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InterviewSolution
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A domain in ferromagnetic iron is in the form of a cube of side length `1 mu 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 55g/mole and is density is `7.9 g//cm^(3)`. Assume that each iron atom has a dipole moment of `9.27 xx 10^(-24) A m^(2)`. |
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Answer» 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 Afagadro number `(6.023 xx 10^(23))` of iron atoms have mass of 55g. Hencem the number of atoms in the 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_("max")`/Domain volume `= 8.0 xx 10^(-13) A m^(2)//10^(-18) m^(3)` `= 8.0 xx 10^(5) Am^(-1)`. |
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