The electric field in a region of space is given by, $\overrightarrow{E}=E_0\widehat{i}+2E_0\widehat{j}$ where $E_0=100N/C$. The flux of the field through a circular surface of radius 0.02m parallel to the Y – Z plane is nearly

$\overrightarrow{E}=E_0\widehat{i}+2E_0\widehat{j}$

Given, E₀ = 100N/C

So, $\overrightarrow{E}=100\widehat{i}+200\widehat{j}$

Radius of circular surface = 0.02m

Area :  $A=\pi r^2=\frac{22}{7}\times 0.02\times 0.02$

$=1.25\times {10}^{-3}\widehat{i}{m}^2$       [Loop is parallel to Y-Z plane]

Now, flux $\left(\varphi \right)=\overrightarrow{E}·\overrightarrow{A}$

$=(100\widehat{i}+200\widehat{j})·\left(1.25\times {10}^{-3}\widehat{i}\right)$

$=125\times {10}^{-3}N{m}^2/C$

= $\text{0}{\text{.125 Nm}}^{\text{2}}\text{/C}$