A square frame of side 10cm and a long straight wire carrying...

$dϕ = \dfrac{{\mu}_{0}i}{2\piy}ady$

$ϕ = \dfrac{{\mu}_{0}ia}{2\pi}\mathop{∫}_{x}^{x + a}\dfrac{dy}{y} = \dfrac{{\mu}_{0}ia}{2\pi}\left[\ln \left(x + a\right) - \ln x\right]$

$\text{E.m.f} =- \dfrac{dϕ}{dt} = - \dfrac{{\mu}_{0}ia}{2\pi}\left[\dfrac{1}{x + a}\dfrac{dx}{dt} - \dfrac{1}{x}\dfrac{dx}{dt}\right]$

$= \dfrac{{\mu}_{0}ia}{2\pi}\dfrac{a . v}{x\left(x + a\right)} = \dfrac{{\mu}_{0}}{2\pi} . \dfrac{i{a}^{2}v}{x\left(x + a\right)} = 2 \times {10}^{- 7} \times \dfrac{1 \times \left({0.1}^{2} \times 10\right)}{0.1 \times 0.2} = 1\muV$

Alternative method:

Consider two sides which are perpendicular to the velocity as case of motional e.m.f.

$E = {E}_{1} - {E}_{2} = \left({B}_{1} – {B}_{2}\right)Vl$

$= \dfrac{{\mu}_{0}I}{2\pi}\left(\dfrac{1}{x} - \dfrac{1}{x + a}\right)Vl$

$= 2 \times {10}^{- 7} \times 1\left(\dfrac{1}{0.1} - \dfrac{1}{0.2}\right) \times 10 \times 0.1$

$= {10}^{- 6}V = 1\muV$