Q. Each atom of an iron bar $(5cm \times 1cm \times 1cm)$ 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 of $6.02 \times 10^{23}$ 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) $75.4\ Am^{2}$
Explanation
(c) The number of atoms per unit volume in a specimen
$n = \frac{\rho N_{A}}{A}$
For iron, $\rho = 7.8 \times 10^{3}kgm^{- 3},$
$N_{A} = 6.02 \times 10^{26}/kgmol,A = 56$
$\Rightarrow n = \frac{7.8 \times 10^{3} \times 6.02 \times 10^{26}}{56} = 8.38 \times 10^{28}m^{- 3}$
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
$= 4.19 \times 10^{23} \times 1.8 \times 10^{- 23} = 7.54\ Am^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.