Answer to Question #93451 in Electricity and Magnetism for MGM

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 1cm^2 in
cross-section?

Answer 1

We can start with magnetization $M$: it is a vector quantity characterizing magnetic state of a macroscopic body. It is defined as magnetic moment within the unit volume:

$$M=nm_{atom}=10^{29}\cdot1.8\cdot10^{-23}=1.8\cdot10^6\text{ A/m}.$$

Magnetic moment of an object is the main quantity which describes magnetic properties of substances:

$$m_{body}=nm_{atom}V=nm_{atom}LA=\\ =MLA=1.8\cdot10^6\cdot0.1\cdot1\cdot10^{-4}=18 \text{ A}\cdot\text{m}^2.$$