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

Given, \(\mathrm{E}_{0}=100 \,\mathrm{N} / \mathrm{c}\)

So, \(\overrightarrow{\mathrm{E}}=100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}\)

Radius of circular surface \(=0.02\, \mathrm{m}\)

Area \(=\pi r^{2}=\frac{22}{7} \times 0.02 \times 0.02\)

\(=1.25 \times 10^{-3} \hat{\mathrm{i}}\, \mathrm{m}^{2}\) [Loop is parallel to \(\mathrm{Y}-\mathrm{Z}\) plane ]

Now, flux \((\phi)=\) \(EA \,cos\) \(\theta\)

\(=(100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}) \cdot 1.25 \times 10^{-3} \mathrm{i} \cos \theta^{\circ}\left[\theta=0^{\circ}\right]\)

\(=125 \times 10^{-3}\, \mathrm{Nm}^{2} / \mathrm{c}\)

\(=0.125\, \mathrm{Nm}^{2} / \mathrm{c}\)