Suppose we define the magnetic field as $B(r) = \frac{1}{c}\int_{V}d^3r' \frac{J(r') \times(r-r')}{|r-r'|^3} $
Show that $\nabla \dot{}B=0$
I tried applying divergence theorem.
$\int_{V} \nabla \dot{} B d^3r' = \int_{S} d^2r'n \dot{}B$
Not exactly sure how to find this result, some insight would be helpful.
Here is a hint for you. Write the $(r - r')/\lvert r - r'\rvert^3$ vector field as the gradient of a well-chosen scalar function. Then, noting that the del operator is being taken with respect to the unprimed coordinates, move it under the integral, and using the fact that $\nabla\cdot(A\times C) = C\cdot(\nabla\times A) - A\cdot(\nabla \times C)$, and that the curl of a gradient is always zero, conclude that your integrand is zero. Conclude that $\nabla \cdot B = 0$.