Question

The domain in ferromagnetic material is in the form of a cube of side 2 mum. Number of atoms in that domain is \(9 \times 10^{10}\) and each atom has a dipole movement of \(9 \times 10^{-24} \, \text{Am}^2\). The magnetisation of the domain is (approximately).

• A. \( 10 \times 10^4 \, \text{Am}^{-1} \)
• B. \( 8 \times 10^4 \, \text{Am}^{-1} \)
• C. \( 12 \times 10^4 \, \text{Am}^{-1} \)
• D. \( 9 \times 10^4 \, \text{Am}^{-1} \)

Hint:

The magnetisation is calculated by dividing the total dipole moment by the volume of the domain.

Answer 1

The Correct Option is A

Magnetisation \( M \) is defined as: \[ M = \frac{\text{Total Dipole Moment}}{\text{Volume of the Domain}} \] Total dipole moment is: \[ \text{Total Dipole Moment} = (\text{Number of atoms}) \times (\text{Dipole moment of each atom}) = (9 \times 10^{10}) \times (9 \times 10^{-24}) = 8.1 \times 10^{-13} \, \text{Am} \] The volume of the domain is: \[ V = (\text{side})^3 = (2 \times 10^{-6})^3 = 8 \times 10^{-18} \, \text{m}^3 \] Thus, the magnetisation is: \[ M = \frac{8.1 \times 10^{-13}}{8 \times 10^{-18}} = 10 \times 10^4 \, \text{Am}^{-1} \]