Solution
= Solution
Far above the source, $\mathcal B\sim\beta z$. Suppose $Q\sim z^p$ and $M\sim z^r$. The first integral equation gives $p-1=r-p$, while the second gives $r-1=1+p-r$. Solving,
$$
\boxed{p=\frac43,\qquad r=\frac53}.
$$
Hence
$$
\boxed{Q\sim z^{4/3},\qquad M\sim z^{5/3}}.
$$