Here
$$
q(x)=\frac{a^2x}{(x^2+d^2)^2},
\qquad
\theta(x)=-\frac{a^2}{2(x^2+d^2)}
$$
after choosing the integration constant so that $\theta(\pm\infty)=0$. The condition $B_z\to0$ at both ends removes the cosine solution. Up to an overall sign and magnitude, one may write
$$
B_z=B_0\sin I(x),\qquad
B_y=B_0\cos I(x),\qquad
I(x)=\frac{a^2}{2(x^2+d^2)}.
$$
The 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 $I_{\max}\leq\pi/2$. Therefore
$$
\boxed{\frac d a\geq\frac1{\sqrt\pi}}
$$
for positive length scales $a,d$. Equality allows $B_y$ to touch zero at $x=0$ without changing sign.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-314/1/d/solution