Physics · Electrostatics

JEE Main 2025 — 8 April, Evening Shift — Question 58

An infinitely long wire has uniform linear charge density $\lambda=2 \mathrm{nC} / \mathrm{m}$ . The net flux through a Gaussian cube of side length $\sqrt{3} \mathrm{~cm}$ , if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be $x \mathrm{Nm}^{2} \mathrm{C}^{-1}$ , where $x$ is[0pt] [Neglect any edge effects and use $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9}$ SI units]

Option A:
$0.72 \pi$
Option B:
$2.16 \pi$
Correct
Option C:
$1.44 \pi$
Option D:
$6.48 \pi$

Answer: B

Step-by-step solution

$\phi=\frac{q_{\text {enclosed }}}{\varepsilon_{0}} \quad I=\sqrt{3} a=3$

$q=\lambda l$ $\phi=\frac{\lambda I}{\varepsilon_{0}}=2 \times 10^{-9} \times 3 \times 10^{-2} \times 4 \pi \times 9 \times 10^{9}$

$=2.16 \pi$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 8 April, Evening Shift
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law

Answer: B

Step-by-step solution

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