In uniform electric field \( \vec{E} = 3\times10^{3}\,\hat{\imath}\ \mathrm{N/C} \), 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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 30\,\mathrm{N\,C^{-1}\,m^{2}} \)
The electric flux through a plane surface in a uniform field is \( \phi = \vec{E}\cdot\vec{A} = EA\cos\theta \), where the area vector \( \vec{A} \) is normal to the surface.
A loop parallel to the yz plane has its normal along the x-axis, the same direction as \( \vec{E} \), so \( \theta = 0^\circ \) and \( \cos\theta = 1 \).
\( A = (0.10\,\mathrm{m})^{2} = 1\times10^{-2}\,\mathrm{m^{2}} \)
\( \phi = 3\times10^{3}\times1\times10^{-2}\times1 = 30\,\mathrm{N\,C^{-1}\,m^{2}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
