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JEE Main 2025 (8 Apr, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2025 (8 Apr, Shift 2)

Q30JEE Main 2025 · 8 APR · SHIFT 2Medium

Electric Flux & Gauss's Law

An infinitely long wire has uniform linear charge density λ=2\lambda = 2 nC/m. The net flux through a Gaussian cube of side length 3\sqrt{3} cm, if the wire passes through any two corners of the cube, that are maximally displaced from each other, would be xx Nm2C−1{}^{2}\text{C}^{-1}, where xx is:

[Neglect any edge effects and use 14πϵ0=9×109\frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} SI units]

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Answer: (d)

  1. The wire runs along a body diagonal: length inside =3×3= \sqrt3\times\sqrt3 cm =3= 3 cm =0.03= 0.03 m.
  2. q=λl=2×10−9×0.03=6×10−11q = \lambda l = 2\times10^{-9}\times0.03 = 6\times10^{-11} C
  3. ϕ=qε0=q×4π×9×109=6×10−11×36π×109=2.16π\phi = \frac{q}{\varepsilon_0} = q\times4\pi\times9\times10^{9} = 6\times10^{-11}\times36\pi\times10^{9} = 2.16\pi
  4. So the answer is (d).
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