}^{1} \:\frac{(y_{0}-y)}{(y_{0}-y)^2 + (x_{0}-x)^2} dxdy = 66.8 \:\mathrm{mT}\:,$ and the y-component of magnetic field $B_y = -\frac{\mu_0 I}{2\pi}\int_{0}^{1}\int_{0}^{1} \:\frac{(x_{0}-x)}{(y_{0}-y)^2 + (x_{0}-x)^2} dxdy = -66.8 \:\mathrm{mT}\:.$

Thus, the magnitude of magnetic field at $\vec{r}_0 = (x_0,y_0) = (2,2) \:\mathrm{m}$ is $B = \sqrt{B_{x}^{2} + B_{y}^{2}} = 94.37 \:\mathrm{mT}\:.$

Think about ampere's law around the center of the square

B 2 π (3/2) root(2) = μ_0 * I

=>B=( μ_0 * I)/(2 π (3/2)*root(2))