Each atom of an iron bar (5cm x 1cm x 1crn) has a magnetic moment 1.8 x 10-23 Am2. Knowing that the density of iron is 7.78 x103 kgm-3, atomic weight is 56 and Avogadro's number is 6.02 x 1023, the magnetic moment of bar in the state of magnetic saturation will be

The number of atoms per unit volume in a specimen, n=ρNAAFor iron, ρ=7.78 x 103 kgm-3,NA=6.02 x 1026 /kgmol, A=56⇒n=7.78 x 103 x 6.02 x 102656=8.38 x 1028 m-3Total number of atoms in the bar isN0=nV=8.38 x 1028 x 5 x 10-2 x 1 x 10-2 x 1 x 10-2N0=4.19 x 1023The saturated magnetic moment of bar=4.19 x 1023 x 1.8 x 10-23=7.54 Am2