Magnetism and Matter

Charge on $\alpha$-particle, $q=2 e$

Time period of $\alpha$-particle

$T=\frac{2 \pi r}{v}$

$\therefore$ Current associated by $\alpha$-particle,

$I=\frac{q}{T}=\frac{2 e}{\frac{2 \pi r}{v}} \Rightarrow I=\frac{e v}{\pi r}$ ..... (i)

Magnetic dipole moment,

$\begin{aligned} & \mu=I \cdot A=\frac{e v}{\pi r} \cdot \pi r^2 \\ & \mu=e v r \end{aligned}$

When paramagnetic substances are kept in a external magnetic field, then it develops feeble magnetisation in the direction of external applied magnetic field i.e., magnetic dipoles are aligned along external magnetic field weakly. Hence, both assertion and reason are true and reason is the correct explanation of assertion.

Vertical component of earth’s magnetic field,

$\begin{aligned} B_V & =B_H \tan \delta \\ & =0.36 \times 10^{-4} \times \tan 60^{\circ} \\ & =0.36 \times 10^{-4} \times \sqrt{3} \\ & =0.622 \times 10^{-4} \mathrm{~T} \\ & =0.622 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \end{aligned}$

$T=2 \pi \sqrt{\frac{I}{M B}}$

or $T^{\prime}=2 \pi \sqrt{\frac{I}{m V^{\prime}}}$

$B=$ Total earth's field intensity

$V=$ Vertical component of $B$

$\begin{array}{rlrl} \frac{T^{\prime}}{T} & =\sqrt{\frac{B}{V}} \\ \Rightarrow \quad & \frac{\left(T^{\prime}\right)^2}{T^2} =\frac{B}{V} \\ & \text { but } \quad V=B \sin \theta \\ \Rightarrow \quad & \frac{\left(T^{\prime}\right)^2}{T^2} =\frac{1}{\sin \theta} \\ \Rightarrow \quad & \sin \theta =\frac{T^2}{\left(T^{\prime}\right)^2} \\ T & =\frac{60}{40}=\frac{3}{2} \mathrm{~s} \end{array}$

Also, and $T^{\prime}=\frac{60}{30}=2 \mathrm{~s}$

$\begin{aligned} & \therefore \quad \sin \theta=0.5625 \\ & \Rightarrow \quad \theta=\sin ^{-1}(0.5625) \\ \end{aligned}$

We know the magnetism of the magnet is achieved due to the spin motion of electrons. Also the spinning electrons possess magnetic dipole moment. This dipole moment is much greater than that due to orbital moment of electrons around the nucleus.