JEE Main 2021 · 17 March · morning · Single correct · +4 / −1
A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current of 5A. The magnetic flux density produced by the solenoid is : (permeability of free space = 4π × 10⁻⁷ H/m)
- (a) πT
- (b) 2 × 10⁻³π T
- (c) 10⁻⁴π T
- (d) π/5 T
Answer and solution
Answer (a) πT
Solution
Physics sir · 4 steps
A core inside the solenoid multiplies the field by μᵣ, so use μ₀ μᵣ n I, not just μ₀ n I.
- Field with a core. Inside a long solenoid in air, B = μ₀ n I. A magnetic core raises this by its relative permeability μᵣ.
B = μ₀ μᵣ n I
- Put in numbers. n = 1000 m⁻¹, μᵣ = 500, I = 5 A. First, μᵣ n I = 500 × 1000 × 5 = 2.5 × 10⁶.
- Multiply by mu0. B = 4π × 10⁻⁷ × 2.5 × 10⁶ = 10π × 10⁻¹ = π T.
- So. The magnetic flux density is π T, option (a).
The idea to keep. For a solenoid, B = μ₀ μᵣ n I, with n in turns per metre. Before you multiply, check whether there is a core: if there is, include its μᵣ.
Answer from the NTA final answer key. Solution by PrepWiser.