Q: An iron bar of length 10 cm and diameter 2 cm is placed in a magnetic field of intensity 1000 Am^{-1} with its length parallel to the direction of the field. Determine the magnetic moment produced in the bar if permeability of its material is 6.3 × 10^{-4} T mA^{-1}

Sol: We know that, μ = μ_{0} (1 + χ)

χ = μ/μ_{0} – 1

χ = (6.3 × 10^{-4})/(4π × 10^{-7} ) – 1 = 500.6

Intensity of magnetisation,

I = χ H = 500.6 × 1000 = 5 × 10^{5} Am^{-1}

∴ Magnetic moment, M = I × V = I × π r^{2} l

= 5 × 10^{5} × 3.14 × (10^{-2})^{2} × (10 × 10^{-2}) = 17.70 A-m^{2}