Question

Each atom of an iron bar (5 cm × 1 cm × 1 cm) 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. $5.74A{m}^2$
• B. $4.75A{m}^2$
• C. $75.4A{m}^2$
• D. $7.54A{m}^2$

Answer 1

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

The number of atoms per unit volume in a specimen,

$n=\frac{\rho N_A}{A}$
For iron, density,

$\rho =7.8\times {10}^{3}kg{m}^{-3},$

$N_A=6.02\times {10}^{23}$

$=56\times {10}^{-3}$

$n=\frac{7.8\times {10}^{3}kg{m}^{-3}\times 6.02\times {10}^{23}/mol}{56\times {10}^{-3}kg/mol}$

$=8.38\times {10}^{28}{m}^{-3}$

Magnetic moment of iron bar in case of magnetic saturation = magnetic moment of each atom × Total no. of atoms in bar
Total number of atoms in the bar is

$N=nV=8.38\times {10}^{28}{m}^{-3}\times (5\times$

${10}^{-2}m\times 1\times {10}^{-2}m\times 1\times {10}^{-2}m)$

$N=4.19\times {10}^{23}$
The saturated magnetic moment of bar

$=4.19\times {10}^{23}\times 1.8\times {10}^{-23}A{m}^2=7.54A{m}^2$
Hence, option (c) is correct.