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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
$0.125\,Nm^2/C$
$0.02\,Nm^2/C$
$0.005\,Nm^2/C$
$3.14\,Nm^2/C$
Solution
$\overrightarrow{\mathrm{E}}=\mathrm{E}_{0} \hat{\mathrm{i}}+2 \mathrm{E}_{0} \hat{\mathrm{j}}$
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 $(\phi)=$ $EA \,cos$ $\theta$
$=(100 \hat{\mathrm{i}}+200 \hat{\mathrm{j}}) \cdot 1.25 \times 10^{-3} \mathrm{i} \cos \theta^{\circ}\left[\theta=0^{\circ}\right]$
$=125 \times 10^{-3}\, \mathrm{Nm}^{2} / \mathrm{c}$
$=0.125\, \mathrm{Nm}^{2} / \mathrm{c}$
