Question

Each atom of an iron bar $(5\, cm \times 1\, cm \times 1\, cm )$ has a magnetic moment $1.8 \times 10^{-23}\, Am ^{2}$. 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. $4.75\,Am^{2}$
• B. $5.74\,Am^{2}$
• C. $7.54\,Am^{2}$
• D. $75.4\,Am^{2}$

Answer 1

Answer: (C)

$7.54\,Am^{2}$

Answer 2

The number of atoms per unit volume in a specimen,
$n=\frac{\rho N_{A}}{A}$
For iron,
$\rho=7.8 \times 10^{3} \,kgm ^{-3}$
$N_{A}=6.02 \times 10^{26} / kg \,mol , A=56$
$\Rightarrow n=\frac{7.8 \times 10^{3} \times 6.02 \times 10^{26}}{56}$
$n=8.38 \times 10^{28} \,m ^{-3}$
Total number of atoms in the bar is
$N_{0}=n V=8.38 \times 10^{28}\times\left(5 \times 10^{-2} \times 1 \times 10^{-2} \times 1 \times 10^{-2}\right)$
$N_{0}=4.19 \times 10^{23}$
The saturated magnetic moment of bar
$=4.19 \times 10^{23} \times 1.8 \times 10^{-23}=7.54 \,Am ^{2}$