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An iron atom has a dipole moment of 9xx1...

An iron atom has a dipole moment of `9xx10^(24)Am^2`. Density of iron is `8g//cm^3` and molecular mass is `55g mole^-1`. If domain size of iron is a cube of side `10^-6m` find (i) number of atoms in the domain (ii) maximum dipole moment (iii) intensity of magnetisation.

Text Solution

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Dipole moment of each atom
`=9xx10^(-24)Am^2`
Density of iron `=8g//c c=8xx10^3kg//m^3`
Molecular mass `=55g=55xx10^-3kg`
Volume of domain, `=(1xx10^-6m)^3=10^(-18)m^3`
Mass of domain = Vol. x density
`=10^(-18)xx8xx10^3=8xx10^(-15)kg`
Number of atoms in the domain
`=("mass of domain" xx "Avogado number")/("molecular mass")`
`=(8xx10^(-15)xx6*023xx10^(23))/(55xx10^(-3))=8*76xx10^(10)`
Maximum dipole moment
`=9xx10^(-24)xx8*76xx10^(10)`
`M=7*88xx10^(-13)Am^2`
Intensity of magnetisation,
`I=M/V=(7*88xx10^(-13))/(10^(-18))=7*88xx10^5Am^-1`
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