$\begin{aligned} \text { Given, } M \text { (intensity of magnetisation) } & =10^6 \mathrm{~A} / \mathrm{m} \\ l(\text { (length) } & =10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m}\end{aligned}$

$$\begin{aligned} & \text { and } \quad I_M=\text { magnetisation current } \\ & \text { We know that } \quad M=\frac{I_M}{l} \\ & \Rightarrow \quad I_M=M \times l \\ & =10^6 \times 0.1=10^5 \mathrm{~A} \end{aligned}$$

We know that

Density of nitrogen $\rho_{\mathrm{N}_2}=\frac{28 \mathrm{~g}}{22.4 \mathrm{~L}}=\frac{28 \mathrm{~g}}{22400 \mathrm{cC}}$

Also, density of copper $\rho_{\mathrm{Cu}}=\frac{8 \mathrm{~g}}{22.4 \mathrm{~L}}=\frac{8 \mathrm{~g}}{22400 \mathrm{cc}}$

Now, comparing both densities

$$\frac{\rho_{\mathrm{N}_2}}{\rho_{\mathrm{Cu}}}=\frac{28}{22400} \times \frac{1}{8}=1.6 \times 10^{-4}$$

Also given $\quad \frac{\chi_{\mathrm{N}_2}}{\chi_{\mathrm{Cu}}}=\frac{5 \times 10^{-9}}{10^{-5}}=5 \times 10^{-4}$

$$\begin{aligned} \text { We know that, } \quad \chi & =\frac{\text { Magnetisation }(M)}{\text { Magnetic intensity }(H)} \\ & =\frac{\text { Magnetic moment }(M) / \text { Volume }(V)}{H} \\ & =\frac{M}{H V}=\frac{M}{H \text { (mass / density) }}=\frac{M \rho}{H m} \end{aligned}$$

$$\begin{aligned} & \therefore \quad \chi \propto \rho \quad\left(\because \frac{M}{H m}=\text { constant }\right) \\ & \text { Hence, } \quad \frac{\chi_{\mathrm{N}_2}}{\chi_{\mathrm{Cu}}}=\frac{\rho_{\mathrm{N}_2}}{\rho_{\mathrm{Cu}}}=1.6 \times 10^{-4} \end{aligned}$$

Thus, we can say that magnitude difference or major difference between the diamagnetic susceptibility of $\mathrm{N}_2$ and Cu .

Susceptibility of magnetic material $\chi=\frac{I}{H}$, where $I$ is the intensity of magnetisation induced in the material and $H$ is the magnetising force.

Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the susceptibility $\chi$ of diamagnetic material is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature is raised, the alignment is disturbed, resulting decrease in susceptibility of both with increase in temperature.

When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.

(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

Let us draw the figure for given situation,

We have to prove that $\boldsymbol{\delta B} \cdot d \mathbf{S}=0$. This is called Gauss's law in magnetisation.

According to question,

Magnetic moment of dipole at origin $O$ is

$$\mathrm{M}=M \hat{k}$$

Let $P$ be a point at distance $r$ from $O$ and $O P$ makes an angle $\theta$ with $z$-axis. Component of $M$ along $O P=M \cos \theta$.

Now, the magnetic field induction at $P$ due to dipole of moment $\mathrm{M} \cos \theta$ is

$$\mathbf{B}=\frac{\alpha_0}{4 \pi} \frac{2 M \cos \theta}{r^3} \hat{\mathbf{r}}$$

From the diagram, $r$ is the radius of sphere with centre at $O$ lying in $y z$-plane. Take an elementary area $d \mathbf{S}$ of the surface at $P$. Then,

$$\begin{aligned} d \mathbf{S} & =r(r \sin \theta d \theta) \hat{\mathbf{r}}=r^2 \sin \theta d \theta \hat{\mathbf{r}} \\ \oint \cdot d \mathbf{S} & =\oint \frac{\alpha_0}{4 \pi} \frac{2 M \cos \theta}{r^3} \hat{\mathbf{r}}\left(r^2 \sin \theta d \theta \hat{\mathbf{r}}\right) \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r} \int_0^{2 \pi} 2 \sin \theta \cdot \cos \theta d \theta \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r} \int_0^{2 \pi} \sin 2 \theta d \theta \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r}\left(\frac{-\cos 2 \theta}{2}\right)_0^{2 \pi} \\ & =-\frac{\alpha_0}{4 \pi} \frac{M}{2 r}[\cos 4 \pi-\cos 0] \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{2 r}[1-1]=0 \end{aligned}$$