An iron rod is placed parallel to magnetic field intensity 1000 A/m. The magnetic flux through the rod is 3 10⁻⁴ Wb and its cross-sectional area is 1.5 cm ^2 . The magnetic permeability of rod in Wb/ _ A-m is

Step-by-step solution

Given magnetic field intensity H = 1000 A/m. Magnetic flux = 3 10⁻⁴ Wb. Cross-sectional area A = 1.5 cm ^2 = 1.5 10⁻⁴ m ^2 . The magnetic induction B is given by B = A . Substituting the values: B = 3 10⁻⁴ 1.5 10⁻⁴ = 2 Wb/m ^2 . The magnetic permeability is given by = B H . = 2 1000 = 2 10⁻³ Wb/(A-m). Answer: 2 10⁻³