Let the magnetic field on the earth be modelled by that of a point magnetic dipole at the centre of the earth. The angle of dip at a point on the geographical equator
A proton has spin and magnetic moment just like an electron. Why then its effect is neglected in magnetism of materials?
The comparison between the spinning of a proton and an electron can be done by comparing their magnetic dipole moment which can be given by
$$M=\frac{e h}{4 \pi m} \text { or } M \propto \frac{1}{m} \quad\left(\because \frac{e h}{4 \pi}=\text { constant }\right)$$
$$\begin{aligned} & \therefore \quad \frac{M_p}{M_e}=\frac{m_e}{m_p} \\ & =\frac{M_e}{1837 M_e} \quad\left(\because M_p=1837 m_e\right) \\ & \Rightarrow \quad \frac{M_p}{M_e}=\frac{1}{1837} \ll<1 \\ & \Rightarrow \quad M_p \ll M_e \end{aligned}$$
Thus, effect of magnetic moment of proton is neglected as compared to that of electron.
A permanent magnet in the shape of a thin cylinder of length 10 cm has $M=10^6 \mathrm{~A} / \mathrm{m}$. Calculate the magnetisation current $I_M$.
$\begin{aligned} \text { Given, } M \text { (intensity of magnetisation) } & =10^6 \mathrm{~A} / \mathrm{m} \\ l(\text { (length) } & =10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m}\end{aligned}$
$$\begin{aligned} & \text { and } \quad I_M=\text { magnetisation current } \\ & \text { We know that } \quad M=\frac{I_M}{l} \\ & \Rightarrow \quad I_M=M \times l \\ & =10^6 \times 0.1=10^5 \mathrm{~A} \end{aligned}$$
Explain quantitatively the order of magnitude difference between the diamagnetic susceptibility of $\mathrm{N}_2\left(\sim 5 \times 10^{-9}\right)$ (at STP) and $\mathrm{Cu}\left(\sim 10^{-5}\right)$.
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 .
From molecular view point, discuss the temperature dependence of susceptibility for diamagnetism, paramagnetism and ferromagnetism.
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.