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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). |
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Answer» Solution :(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") = (8.647xx 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)` |
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