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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). |
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Answer» SOLUTION :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)` |
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