A domain in ferromagnetic iron is in the form of a cube of side length `1mu` m. 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 `7.9 g/(cm)^(3)`. Assume that each iron atom has a dipole moment of `9.27 xx 10^(-24) "Am"^(2)`.

Explanation

(a ) Density of given substance is `rho = (M.)/( V) … (1)`
No. of moles of atoms in above substance is,
`mu= (M.)/( M_0) = (N)/( N_A) rArr M. = (NM_(0) )/( N_A) …(2)`
Where M. = total mass of atoms in volume V
`M_0=` molar mass of atoms
`N=` total no. of atoms in volume V
`N_A=` Avogadro number
From equation (1) and (2),
`rho = (NM_0)/( VN_A)`
`therefore N= (rho VN_A)/( M_0)`
therefore `N= (rho l^(3) N_(A) )/( M_0)`
(For a cube of side length l, its volume is V `=I^(3)`)
`therefore N= ((7.9 xx 10^(3) ) (10^(-6) )^(3) (6.02 xx 10^(23) ))/( (55 xx 10^(-3) ))`
`therefore N= 8.647 xx 10^(10)` atoms
(b) If `overset(to) (m_b)` is the induced magnetic dipole moment of one atomic dipole then when N such atomic dipoles align in same direction then we have maximum possible induced magnetic dipole moment,
`overset(to)((m_b) )_("max") = Noverset(to)( m_b)`
Taking magnitudes,
`(m_b))_("max") = Nm_b`
`therefore (m_b)_("max") = (8.647 xx 10^(10) ) (9.27 xx 10^(-24) )`
`=8.016 xx 10^(-13) "Am"^(2)`
(c) Corresponding maximum magnetisation obtained will be,
`M_("max") = ((m_b)_("max") )/( V)`
`= (8.016 xx 10^(-13) )/( 10^(-18) )" "( because V- l^(3) = (10^(-6) )^(3) = 10^(-18) m^(3) )`
`= 8.016 xx 10^(5) "Am"^(-1)`