Question

Each atom of an iron bar of size$(5\text{cm}\times 1\text{cm}\times 1\text{cm})$ has a magnetic moment $1.8\times {10}^{-23}{\text{Am}}^2$. What will be the magnetic moment of the bar in the state of magnetic saturation?
$[{\rho }_{\text{iron}}=7.8\times {10}^3{\text{kgm}}^{-3},A_{\text{atomic weight}}=56]$
$[N_{\text{Avogadros Number}}=6.02\times {10}^{23}{\text{mol}}^{-1}]$

• A. $7.54{\text{Am}}^2$
• B. $6.54{\text{Am}}^2$
• C. $5.54{\text{Am}}^2$
• D. $4.54{\text{Am}}^2$

Answer 1

The correct option is A $7.54{\text{Am}}^2$

Given:
$V=(5\times 1\times 1)=5{\text{cm}}^3=5\times {10}^{-6}{\text{m}}^3$
$M=1.8\times {10}^{-23}{\text{Am}}^2$
${\rho }_{\text{Iron}}=7.8\times {10}^3{\text{kgm}}^{-3}$
$A_{\text{atomic weight}}=56\text{g}=56\times {10}^{-3}\text{kg}$
$N_{\text{Avogadros Number}}=6.02\times {10}^{23}{\text{mol}}^{-1}$

The total number of atoms in a specimen is,

$n=\frac{{\rho }_{\text{iron}}V}{A_{\text{atomic weight}}}{N}_{\text{Avogadros Number}}$

$n=\frac{7.8\times {10}^3\times 5\times {10}^{-6}\times 6.02\times {10}^{23}}{56\times {10}^{-3}}$

$n=4.19\times {10}^{23}$

Saturated magnetic moment
$\text{=Magnetic moment of each atom}\times \text{total number of atoms in the ba}$

$=1.8\times {10}^{-23}\times 4.19\times {10}^{23}$

$=7.54{\text{Am}}^2$

Hence, option $\left(A\right)$ is correct.