Question

A domain in ferromagnetic iron is in the form of a cube of side length $1\mu m$ . The molecular mass of iron is $56g$/$mole$ and its density is $8g$/$c{m^3}$. Assume that each iron atom has a dipole moment of $9.1 \times {10^{ - 23}}A{m^2}$. Take Avogadro number $6 \times {10^{23}}$.
(i)Number of atoms in domain$ = 8.2 \times {10^{12}}$
(ii)Maximum possible dipole moment of the domain$ = 7.8 \times {10^{ - 12}}A{m^2}$
(iii)Maximum magnetisation of domain is$ = 7.8 \times {10^6}A$/$m$
Which of the following options is correct?
A. all are correct
B. only (i) & (ii) are correct
C. only (ii) & (iii) are correct
D. only (i) & (iii) are correct

Answer 1

Hint: We can find the number of atoms in domain using the mass of domain, molecular mass of iron and Avogadro number. Maximum possible dipole moment of the domain ${D_{\max }}$ is achieved when all the atomic dipole moments are perfectly aligned (which is actually not practically possible). We can obtain the maximum magnetisation of domain ${M_{\max }}$ using ${D_{\max }}$ and the volume of the domain $V$.

Complete Step by step answer:
We can use the formula $V = {l^3}$ to calculate the volume of the domain $V$ where $l$ is the length of the side of the cube.
The formula $m = V \times d$ is used to calculate the mass of the domain $m$where $V$ is the volume of the domain and $d$ is the density of the iron atom.
The formula ${M_{\max }} = \dfrac{{{D_{\max }}}}{V}$ is used to calculate the maximum magnetisation of domain ${M_{\max }}$ where ${D_{\max }}$ is the maximum possible dipole moment of the domain and $V$ is the volume of the domain.
The length of the side of the cube $l$ is given as $1\mu m$ which is equal to ${10^{ - 6}}m$.
Using the formula $V = {l^3}$, we get volume $V = {({10^{ - 6}})^3}{m^3}$
$ \Rightarrow V = {10^{ - 18}}{m^3} = {10^{ - 12}}c{m^3}$
The density of the iron atom $d$ is given as $8g$/$c{m^3}$.
Using the formula $m = V \times d$ , we get mass of domain $m = ({10^{ - 12}}) \times 8g$
$ \Rightarrow m = 8 \times {10^{ - 12}}g$
It is given that $56g$of iron contains $6 \times {10^{23}}$ iron atoms (Avogadro number).
$ \Rightarrow $Number of atoms in domain$ = \dfrac{{6 \times {{10}^{23}} \times 8 \times {{10}^{ - 12}}}}{{56}} = 8.6 \times {10^{10}}$ atoms.
$\therefore $ (i) is not correct.
The dipole moment of each iron atom is given as $9.1 \times {10^{ - 23}}A{m^2}$ .
As maximum possible dipole is achieved when all the atomic dipole moments are perfectly aligned,
$ \Rightarrow {D_{\max }} = (8.6 \times {10^{10}}) \times 9.1 \times {10^{ - 23}}A{m^2} = 7.8 \times {10^{ - 12}}A{m^2}$
$\therefore $ (ii) is correct.
Using the formula ${M_{\max }} = \dfrac{{{D_{\max }}}}{V}$ , we get the maximum magnetisation of domain ${M_{\max }} = \dfrac{{7.8 \times {{10}^{ - 12}}}}{{{{10}^{ - 18}}}}A{m^{ - 1}} = 7.8 \times {10^6}A{m^{ - 1}}$.
$\therefore $ (iii) is correct.

Hence, option C is correct (only (ii) & (iii) are correct).

Note: It should be remembered when the maximum dipole will be achieved that is when all the dipole moments are perfectly aligned. Candidates can commit mistakes in writing the proper units.