As the magnetic field will be variable with distance from long straight wire, so the flux through square loop can be calculated by integration.


Let us assume a width $d r$ of the square loop at a distance $r$ from straight wire
$$
\begin{aligned}
B &=\frac{\mu_0 2 I}{4 \pi r} \\
\phi &=B a d r=\frac{\mu_0 2 I}{4 \pi r} a d r
\end{aligned}
$$
Total flux associated with square loop
$$
\begin{aligned}
\phi &=\int d \phi=\frac{\mu_0}{4 \pi} 2 I a \int_x^{x+a} \frac{d r}{r} \\
\text { or } \phi &=\frac{\mu_0}{4 \pi} 2 I d\left[\log _e r\right]_x^{x+a} \\
\text { or } \phi &=\frac{\mu_0}{4 \pi} 2 I a\left[\log _e \frac{x+a}{x}\right] \\
\text { or } \phi &=\frac{\mu_0 I a}{2 \pi} \log _e(1+a / x)
\end{aligned}
$$
(b) The square loop is moving right with a constant speed $v$, the instantaneous flux can be taken as
$$
\begin{aligned}
&\phi=\frac{\mu_0 I a}{2 \pi} \log _e(1+a / x) \\
&\text { Induced Emf } e=-\frac{d \phi}{d t}=-\frac{d \phi}{d x} \frac{d x}{d t}=-v \frac{d \phi}{d x} \\
&e=-\frac{\mu_0 I a v}{2 \pi} \frac{d \log _e(1+a / x)}{d x} \\
&e=-\frac{\mu_0 I a v}{2 \pi} \frac{1}{\left(1+\frac{a}{x}\right)^2}\left[-a / x^2\right] \\
&\text { or } e=\frac{\mu_0}{2 \pi} \frac{a^2 v}{x(x+a)} I \\
&\text { or } e=2 \times 10^{-7} \frac{[0.1]^2 10 \times 50}{0.2[0.2+0.1]}=1.67 \times 10^{-5} \mathrm{~V}
\end{aligned}
$$