Essential difference between electrostatic shielding by a conducting shell and magnetostatic shielding is due to
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 .