A square frame of resistance $\mathcal{l}n\left(2\right)\Omega$  and $a=20cm$  and a long straight wire carrying a current $\text{I=10A}$  are located in the same plane. The frame is rotated through an angle of  ${120}^0$ about the side PQ. The amount of charge flown through the loop during this time is  $q\times {10}^{-7}$ coulomb, find the value q.

After moving the loop about side PQ through  ${120}^0,$ the normal on the side PS from the wire will be passing through midpoint C and the net magnetic flux through the loop in the new position will be zero
$\therefore$  Changes is flux  $\Delta f=$ Flux through the loop in the initial position.
Flux through area element dA
 
 $d\varphi =\overrightarrow{\text{B}}.\text{d\hspace{0.17em}}\overrightarrow{\text{A}}\frac{{\mu }_{0}I}{2\pi x}a.dx$
 $\varphi =\underset{a}{\overset{2a}{\int }}\frac{{\mu }_{0}Iadx}{2\pi x}=\frac{{\mu }_{0}Ia}{2\pi }\mathcal{l}n\left(2\right)$
$\therefore$  Charge flown through the square loop $=\frac{\Delta \varphi }{R}=\frac{{\mu }_{0}Ia}{2\pi R}1n\left(2\right)$