A conducting square frame of side a and a long straight wire...

The potential difference across AB is
${V}_{A} - {V}_{B} = {B}_{1}a . V .$
$⇒ \dfrac{{\mu}_{i}}{2\pi\left(x - \dfrac{a}{2}\right)}aV$
The potential difference across CD is
${V}_{C} - {V}_{D} = {B}_{2}a . V$
${B}_{2} = \dfrac{{\mu}_{0}i}{2\pi\left(x + \dfrac{a}{2}\right)}$
${V}_{C} - {V}_{D} = \dfrac{{\mu}_{0}i}{2\pi\left(x + \dfrac{a}{2}\right)}aV$
Net Potential difference $= \dfrac{{\mu}_{i}aV}{2\pi}\left(\dfrac{1}{x - \dfrac{a}{2}} - \dfrac{1}{x + \dfrac{a}{2}}\right)$
$( {V}_{A} - {V}_{B} ) - ( {V}_{C} - {V}_{D} ) = \dfrac{\muia}{2\pi} ( \dfrac{2a}{{x}^{2} - \dfrac{{a}^{2}}{4}} ) ∝ \dfrac{1}{4{x}^{2} - {a}^{2}} ∝ \dfrac{1}{\left(2x + a\right) ( 2x - a )}$