Question

A domain in a magnet is in the form of a cube of side length $1\mu \text{m}$. If it contains $8\times {10}^{10}$ atoms and each atomic dipole has a dipole moment of $9\times {10}^{-24}{\text{A-m}}^2$ and all of this dipoles are aligned in the same direction, then the magnetisation of the domain is:

• A. $7.2\times {10}^5{\text{Am}}^{-1}$
• B. $7.2\times {10}^3{\text{Am}}^{-1}$
• C. $3,6\times {10}^5{\text{Am}}^{-1}$
• D. $1.8\times {10}^3{\text{Am}}^{-1}$

Answer 1

The correct option is A $7.2\times {10}^5{\text{Am}}^{-1}$

We know that magnetisation is given by,

$I=\frac{M_{net}}{\text{volume}}$

Here,

$M_{net}=8\times {10}^{10}\times 9\times {10}^{-24}{\text{A-m}}^2$

$M_{net}=72\times {10}^{-14}{\text{A-m}}^2$

Volume of the cube,

$V=({10}^{-6}{)}^3={10}^{-18}{\text{m}}^3$

$\therefore I=\frac{72\times {10}^{-14}}{{10}^{-18}}=7.2\times {10}^5{\text{Am}}^{-1}$

Hence, option $\left(A\right)$ is the correct answer.