A long solenoid has 1000 turns per metre and carries a current of 1 A . It has a soft iron core of $\alpha_r=1000$. The core is heated beyond the Curie temperature, $T_c$.
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 $I_M$.
$\begin{aligned} \text { Given, } M \text { (intensity of magnetisation) } & =10^6 \mathrm{~A} / \mathrm{m} \\ l(\text { (length) } & =10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m}\end{aligned}$
$$\begin{aligned} & \text { and } \quad I_M=\text { magnetisation current } \\ & \text { We know that } \quad M=\frac{I_M}{l} \\ & \Rightarrow \quad I_M=M \times l \\ & =10^6 \times 0.1=10^5 \mathrm{~A} \end{aligned}$$