= Solution
Assume uniform charge per unit <length> and the same angular <speed> $\Omega$. A charge element at distance $r$ has $\mathbf v=\boldsymbol\Omega\times\mathbf r$, so the <Biot-Savart law> gives the <centre field of a rigidly rotating charged loop> as $dB=\mu_0\Omega\,dq/(4\pi r)$. For one square side, with $b=a/2$,
$$
\int_{-b}^{b}\frac{dx}{\sqrt{x^2+b^2}}=2\ln(1+\sqrt2).
$$
Using $dq=Q\,dx/(4a)$ on each side gives
$$
B_{\square}=\frac{\mu_0Q\Omega}{2\pi a}\ln(1+\sqrt2),\qquad
B_{\circ}=\frac{\mu_0Q\Omega}{2\pi a}.
$$
Thus \b[$B_{\square}/B_{\circ}=\ln(1+\sqrt2)\simeq0.8814$]: the circular field is larger by $1/\ln(1+\sqrt2)\simeq1.135$. This is a rigidly rotating charge distribution, not a current constrained to the sides of a stationary square.
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