A domain in ferromagnetic iron is in the form of a cube of side length 1µm. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The molecular mass of iron is 55 g/mole and its density is `7.9g//cm^(3)`. Assume that each iron atom has a dipoe moment of `9.27xx10^(-24)A m^(2)`.

Explanation

The volume of the cubic domain is
`V=(10^(-6) m)^(3) =10^(-12) cm^(3)`
`mass =volume ×x density =79g cm^(-3) x 10^(-12) cm^(3) = 7.9 xx 10^(-12)`g
It is given that Avagadro number `(6.023 xx 10^(23))` of iron atoms have a mass of 55 g. Hence, the number of atoms in the domain is
`N=7.9xx10^(-12)xx6.023xx10^(23)/55=8.65xx10^(10)atoms`
The maximum possible dipole moment `M_(max)` is achieved for the (unrealistic) case when all the atomic moments are perfectly aligned. Thus
`M_(max)=(8.65xx10^(10))xx(9.27xx10^(-24))=8.0xx10^(-13)Am^(2)`
The consequent magnetisation is `I_(max) =M_(mass)// Domain` volume`=8.0xx10^(5)Am^(-1)`