A square loop of side 10 cm and resistance 0 . 7 Ω is placed vertically in the east-west plane. A uniform…

The area vector for the square loop is given by

$\overrightarrow{\mathrm{A}}=(0.1{)}^2 \hat{\mathrm{j}} {\mathrm{m}}^2$

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

$\overrightarrow{\mathrm{B}}=\frac{0.2}{\sqrt{2}}\hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}}\hat{\mathrm{j}}$

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

$\begin{array}{rcl}e& =& \frac{\Delta \varphi }{\Delta t}\\ & =& \frac{\overrightarrow{B}·\overrightarrow{A}-0}{1}\\ & =& \frac{\left(\frac{0.2}{\sqrt{2}}\hat{\mathrm{i}}+\frac{0.2}{\sqrt{2}}\hat{\mathrm{j}}\right)·\left((0.1{)}^2 \hat{\mathrm{j}} \right)-0}{1} \mathrm{V}\\ & =& \sqrt{2}\times {10}^{-3} \mathrm{V}\end{array}$

Hence, $x=2$.