Question

A domain in ferromagnetic iron in the form of cube is having $5\times {10}^{10}$ atoms. If the side length of this domain is $1.5\mu m$ and each atom has a dipole moment of $8\times {10}^{-24}$A $m^2$, then magnetisation of domain is?

• A. $11.8\times {10}^5$A $m^{-1}$
• B. $1.18\times {10}^4$A $m^{-1}$
• C. $11.8\times {10}^4$A $m^{-1}$
• D. $1.18\times {10}^5$A $m^{-1}$

Answer 1

Answer: (D)

$1.18\times {10}^5$A $m^{-1}$

The volume of the cubic domain is
$V=(1.5\times {10}^{-6}m{)}^3=3.38\times {10}^{-18}{m}^3=3.38\times {10}^{-12}c{m}^3$
Number of atoms in domain (N)$=5\times {10}^{10}$atoms
Since each iron atom has a dipole moment(m)
$=8\times {10}^{-24}$A $m^2$
$m_{max}=$total number of dipole moment for all atoms
$=N\times m$
$=5\times {10}^{10}\times 8\times {10}^{-24}$
$=40\times {10}^{-14}=4\times {10}^{-13}$A $m^2$
Now the consequent magnetisation
$M_{max}=\frac{m_{max}}{Domainvolume}=\frac{4\times {10}^{-13}A{m}^2}{3.38\times {10}^{-18}{m}^3}$
$=1.18\times {10}^5$A $m^{-1}$.