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InterviewSolution
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A domain in ferromagnetic iron is in the form of a cube of side length `10^-4m`. 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 `55g//"mole"`, and its density is `7*9g//cm^3`. Assume that each iron atom has a dipole moment of `9*27xx10^(-24)Am^2`. |
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Answer» The volume of the cube domain is `V=(10^(-6) m)^(3)=10^(-18)m^(3)=10^(-12)cm^(3)` Its mass is volumexx density `=7.9g cm^(-3)xx10^(-12)=7.9xx10^(-12)g` it is given that Avagadro number (6.023xx`10^(23)`) of iron atoms have a mass of 55g. Hence, the number of atoms in the domain is `N=(7.9xx10^(-12)xx6.0233xx10^(23))/(55)` the maximum possible dipole moment `m_("max")` is achieved for the (unrealisec) when all the atomic moments are perfectly alligned. Thus, `m_("max")=(8.65xx10^(10))xx(9.27xx10^(24))` `=8.0xx10^(-13)A m^(2)` the consequent magnetisation is `M_("max")=m_("max")//` Domain volume `=8.0xx10^(-113)Am^(2)//10^(-18)m^(3)` `=8.0xx10^(5)Am^(-1)` |
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