JEE Mains 2025 Physics: Gauss Law Cube Flux Problem (8th April Shift 2 Solution)
The JEE Mains 2025 Physics paper included an interesting problem based on Gauss’s Law and Electric Flux. In this post, we’ll solve the 8th April Shift 2 cube diagonal wire problem step by step, and provide a detailed video explanation to help you understand the concept clearly.
🎥 Watch the Full Video Solution
This video explains the complete step-by-step solution. Watch it first before reading the breakdown below.
Question
JEE Mains 2025 Physics – 8th April Shift 2
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 Nm²C⁻¹, where x is:
- (1) 0.72π
- (2) 2.16π
- (3) 1.44π
- (4) 6.48π
[Neglect edge effects and use 1/(4πε₀) = 9 × 10⁹ in SI units]
Detailed Step-by-Step Solution
Step 1: Identify the Concept
This problem involves applying Gauss’s Law to calculate the electric flux through a closed surface when an infinite wire passes through it.
Step 2: Determine the Length of Wire Inside the Cube
The wire passes through the body diagonal of the cube. The length of the body diagonal is:
Ldiagonal = √3 × side
Given side = √3 cm = √3 × 10-2 m:
Linside = √3 × √3 × 10-2 m = 3 × 10-2 m
Step 3: Calculate the Charge Enclosed
Use the linear charge density:
qenclosed = λ × Linside
= (2 × 10-9 C/m) × (3 × 10-2 m) = 6 × 10-11 C
Step 4: Apply Gauss’s Law
The electric flux is:
Φ = qenclosed / ε₀
We know:
1/(4πε₀) = 9 × 109 ⇒ ε₀ = 1 / (36π × 109)
So,
Φ = (6 × 10-11 C) × (36π × 109)
= 216 × π × 10-2 Nm²/C
= 2.16π Nm²/C
Step 5: Final Answer
The correct option is:
Option (2): 2.16π Nm²/C
Conclusion
This problem demonstrates how to use Gauss’s Law to calculate the flux through a cube when an infinite wire passes diagonally through it. Mastering such problems is essential for scoring well in JEE Mains Physics.
