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 $= \dfrac{{\mu}_{i}aV}{2\pi} \left(\dfrac{1}{x - \dfrac{a}{2}} - \dfrac{1}{x + \dfrac{a}{2}}\right)<br /><br />$( {V}_{A} - {V}_{B} ) - ( {V}_{C} - {V}_{D} ) = $\dfrac{\muia}{2\pi} ( \dfrac{2a}{x^{2} - \dfrac{a^{2}}{4}} )<br /><br />$∝ $\dfrac{1}{4x^{2} - a^{2}}$∝ $\dfrac{1}{\left(2x + a\right) ( 2x - a )}$