JEE Main 2021 · 20 July · evening · Single correct · +4 / −1
The magnetic susceptibility of a material of a rod is 499. Permeability in vacuum is 4π × 10⁻⁷ H/m. Absolute permeability of the material of the rod is :
- (a) 4π × 10⁻⁴ H/m
- (b) 2π × 10⁻⁴ H/m
- (c) 3π × 10⁻⁴ H/m
- (d) π × 10⁻⁴ H/m
Answer and solution
Answer (b) 2π × 10⁻⁴ H/m
Solution
Physics sir · 3 steps
Add 1 to the susceptibility to get the relative permeability, then multiply by μ₀. That is all it takes.
- Relative permeability. Susceptibility χ and relative permeability μᵣ are linked by μᵣ = 1 + χ. Here μᵣ = 1 + 499 = 500.
μ = μ₀ μᵣ = μ₀ (1 + χ)
- Put in numbers. μ = 4π × 10⁻⁷ × 500 = 2000π × 10⁻⁷ H/m. Since 2000 × 10⁻⁷ = 2 × 10⁻⁴, this is 2π × 10⁻⁴ H/m.
- So. The absolute permeability of the rod is 2π × 10⁻⁴ H/m, which is option (b).
The idea to keep. Susceptibility measures the extra magnetisation on top of vacuum, so μᵣ = 1 + χ and μ = μ₀ (1 + χ). Always add the 1 before multiplying by μ₀.
Answer from the NTA final answer key. Solution by PrepWiser.