Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 314 / 1 / d / Solution

Hereafter choosing the integration constant so that $\theta (\pm \infty )=0$. The condition $B_z\to 0$ at both ends removes the cosine solution. Up to an overall sign and magnitude, one may writeThe angle ranges from $0$ to $I_{\max }=a^2/(2d^2)$. Neither nonzero component changes sign when the entire range lies in the first quadrant, namely $I_{\max }\le \pi /2$. Thereforefor positive length scales $a,d$. Equality allows $B_y$ to touch zero at $x=0$ without changing sign.