Question

If the number of atoms in the domain in ferromagnetic iron, in the form of a cube of side length 1μm, is 8.65 X 10^10 atoms and dipole moment of each iron atom is 9.27 X 10^-24 Am^2, what is the maximum Magnetization of the domain?

(a) 6 X 10^5 A/m

(b) 7 X 10^5 A/m

(c) 8 X 10^5 A/m

(d) 9 X 10^5 A/m

The question was asked by my school principal while I was bunking the class.

My doubt stems from Classification of Magnetic Materials in portion Magnetic Properties of Materials of Engineering Physics II

Answer 1

The correct choice is (c) 8 X 10^5 A/m

Best explanation: Now, we know the maximum dipole moment = N X m

Mmax = 8.65 X 10^10 X 9.27 X 10^-24

= 8 X 10^-13 Am^2

Volume = (10^-6)^3 = 10^-18 m^3

Therefore, Magnetization = Mmax/ Volume

= 8 X 10^-13 Am^2/10^-18 m^3

= 8 X 10^5 A/m.