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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2025 · 8 April · Evening Shift · Physics · Electrostatics

An infinitely long wire has uniform linear charge density λ=2 nC /...

An infinitely long wire has uniform linear charge density λ=2nC/m\lambda=2 \mathrm{nC} / \mathrm{m}. The net flux through a Gaussian cube of side length 3 cm\sqrt{3} \mathrm{~cm}, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be xNm2C−1\mathrm{xNm}^2 \mathrm{C}^{-1}, where x is :
[Neglect any edge effects and use 14πε0=9×109\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 SI units]An infinitely long wire has uniform linear charge density λ=2 nC / m . The net flux through a Gaussian cube of side

Options

  1. A0.72π0.72 \pi
  2. B1.44π1.44 \pi
  3. C6.48π6.48 \pi
  4. D2.16π2.16 \pi

Correct answer

D. 2.16π2.16 \pi

Step-by-step solution

An infinitely long wire has uniform linear charge density λ=2 nC / m . The net flux through a Gaussian cube of side
ϕ=qencε0=λ⋅3aε0=2×10−9×3×3×10−2×36π×109Nm2C−1=2.16πNm2C−1\begin{aligned} & \phi=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_0}=\frac{\lambda \cdot \sqrt{3} \mathrm{a}}{\varepsilon_0} \\ & =2 \times 10^{-9} \times \sqrt{3} \times \sqrt{3} \times 10^{-2} \times 36 \pi \times 10^9 \mathrm{Nm}^2 \mathrm{C}^{-1} \\ & =2.16 \pi \mathrm{Nm}^2 \mathrm{C}^{-1}\end{aligned}
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