A domain in ferromagnetic iron in the form of a cube of side length `1mum`. Estimate the number of iron atoms in the domain and the maximum possible dipole moment and magnetisation of the domain. The atomic mass of iron 55g/mole and its density is `7.9 g/cm^(3)`. Assume that each iron has a dipole moment of `9.27 xx 10^(-24) A m^(2)`.

Explanation

The volume of the cubic domain is `V=(10^(-6) m)^(3)=10^(-18) m^(3)=10^(-12) cm^(3)`
Its mass is volume `xx " density "=7.9 g cm^(-3) xx 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 55g. Hence the number of atoms in hte domain is `N=(7.9 xx 10^(-12) xx 6.023 xx 10^(23))/(55)`
`=8.65 xx 10^(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")=(8.65 xx 10^(10)) xx (9.27 xx 10^(-24))`
`=8.0 xx 10^(-13)Am^(2)`
The consequent magnetisation is
`M_("max")=m_("maxx")//"Domain volume"`
`=8.0 xx 10^(-13) Am^(2)//10^(-18) m^(3)`
`=8.0 xx 10^(5) Am^(-1)`