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 $7.78\times {10}^{3}k{g}^{-3}m$, 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

The correct option is C $7.54A{m}^2$

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