JEE Main 2025 Apr 8 Shift 2, Physics Q7: Gauss's Law

JEE Main2025Apr 8, Shift 2Physics

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

[Neglect any edge effects and use $\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9}$ SI units]

A
$0.72 π$
B
$1.44 π$
C
$6.48 π$
D
$2.16 π$

Solution

The wire enters at one vertex of the cube and exits at the diagonally opposite vertex; the length of wire enclosed is the space diagonal, $ℓ = a 3$ .

Enclosed charge: $q_{enc} = λ \cdot a 3$ .

With $a = 3 \times 1 0^{- 2}$ m, $a 3 = 3 \times 1 0^{- 2}$ m, so $q_{enc} = 2 \times 1 0^{- 9} \times 3 \times 1 0^{- 2} = 6 \times 1 0^{- 11}$ C.

By Gauss's law, $ϕ = \frac{q _{enc}}{ε _{0}} = 4 π \cdot \frac{1}{4 π ε _{0}} \cdot q_{enc} = 4 π \cdot 9 \times 1 0^{9} \cdot 6 \times 1 0^{- 11}$ .

$ϕ = 4 π \times 0.54 = 2.16 π$ Nm $^{2}$ C $^{- 1}$ , matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q7: Gauss's Law

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