Past exam of the mathematics course of the University of Cambridge / 2004 / ib / Paper 2 / 18B / Solution

Put $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$. The magnetic field is the curl of the magnetic vector potential. Since $\mathbf{J}({\mathbf{r}}^{\prime })$ does not depend on the differentiated variable $\mathbf{r}$,Taking the curl under the source integral consequently gives the Biot-Savart law,Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable $R^{-2}$ derivative kernel gives the same formula at interior points.

For the loop, choose positive current circulating counterclockwise as viewed from the positive $z$-axis. Parametrize its source position and directed line element byAt $(0,0,z)$ the denominator is the constant $(a^2+z^2{)}^{3/2}$, whileThe transverse terms integrate to zero around the loop. The Biot-Savart law therefore yieldsThe magnetic field is axial, with the stated magnitude for $I>0$; reversing the electric current reverses its direction.