In uniform electric field $\vec E =3×10^3\hat i N/C$, a square loop of side 10 cm is placed such that plane of square loop is parallel to the yz plane. The flux linked with the square loop will be:

$30 NC^{-1} m^2$

The correct answer is Option (2) → $30 NC^{-1} m^2$

The electric flux through a surface is -

$\phi_E=\vec E.\vec A=EA\cos θ$

$=(3×10^3)(0.1)^2(\cos 0°)$

$=30 NC^{-1} m^2$