A square wire loop of .side a and a straight infinite conductor are placed in the same plane with two sides of the square parallel to the conductor. The inductance and resistance of loop are L and R respectively. Now the frame {loop) is rotated by $180°$ about axis $\mathrm{XX}$ as shown. The electric charge that flows in the square loop is 

$\frac{{\mu }_0{\mathrm{I}}_{\mathrm{a}}}{2\pi \mathrm{R}}\ln \left(\frac{2\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\times \alpha \mathit{.}$ Determine the value $\mathrm{\alpha } ?$

Let the flux linked with loop in initial and final configurations are ${\varphi }_{\mathrm{i}}\text{ and }{\varphi }_{\mathrm{f}}$ respectively. 

From line diagram shows initial position of loop while dotted line shows final position of loop. Consider a strip of loop of .width dx at a distance of x from wire as shown in figure. Flux linked with this elemental strip is, 

$\begin{array}{l}\mathrm{d}\varphi =\frac{{\mu }_{0}I}{2\pi x}\times adx\\ {\varphi }_i={\int }_b^{a+b} \frac{{\mu }_{0}I}{2\pi x}adx=\frac{{\mu }_{0}la}{2\pi }\ln \left(\frac{a+b}{b}\right)\\ {\varphi }_{\mathrm{f}}={\int }_{a+b}^{2a+b} \frac{{\mu }_{0}la}{2\pi x}dx\mathit{ }\mathit{cos}\mathit{ }{\mathit{180}}^{\mathit{0}}\\ =-\frac{{\mu }_{0}Ia}{2\pi }\ln \left(\frac{2a+b}{a+b}\right)\end{array}$

By writing $\mathrm{KVL}$ equation fat loop, 

$iR=e-\frac{Ldi}{dt}$ where i is induced current in loop at any time t and e is induced emf. 

$\begin{array}{l}R\left(idt\right)=edt-Ldi\\ \Rightarrow \int \text{ Rdq }=-\int \frac{d\varphi }{dt}dt-L\int di\\ \Rightarrow R\times q=-\left[{\varphi }_f-{\varphi }_i\right]+L\left[i_{initial}-i_{\text{final }}\right]\\ \Rightarrow q=\frac{{\varphi }_i-{\varphi }_f}{R}\text{ as }{i}_{\text{initial }}=i_{\text{final }}=0\\ \text{ So, }q=\frac{{\mu }_{0}Ia}{2\pi R}\ln \frac{\mathit{(}\mathit{2}a+b\mathit{)}}{b}\end{array}$