# Curl of a function of radius-vector [closed]

Good morning, Everybody!

I want to calculate magnetic field from vector potential explicitly for common case in order to make some approximations. Under sign of integral I have that equation

$\frac{\partial }{{\partial {\mkern 1mu} \vec R}} \times \frac{{\vec J(\vec r,\vec R)}}{{\left| {\vec R - \vec r} \right|}} = \det \left( {\begin{array}{*{20}{c}}{\vec i}&{\vec j}&{\vec k}\\{\frac{\partial }{{\partial X}}}&{\frac{\partial }{{\partial Y}}}&{\frac{\partial }{{\partial Z}}}\\{\frac{{{J_X}(\vec r,\vec R)}}{{\left| {\vec R - \vec r} \right|}}}&{\frac{{{J_Y}(\vec r,\vec R)}}{{\left| {\vec R - \vec r} \right|}}}&{\frac{{{J_Z}(\vec r,\vec R)}}{{\left| {\vec R - \vec r} \right|}}}\end{array}} \right)$

Let's consider the ${\vec i }$- component:

$\frac{\partial }{{\partial Y}}\frac{{{J_Z}(\vec r ,\vec R )}}{{\left| {\vec R - \vec r } \right|}} - \frac{\partial }{{\partial Z}}\frac{{{J_Y}(\vec r ,\vec R )}}{{\left| {\vec R - \vec r } \right|}}$

Now we are going to look just at the derivative of $Y$: $\frac{\partial }{{\partial Y}}\frac{{{J_Z}(\vec r ,\vec R )}}{{\left| {\vec R - \vec r } \right|}} = \frac{1}{{\left| {\vec R - \vec r } \right|}}\frac{\partial }{{\partial Y}}{J_Z}(\vec r ,\vec R ) + {J_Z}(\vec r ,\vec R )\frac{\partial }{{\partial Y}}\frac{1}{{\left| {\vec R - \vec r } \right|}} = \frac{1}{{\left| {\vec R - \vec r } \right|}}\frac{{\partial {J_Z}(\vec r ,\vec R )}}{{\partial \vec R }}\frac{{\vec R }}{{\partial Y}} + {J_Z}(\vec r ,\vec R )\left( { - \frac{{\vec R - \vec r }}{{{{(\vec R - \vec r )}^3}}}} \right)\frac{{\vec R }}{{\partial Y}}$

But when this expression is multiplied by ${\vec i }$

we get $\vec i \cdot \frac{{\partial \vec R }}{{\partial Y}} = \vec i \cdot \vec j = 0$

Why do I get that magnetic field is zero for any vector potential?