A conducting square frame of side a and a long straight wire carrying current I are located in the same…

The potential difference across AB is
$V_A-V_B=B_1 a.V.$
$\Rightarrow \frac{{\mu }_i}{2\pi \left(x-\frac{a}{2}\right)} aV$
The potential difference across CD is
$V_C-V_D=B_2 a.V$
$B_2=\frac{{\mu }_{0}i}{2\pi \left(x+\frac{a}{2}\right)}$
$V_C-V_D= \frac{{\mu }_{0}i}{2\pi \left(x+\frac{a}{2}\right)} aV$
Net Potential difference $=\frac{{\mu }_i aV}{2 \pi } \left(\frac{1}{x - \frac{a}{2}}- \frac{1}{x + \frac{a}{2}}\right)$
$\left(V_A-V_B\right)-\left(V_C-V_D\right)=\frac{\mu ia}{2\pi }\left(\frac{2a}{x^2-\frac{a^2}{4}}\right)\propto \frac{1}{4x^2-a^2} \propto \frac{1}{\left(2x+a\right)(2x-a)}$