A rectangular conducting loop of length 4cm and width 2cm is in the...

The two sides of the rectangular loop which are perpendicular to the wire would not contribute in induced emf.

Angle of velocity vector with x-axis is $tan\alpha = \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}} = \dfrac{1}{\sqrt{3}}$. Therefore, $\alpha = 30 °$.

For parallel sides:

$\overset{\to}{E} = \overset{\to}{B} \times \overset{\to}{v}$

$= \dfrac{{\mu}_{0}I}{2\pix} \times v$

$⇒$ Net emf $= \left({E}_{1}cos{60}^{o} - {E}_{2}cos{60}^{o}\right) \times$width

$= \dfrac{1}{2} \times \dfrac{2}{100} \times \dfrac{{\mu}_{0}Iv}{2\pi}\left[\dfrac{1}{\dfrac{4}{100}} - \dfrac{1}{\dfrac{8}{100}}\right]$

$= \dfrac{1}{100} \times \left(2 \times {10}^{- 7}\right) \times 10 \times v \times \dfrac{100}{8}$

$= 2 . 5v \times {10}^{- 7}$

Now, net EMF $= i \times R$

$⇒ v = \dfrac{\left(10 \times {10}^{- 6}\right) \times 0 . 1}{2 . 5 \times {10}^{- 7}} = 4m{s}^{- 1}$