Question

A domain in ferromagnetic

iron is in the form of cube of side 1 µm.

Estimate the number of iron atoms in the

domain, maximum possible dipole moment

and magnetisation of the domain. The

molecular mass of iron is 55 g/mole and

density is 7.9 g/cm3

. Assume that each iron

atom has a dipole moment of 9.27 × 10-24

Am2​

Answer 1

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Solution: The volume of the cubic domain

V = (10-6)3

= 10-18 m3

= 10-12 cm3

mass = volume × density

= 7.9 × 10-12 g.

An Avogadro number (6.023 x 1023) of ironatoms has mass 55 g.The number of atoms in the domain N

N 7 9 10 6 023 10/55

8 56 1012 2310

The maximum possible dipole moment mmax

is achieved for the case when all the atomic

moments are perfectly aligned (though this

will not be possible in reality).

mmax = 8.65 × 1010 × 9.27 × 10-24

= 8 × 10-13 Am2

Magnetisation M = mmax / domain volume

= 8 × 105 Am-1

$\huge\star\mathfrak\blue{{Follow\: me-}}$

Answer 2

Here, length of cubic domain

=l=10−4m

Volume of domain =V=l3=(10−4)3

=10−12m3=10−6cm3

Mass of domain =Volumexxdensity

=10−6×7⋅9gram.

It is given that 55gram of iron contain 6⋅023×1023 iron atoms (Avogadro's number)

∴ Number of atoms in the domain

=6⋅023×1023×7⋅9×10−655

=8⋅65×1016 atoms.

Maximum possible dipole moment is achieved when all the atomic dipole moments are perfectly aligned (which of course is unrealistic).

∴Mmax=(8⋅65×1016)×9⋅27×10−24

=8⋅0×10−7Am2

Max. Intensity of magnetisation

Imax=MmaxVolume of domain=8⋅0×10−710−12

=8×105Am−1