A square loop of side 10cm and resistance 0 . 7Ω is placed...

The area vector for the square loop is given by

\overset{\to}{A} = ( 0 . 1{)}^{2}\overset{^}{j}{m}^{2}

According to the above diagram, the magnetic field is given by

\overset{\to}{B} = \dfrac{0 . 2}{\sqrt{2}}\overset{^}{i} + \dfrac{0 . 2}{\sqrt{2}}\overset{^}{j}

Hence, the magnitude of induced emf can be calculated as follows

\begin{cases}e & = & \dfrac{\Deltaϕ}{\Deltat} \\ & = & \dfrac{\overset{\to}{B} \cdot \overset{\to}{A} - 0}{1} \\ & = & \dfrac{\left(\dfrac{0 . 2}{\sqrt{2}}\overset{^}{i} + \dfrac{0 . 2}{\sqrt{2}}\overset{^}{j}\right) \cdot \left(( 0 . 1{)}^{2}\overset{^}{j}\right) - 0}{1}V \\ & = & \sqrt{2} \times {10}^{- 3}V\end{cases}

Hence, $x = 2$.