$(a)$ 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.

$(a)$ Take a small element dy in the loop at a distance y from the long straight wire (as shown in the given figure).

Magnetic flux associated with element $d y,$

$d \phi=B d A$

Where, $dA =$ Area of element $dy = a dy$

$B =$ Magnetic field at distance y $=\frac{\mu_{0} I}{2 \pi y}$

$I =$ Current in the wire $\mu_{0}=$ Permeability of free space $=4 \pi \times 10^{-7}$

$\therefore d \phi=\frac{\mu_{0} I a}{2 \pi} \frac{d y}{y}$

$\phi=\frac{\mu_{0} I_{a}}{2 \pi} \int \frac{d y}{y}$

y tends from $x$ to $a+x$

$\therefore \phi=\frac{\mu_{0} I a}{2 x} \int\limits_{x}^{a+x} \frac{d y}{y}$

$=\frac{\mu_{0} I_{a}}{2 \pi}\left[\log _{e} y\right]_{x}^{a+x}$

$=\frac{\mu_{0} I a}{2 \pi} \log _{e}\left(\frac{a+x}{x}\right)$

For mutual inductance $M ,$ the flux is given as

$\phi=M I$

$\therefore M I=\frac{\mu_{0} I a}{2 \pi} \log _{e}\left(\frac{a}{x}+1\right)$

$M=\frac{\mu_{0} a}{2 \pi} \log _{e}\left(\frac{a}{x}+1\right)$

$(b)$ $Emf$ induced in the loop,

$e=B$ 'av $=\left(\frac{\mu_{0} I}{2 \pi x}\right) a v$

Given, $I=50 \,A$

$x =0.2\, m$

$a =0.1 \,m$

$v =10 \,m / s$

$e=\frac{4 \pi \times 10^{-7} \times 50 \times 0.1 \times 10}{2 \pi \times 0.2}$

$e=5 \times 10^{-5}\, V$