Question

A domain in ferromagnetic iron is in the form of a cube of side length 2$\mu$ m then the number of iron atoms in the domain are (Molecular mass of iron = $55gmo{l}^{-1}$ and density = $7.92gc{m}^{-3}$), maximum value of magnetisation of the given domain is (Dipole moment of an iton atom $9.27\times {10}^{-24}A{m}^2)$

• A. $8.0\times {10}^{5}A{m}^{-1}$
• B. $6.0\times {10}^{4}A{m}^{-1}$
• C. $8.0\times {10}^{3}A{m}^{-1}$
• D. $6.0\times {10}^{3}A{m}^{-1}$

Answer 1

The correct option is A $8.0\times {10}^{5}A{m}^{-1}$

Net dipole moment, $M_N=N\times m=6.92\times {10}^{11}\times 9.27\times {10}^{-24}=6.4\times {10}^{-12}A{m}^2$
Net magnetisation =$\frac{6.4\times {10}^{-12}}{8\times {10}^{-18}}=8\times {10}^{5}A{m}^{-1}$