91Ó°ÊÓ

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:

$\mathrm{Δ³¾}=−\frac{{\mathrm{e}}^2{\mathrm{r}}^2}{4{\mathrm{m}}_{\mathrm{e}}}\mathrm{B}$

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

Rewrite the expression as:

$\mathrm{M}=\frac{\mathrm{Δ³¾}}{\mathrm{V}}=−\frac{{\mathrm{e}}^2{\mathrm{r}}^2}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{V}}\mathrm{B}$ ….. (i)

Its been known from equation 6.29 that,

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

Here $\mathrm{χ}$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,

$\mathrm{M}=\frac{{\mathrm{χ}}_{\mathrm{m}}}{{\mathrm{μ}}_0(1+{\mathrm{χ}}_{\mathrm{m}})}$

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

$\begin{array}{c}{\mathrm{χ}}_{\mathrm{m}}=\frac{\mathrm{M}}{\mathrm{H}}\\ =−\frac{{\mathrm{e}}^2{\mathrm{r}}^2\mathrm{B}}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{VH}}\\ =−\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{μ}}_0\mathrm{H}}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{VH}}\\ =−\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{μ}}_0}{4{\mathrm{m}}_{\mathrm{e}}\mathrm{V}}\end{array}$ …..(¾±¾±)

From equation 6.30, it can also be found that,

$\mathrm{B}={\mathrm{μ}}_0\text{​}(\mathrm{H}+\mathrm{M})={\mathrm{μ}}_0\text{​}(1+{\mathrm{χ}}_{\mathrm{m}}\text{​})\mathrm{H}$

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

Let’s say there is a spherical substance,

Its volume (V) is written as:

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

$\begin{array}{c}{\mathrm{χ}}_{\mathrm{m}}=−\frac{{\mathrm{e}}^2{\mathrm{r}}^2{\mathrm{μ}}_0}{4{\mathrm{m}}_{\mathrm{e}}\frac{4}{3}{\mathrm{Ï€°ù}}^3}\\ =−\frac{4\mathrm{Ï€}}{{\mathrm{μ}}_0}\text{​â¶Ä‹}\left(\frac{3{\mathrm{e}}^2}{4{\mathrm{m}}_{\mathrm{e}}\text{​}\mathrm{r}}\text{​}\right)\end{array}$

If we use$r=1^{−10}m$

Then,

$\begin{array}{c}{\mathrm{χ}}_{\mathrm{m}}=−\frac{4\mathrm{Ï€}}{{\mathrm{μ}}_0}\text{​â¶Ä‹}\left(\frac{3{\mathrm{e}}^2}{4{\mathrm{m}}_{\mathrm{e}}\text{​}\mathrm{r}}\text{​}\right)\\ {\mathrm{χ}}_{\mathrm{m}}=−\left({10}^{−7}\right)\left(\frac{3{(1.6×{10}^{−19})}^2}{4(9.1×{10}^{−31})\left({10}^{−10}\right)}\right)\\ {\mathrm{χ}}_{\mathrm{m}}=−2×{10}^{−5}\end{array}$

From table 6.1, ${\mathrm{χ}}_m\text{​}=−1×{10}^{−5}$, 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 $(\mathrm{Δ}m~r^2)$. Here competing paramagnetic materials have been neglected.