$(a)$ Obtain an expression for the mutual inductance between a long straight wire and a square loop of side $a$ as shown in Figure.
$(b)$ Now assume that the straight wire carries a current of $50\; A$ and the loop is moved to the right with a constant velocity, $v=10\; m / s$. Calculate the induced $emf$ in the loop at the instant when $x=0.2\; m$. Take $a=0.1\; m$ and assume that the loop has a large resistance.

(N/A) Consider a small element $dy$ in the loop at a distance $y$ from the long straight wire.
The magnetic flux associated with the element $dy$ is given by $d\phi = B dA$.
Here, $dA = a dy$ is the area of the element.
The magnetic field $B$ at distance $y$ is $B = \frac{\mu_0 I}{2\pi y}$.
Thus, $d\phi = \left(\frac{\mu_0 I}{2\pi y}\right) a dy = \frac{\mu_0 I a}{2\pi} \frac{dy}{y}$.
Integrating from $y = x$ to $y = x + a$:
$\phi = \int_{x}^{x+a} \frac{\mu_0 I a}{2\pi} \frac{dy}{y} = \frac{\mu_0 I a}{2\pi} [\ln y]_{x}^{x+a} = \frac{\mu_0 I a}{2\pi} \ln\left(\frac{x+a}{x}\right) = \frac{\mu_0 I a}{2\pi} \ln\left(1 + \frac{a}{x}\right)$.
Since $\phi = MI$, the mutual inductance is $M = \frac{\mu_0 a}{2\pi} \ln\left(1 + \frac{a}{x}\right)$.
$(b)$ The induced $emf$ is given by $e = |\frac{d\phi}{dt}|$. Alternatively, $e = (B_1 - B_2)av$, where $B_1 = \frac{\mu_0 I}{2\pi x}$ and $B_2 = \frac{\mu_0 I}{2\pi (x+a)}$.
$e = \frac{\mu_0 I a v}{2\pi} \left(\frac{1}{x} - \frac{1}{x+a}\right) = \frac{\mu_0 I a v}{2\pi} \left(\frac{a}{x(x+a)}\right)$.
Substituting values: $I = 50\; A$, $x = 0.2\; m$, $a = 0.1\; m$, $v = 10\; m/s$, $\mu_0 = 4\pi \times 10^{-7}\; T\cdot m/A$.
$e = \frac{(4\pi \times 10^{-7}) \times 50 \times 0.1 \times 10}{2\pi} \times \left(\frac{0.1}{0.2(0.2+0.1)}\right) = (2 \times 10^{-7} \times 50) \times \left(\frac{0.1}{0.2 \times 0.3}\right) = 10^{-5} \times \left(\frac{0.1}{0.06}\right) = 1.67 \times 10^{-5}\; V$.