To solve this problem, we need to determine the emf induced in the square frame as it moves with velocity $V$ near a long straight wire carrying current $I. \\$The magnetic field $B$ due to a long straight wire at a distance $r$ is given by:$\\ B = \frac{\mu_0 I}{2\pi r} \\$Consider the position of the square frame. The left side of the frame is at a distance $x$ from the wire, and the right side is at a distance $x + a. \\$The emf induced in the frame is due to the change in magnetic flux as it moves. The flux $\Phi$ through the frame is:$\\ \Phi = \int_{x}^{x+a} B \cdot a \, dr = \int_{x}^{x+a} \frac{\mu_0 I}{2\pi r} \cdot a \, dr \\$Calculate the integral:$\\ \Phi = \frac{\mu_0 I a}{2\pi} \int_{x}^{x+a} \frac{1}{r} \, dr \\$The integral of $\frac{1}{r}$ is $\ln(r),$ so:$\\ \Phi = \frac{\mu_0 I a}{2\pi} \left[ \ln(r) \right]_{x}^{x+a} \\ \Phi = \frac{\mu_0 I a}{2\pi} \left( \ln(x+a) - \ln(x) \right) \\ \Phi = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right) \\$The emf $\varepsilon$ induced is the rate of change of flux:$\\ \varepsilon = -\frac{d\Phi}{dt} \\$Since the frame moves with velocity $V,$ we have $\frac{dx}{dt} = V. \\$Differentiate the expression for $\Phi: \\ \varepsilon = -\frac{\mu_0 I a}{2\pi} \cdot \frac{d}{dt} \left( \ln\left(\frac{x+a}{x}\right) \right) \\ \varepsilon = -\frac{\mu_0 I a}{2\pi} \cdot \left( \frac{1}{x+a} \cdot \frac{d(x+a)}{dt} - \frac{1}{x} \cdot \frac{dx}{dt} \right) \\ \varepsilon = -\frac{\mu_0 I a}{2\pi} \cdot \left( \frac{V}{x+a} - \frac{V}{x} \right) \\ \varepsilon = -\frac{\mu_0 I a V}{2\pi} \cdot \left( \frac{x - (x+a)}{x(x+a)} \right) \\ \varepsilon = \frac{\mu_0 I a V}{2\pi} \cdot \frac{a}{x(x+a)} \\$The emf is proportional to $\frac{1}{x(x+a)}. \\$Therefore, the correct option is $\frac{1}{(2x-a)(2x+a)}. \\$This corresponds to Option 3.