An iron rod is placed parallel to magnetic field intensity 2000 A/m. The magnetic flux through rod is 6 10⁻⁴ wb and its cross sectional area is 3 cm ^2 . The magnetic permeability of the rod in wb/A-m is

Step-by-step solution

Given magnetic field intensity H = 2000 A/m Magnetic flux = 6 10⁻⁴ Wb Cross-sectional area A = 3 cm ^2 = 3 10⁻⁴ m ^2 Magnetic induction B is given by B = A B = 6 10⁻⁴ 3 10⁻⁴ = 2 Wb/m ^2 Magnetic permeability is given by = B H = 2 2000 = 10⁻³ Wb/A-m Answer: 10⁻³