Figure shows instantaneous position of a square metallic loop of side ‘$a$’ moving with constant velocity $V$ in the magnetic field of infinitely long wire carrying current $i$. Resistance per unit length of square metallic loop is $r$. At this instant

 

a) $B=\frac{{\mu }_{0}i}{2\pi x}$

$d\varphi =BdA=\frac{{\mu }_{0}i}{2\pi x}adx$

(d)  $\varphi =\frac{{\mu }_{0}ia}{2\pi }{\int }_a^{2a} \frac{1}{x}dx=\frac{{\mu }_{0}ia}{2\pi }\mathrm{In}2$

$B=\frac{{\mu }_{0}i}{2\pi x};emfinBC=Blv=\frac{{\mu }_{0}i}{2\pi \left(2a\right)}aV$

Induced current is from C to B, from Lenz's law current in the loop is clockwise direction.

$V_C-V_B=e+iR$

$=\frac{{\mu }_{i}iV}{4\pi }+\frac{{\mu }_{0}iV}{16\pi ar}\left(ar\right)=\frac{5{\mu }_{0}iV}{16\pi }$

b) But $\varphi =MI\Rightarrow M=\frac{{\mu }_{0}a}{2\pi }In2$

c) At the instant, let  Y be distance of AD from long conduction. At that instant

$\varphi =\frac{{\mu }_{0}ia}{2\pi }{\left[Iny\right]}_y^{y+a}=\frac{{\mu }_{0}ia}{2\pi }In\left[1+\frac{a}{y}\right]$

induced emf $e=-\frac{d\varphi }{dt}$

$\Rightarrow ;e=-\frac{{\mu }_{0}ia}{2\pi }\left(\frac{y}{y+a}\right)\left(\frac{-a}{y^2}\right)\frac{dy}{dt}aty=a$

$e=\frac{{\mu }_{0}ia}{2\pi }\times \frac{a}{2a}\times \frac{a}{a^2}V=\frac{{\mu }_{0}iV}{4\pi }$