Question

Each atom of an iron bar $(5cm\times 1cm\times 1cm)$ has a magnetic moment $1.8\times 10^{-23}A{m}^2$. Knowing that the density of iron is $7.78\times 10^{3}kg{m}^{-3}$, atomic weight is $56$ and Avogadro's number is $6.02\times 10^{23}$ the magnetic moment of bar in the state of magnetic saturation will be

• A. $4.75A{m}^2$
• B. $5.74A{m}^2$
• C. $7.54A{m}^2$
• D. $75.4A{m}^2$

Answer 1

Answer: (C)

The number of atoms per unit volume in a specimen,
$n=\frac{\rho N_A}{A}$
For iron,
$\rho =7.8\times 10^{3}kg{m}^{-3}$
$N_A=6.02\times 10^{26}/kgmol,A=56$
$\Rightarrow n=\frac{7.8\times 10^3\times 6.02\times 10^{26}}{56}$
$n=8.38\times 10^{28}{m}^{-3}$
Total number of atoms in the bar is
$N_0=nV=8.38\times 10^{28}\times \left(5\times 10^{-2}\times 1\times 10^{-2}\times 1\times 10^{-2}\right)$
$N_0=4.19\times 10^{23}$
The saturated magnetic moment of bar
$=4.19\times 10^{23}\times 1.8\times 10^{-23}=7.54A{m}^2$