On the basis of the naïve model presented in Sect. 6.1.3, estimate the magnetic susceptibility of a diamagnetic metal such as copper. Compare your answer with the empirical value in Table 6.1, and comment on any discrepancy.

The properties of attraction and repulsion have been produced by moving electric charges or magnetic materials like a bar magnet. When poles come into the vicinity, they repel each other, unlike poles that attract each other.

Consider the formula for the change in dipole moment as:

$\mathrm{\Delta m}=-\frac{{\mathrm{e}}^2{\mathrm{r}}^2}{4{\mathrm{m}}_{\mathrm{e}}}\mathrm{B}$

Now to get Magnetisation (M), we have to divide the above equation by volume (V)

Rewrite the expression as:

$\mathrm{M}=\frac{\mathrm{\Delta m}}{\mathrm{V}}=-\frac{{\mathrm{e}}^2{\mathrm{r}}^2}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{V}}\mathrm{B}$ ….. (i)

Its been known from equation 6.29 that,

$\mathrm{M}={\mathrm{\chi }}_{\mathrm{m}}\mathrm{H}$

Here $\chi$is the magnetic susceptibility, and H is the magnetic field intensity.

On further solving equation 6.29, it is being obtained from equation 6.30 that,

$\mathrm{M}=\frac{{\mathrm{\chi }}_{\mathrm{m}}}{{\mathrm{\mu }}_0(1+{\mathrm{\chi }}_{\mathrm{m}})}$

Now,$\mathrm{B}={\mathrm{\mu }}_0\mathrm{H}$therefore, from equation (i) write as:

$\begin{array}{c}{\mathrm{\chi }}_{\mathrm{m}}=\frac{\mathrm{M}}{\mathrm{H}}\\ =-\frac{{\mathrm{e}}^2{\mathrm{r}}^2\mathrm{B}}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{VH}}\\ =-\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{\mu }}_0\mathrm{H}}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{VH}}\\ =-\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{\mu }}_0}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{V}}\end{array}$ …..(ii)

From equation 6.30, it can also be found that,

$\mathrm{B}={\mathrm{\mu }}_0\text{}(\mathrm{H}+\mathrm{M})={\mathrm{\mu }}_0\text{}(1+{\mathrm{\chi }}_{\mathrm{m}}\text{})\mathrm{H}$

But as ${\chi }_m$<<1, therefore, $\text{}(1+{\chi }_m\text{})$has been reduced to 1, We get $B={\mu }_{0}H$.

Let’s say there is a spherical substance,

Its volume (V) is written as:

Therefore, from equation (ii), we can write:

$\begin{array}{c}{\mathrm{\chi }}_{\mathrm{m}}=-\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{\mu }}_0}{4{\mathrm{m}}_{\mathrm{e}}\frac{4}{3}{\mathrm{\pi r}}^3}\\ =-\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\text{}\left(\frac{3{\mathrm{e}}^2}{4{\mathrm{m}}_{\mathrm{e}}\text{}\mathrm{r}}\text{}\right)\end{array}$

If we use$r=1^{-10}m$

Then,

$\begin{array}{c}{\mathrm{\chi }}_{\mathrm{m}}=-\frac{4\mathrm{\pi }}{{\mathrm{\mu }}_0}\text{}\left(\frac{3{\mathrm{e}}^2}{4{\mathrm{m}}_{\mathrm{e}}\text{}\mathrm{r}}\text{}\right)\\ {\mathrm{\chi }}_{\mathrm{m}}=-\left({10}^{-7}\right)\left(\frac{3{(1.6\times {10}^{-19})}^2}{4(9.1\times {10}^{-31})\left({10}^{-10}\right)}\right)\\ {\mathrm{\chi }}_{\mathrm{m}}=-2\times {10}^{-5}\end{array}$

From table 6.1, ${\chi }_m\text{}=-1\times {10}^{-5}$, However, here it has been used, only one electron per atom and a very crude value for r. As the orbital radius is smaller for the inner electrons, they count for less $(\Delta m~r^2)$. Here competing paramagnetic materials have been neglected.