$$
d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}\times \vec{r}}{r^3}.
$$
If we change the $ d\vec{\ell}\times \vec{r} $ with the $ d\vec{\ell}\times$r . sin(Ө) = $ d\vec{\ell}\times$R in order to find the magnitude, then it means magnetic field falls off as the inverse cube of the distance not the inverse square, even the college physics books write it as an inverse square law.
Can we write the Biot-Savart formula with R, like this:
$$
d\vec{B}~=~\frac{\mu_0}{4\pi}\frac{I d\vec{\ell}.R}{r^3}.
$$
Best Answer
The distance on top, $R$ or $r$ effectively cancels the $r^3$ on the bottom to give overall a $r^{-2}$ factor, i.e. inverse square. Often you'll find expressions like this in EM, with things like: $$ {\vec r \over r} = \hat{r} $$ so that $$ {\vec r \over r^3} = {\hat{r} \over r^2} $$