Answer to Question #85662 in Electricity and Magnetism for Swati Verma

Question

Metallic iron contains approximately 10^(29) atoms per cubic meter.The magnetic moment of each iron atom is 1.8×10^(-23) Am(-1).If all the dipoles were perfectly aligned what would be the magnetisation M and the magnetic moment of a bar 10 cm long and 1 cm^(2) in cross-section?

Answer 1

n= 〖10〗^29 1/m^3

μ_0=1.8×〖10〗^(-23) A/m

l=0.1 m

S=〖10〗^(-4) m^2

The magnetic moment of a bar iron μ=μ_0 N, N=nV →nSl, then μ=μ_0 nSl

magnetization M=μ/Sl,

μ=1.8×〖10〗^(-23)∙1.8×〖10〗^(-23) ∙0.1 ∙〖10〗^(-4)=18 A⁄m

M=18/(0.1∙〖10〗^(-4) )=1.8×〖10〗^6