Past exam of the mathematics course of the University of Cambridge / 2026 / ib / Paper 1 / 14D (Electromagnetism)

Using $\nabla |x-x\prime {|}^{-1}=-\nabla \prime |x-x\prime {|}^{-1}$, the divergence is a line integral of a total derivative, zero for a closed wire or suitable vanishing endpoints. Taking the curl and moving it under the integral gives the stated Biot–Savart formula.

The negative-$z$ segment integrates to the stated azimuthal field. The negative-$y$ segment contributesAdding this to the first field is the complete answer. On $x=0,y=1$, the field tends to $0$ as $z\to +\infty$ and to $-{\mu }_{0}I\hat{x}/(2\pi )$ as $z\to -\infty$.