Biot--Savart law has the form $$ \begin{equation} \mathbf{B} (\mathbf{x}) = \frac{ \mu_0 i }{ 4 \pi } \int \frac{ d\mathbf{l} \times \mathbf{r} }{ \lvert \mathbf{r} \rvert^3 }, \end{equation} $$ where \( d\mathbf{l} \) is a vector whose magnitude is the length of the differential element of the wire in the direction of conventional current; and \( \mathbf{r} = \mathbf{x} - \mathbf{l} \) is the displacement vector from the wire element \( \mathbf{l} \) to the point \( \mathbf{x} \) at which the field is being computed; \( i \) is the magnitude of the current. This could also be written as $$ \begin{equation} \mathbf{B} (\mathbf{x}) = \frac{ \mu_0 i }{ 4 \pi } \int \frac{ d\mathbf{l} \times \hat{ \mathbf{r} } }{ \lvert \mathbf{r} \rvert^2 }, \end{equation} $$ where \( \hat{ \mathbf{r} } \) means the unit vector of that direction.
For a single circular with radius \( R \) loop like this (figure from here), located at \( x-y \) plane. Any point on it has coordinates \( \mathbf{l} = (R\cos \theta, R\sin \theta, 0) \). Then the magnetic field, at point \( \mathbf{x} = (x, y, z) \), it results in is $$ \mathbf{B}(x, y, z) = \int \begin{bmatrix} dB_x \\ dB_y \\ dB_z \end{bmatrix} = \int_{0}^{2\pi} \frac{ \mu_0 i }{ 4 \pi } \frac{R d\theta}{ r^3 } \begin{bmatrix} z \cos \theta \\ z \sin \theta \\ R - x \cos \theta - y \sin \theta \end{bmatrix}, $$ where \( r = \sqrt{ (x - R \cos \theta)^2 + (y - R \sin \theta)^2 + z^2 } \).
The left panel draws magnetic field on \( x-z \) plane, with vertical-axis to be \( z \), horizontal to be \( x \). The right panel draws field on \( (x, y, 0.1 ) \) plane. With with vertical-axis to be \( y \), horizontal to be \( x \).