Cross Product for Biot-Savart Derivation of Current Loop

Question

Biot-Savart's law can be used to determine the magnetic field produced by a figure at a point. Introductory physics texts integrate $dB$ to obtain $B$ where

$$dB = \frac{I\mu_{0}}{4\pi r^2} dl \times \frac{r}{\lVert \mathbf{r} \rVert}$$

where $I$ and $\mu_{0}$ are constants.

Here, $dl$ and $r$ ($\frac{r}{\lVert \mathbf{r} \rVert}$ might as well be $r$ for this comparison) make an angle. The angle of interest is formed by the intersection of $r$ (in the diagram) and $dl$. My textbooks claim that the angle is $90$. I don't see how.

Answer 1

take a look at the point were the vector 'r' meets the circular loop....at that point this 'r' is perpendicular to the tangent of that circular loop at that point.... in the law the term 'dl ' is the current carrying element ..... here it's that tangent now it's clear that the angle is 90 ...and it's true for all the points on that loop

Answer 2

Up to a rotation around the $\hat x$ axis, you can always represent the punctual situation as you did in your drawing.

Now: $\hat r$ lies in the plane of the sheet of paper (the one spanned by the radius $\vec R$ and by $\hat x$), while $d\vec l$ is perpendicular to the sheet of paper. Hence, to all the vectors lying in it.