Now, suppose that the angle between $\mathbf{M}$ and $B$ is $\theta$.

Torque on magnetic dipole moment $\mathbf{M}$ in magnetic field $\mathbf{B}$,

$$\tau^{\prime}=M B \sin \theta$$

$$\begin{aligned} &\text { Two motions will be identical, if }\\ &\begin{aligned} & p E \sin \theta =M B \sin \theta \\ \Rightarrow\quad & p E =M B \\ \text { But, } \quad & E =c B\quad \text{.... (i)} \end{aligned} \end{aligned}$$

$\therefore$ Putting this value in Eq. (i),

$$\begin{aligned} p c B & =M B \\ \Rightarrow\quad p & =\frac{M}{c} \end{aligned}$$

Given, $I=$ moment of inertia of the bar magnet

$m=$ mass of bar magnet

$l=$ length of magnet about an any passing through its centre and perpendicular to its length

$M=$ magnetic moment of the magnet

$B=$ uniform magnetic field in which magnet is oscillating, we get time period of oscillation is,

$$\begin{aligned} & T=2 \pi \sqrt{\frac{I}{M B}} \\ \text{Here,}\quad & I=\frac{m l^2}{12} \end{aligned}$$

When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is

$I^{\prime} =\frac{m}{2} \frac{(l / 2)^2}{12}=\frac{m l^2}{12} \times \frac{1}{8}=\frac{I}{8}$

$$\begin{aligned} \text{Magnetic dipole moment}\quad M^{\prime} & =M / 2 \end{aligned}$$

$$\begin{aligned} \text{ Its time period of oscillation is}\quad T^{\prime} & =2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} B}}=2 \pi \sqrt{\frac{I / 8}{(M / 2) B}}=\frac{2 \pi}{2} \sqrt{\frac{I}{M B}} \\ T^{\prime} & =\frac{T}{2} \end{aligned}$$

Consider a magnetic field line of B through the bar magnet as given in the figure below.

The magnetic field line of B through the bar magnet must be a closed loop.

Let $C$ be the amperian loop. Then,

$$\int_Q^P H \cdot d l=\int_Q^P \frac{B}{m_0} \cdot d l$$

We know that the angle between $\mathbf{B}$ and $\mathbf{d l}$ is less than $90^{\circ}$ inside the bar magnet. So, it is positive.

i.e., $$\int_Q^P \mathrm{H} \cdot \mathrm{dl}=\int_Q^P \frac{\mathrm{~B}}{\alpha_0} \cdot \mathrm{dl}>0$$

Hence, the lines of B must run from south pole $(S)$ to north pole $(N)$ inside the bar magnet. According to Ampere's law,

$$\begin{array}{ll} \therefore & \oint_\limits{P Q P} \mathrm{H} . \mathrm{dl}=0 \\ \therefore & \oint_\limits{P Q P} \mathrm{H} \cdot \mathrm{dl}=\int_P^Q \mathrm{H} \cdot \mathrm{dl}+\int_Q^P \mathrm{H} \cdot \mathrm{dl}=0 \\ \text { As } & \int_Q^P \mathrm{H} . \mathrm{dl}>0, \mathrm{so}, \int_P^Q \mathrm{H} . \mathrm{dl}<0 \quad\text{(i.e., negative)} \end{array}$$

It will be so if angle between $\mathbf{H}$ and $\mathbf{d l}$ is more than $90 \Upsilon$, so that $\cos \theta$ is negative. It means the line of H must run from N -pole to S -pole inside the bar magnet.

From $P$ to $Q$, every point on the $z$-axis lies at the axial line of magnetic dipole of moment $\mathbf{M}$. Magnetic field induction at a point distance $z$ from the magnetic dipole of moment is

$$|\mathbf{B}|=\frac{\alpha_0}{4 \pi} \frac{2|\mathbf{M}|}{z^3}=\frac{\alpha_0 M}{2 \pi z^3}$$

(i) Along $z$-axis from $P$ to $Q$.

$$\begin{aligned} \int_P^Q \mathrm{~B} . \mathrm{dl} & =\int_P^Q \mathrm{~B} \cdot \mathrm{dl} \cos 0 \mathrm{Y}=\int_a^R \mathrm{~B} d z \\ & =\int_a^R \frac{\alpha_0}{2 \pi} \frac{M}{z^3} d z=\frac{\alpha_0 M}{2 \pi}\left(\frac{-1}{2}\right)\left(\frac{1}{R^2}-\frac{1}{a^2}\right) \\ & =\frac{\alpha_0 M}{4 \pi}\left(\frac{1}{a^2}-\frac{1}{R^2}\right) \end{aligned}$$

(ii) Along the quarter circle $Q S$ of radius $R$ as given in the figure below

The point $A$ lies on the equatorial line of the magnetic dipole of moment $M \sin \theta$. Magnetic field at point $A$ on the circular arc is

$$\begin{aligned} B & =\frac{\alpha_0}{4 \pi} \frac{M \sin \theta}{R^3} ; \mathrm{dl}=R d \theta \\ \therefore \quad \int^{\mathrm{B} \cdot \mathrm{dl}} & =\int \mathrm{B} \cdot \mathrm{dl} \cos \theta=\int_0^{\frac{\pi}{2}} \frac{\alpha_0}{4 \pi} \frac{M \sin \theta}{R^3} R d \theta \\ \text { Circular arc }=\frac{\alpha_0}{4 \pi} \frac{M}{R}(-\cos \theta)_0^{\pi / 2} & =\frac{\alpha_0}{4 \pi} \frac{M}{R^2} \end{aligned}$$

(iii) Along $x$-axis over the path $S T$, consider the figure given ahead

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance $x$ from the dipole is

$$\begin{aligned} B & =\frac{\alpha_0}{4 \pi} \frac{M}{x^3} \\ \int_S^{\top} \mathrm{B} \cdot \mathrm{dl} & =\int_R^a-\frac{\alpha_0 \mathbf{M}}{4 \pi x^3} \cdot \mathrm{dl}=0\left[\because \text { angle between }(-\mathbf{M}) \text { and } \mathrm{dl} \text { is } 90^{\circ}\right] \end{aligned}$$

(iv) Along the quarter circle $T P$ of radius a. Consider the figure given below

From case (ii), we get line integral of $\mathbf{B}$ along the quarter circle $T P$ of radius $a$ is circular arc TP

$$\begin{aligned} \int \mathrm{B} . \mathrm{dl} & =\int_{\pi / 2}^0 \frac{\alpha_0}{4 \pi} \frac{M \sin \theta}{a^3} \mathrm{ad} \theta \\ & =\frac{\propto_0}{4 \pi} \frac{M}{a^2} \int_{\pi / 2}^0 \sin \theta \mathrm{~d} \theta=\frac{\propto_0}{4 \pi} \frac{M}{a^2}[-\cos \theta]_{\pi / 2}^0 \\ & =\frac{-\propto_0}{4 \pi} \frac{M}{\mathrm{a}^2} \\ \therefore\quad\oint_\limits{P Q S T} \mathrm{~B} . \mathrm{dl} & =\int_P^Q \mathrm{~B} \cdot \mathrm{dl}+\int_Q^S \mathrm{~B} \cdot \mathrm{dl}+\int_S^T \mathrm{~B} \cdot \mathrm{dl}+\int_T^P \mathrm{~B} \cdot \mathrm{dl} \\ & =\frac{\alpha_0 M}{4}\left[\frac{1}{\mathrm{a}^2}-\frac{1}{R^2}\right]+\frac{\propto_0}{4 \pi} \frac{M}{R^2}+0+\left(-\frac{\propto_0}{4 \pi} \frac{M}{\mathrm{a}^2}\right)=0 \end{aligned}$$

As $I$ and $H$ both have same units and dimensions, hence, $\chi$ has no dimensions. Here, in this question, $\chi$ is to be related with $e, m, v, R$ and $\alpha_0$. We know that dimensions of $\propto_0=\left[\mathrm{ML} \theta^{-2}\right]$

From Biot-Savart's law,

$$\begin{aligned} &\begin{aligned} & d B=\frac{\propto_0}{4 \pi} \frac{I d l \sin \theta}{r^2} \\ \Rightarrow \quad \propto_0 & =\frac{4 \pi r^2 d B}{I d l \sin \theta}=\frac{4 \pi r^2}{I d l \sin \theta} \times \frac{f}{q v \sin \theta} \quad \quad\left[\because d B=\frac{F}{q v \sin \theta}\right] \\ \therefore \quad \text { Dimensions of } \propto_0 & =\frac{\mathrm{L}^2 \times\left(\mathrm{ML}^{-2}\right)}{\left(\mathrm{QT}^{-1}\right)(\mathrm{L}) \times 1 \times(\mathrm{Q})\left(\mathrm{LT}^{-1}\right) \times(1)}=\left[\mathrm{MLQ}^{-2}\right] \end{aligned}\\ &\text { where } Q \text { is the dimension of charge. } \end{aligned}$$

As $\chi$ is dimensionless, it should have no involvement of charge $Q$ in its dimensional formula. It will be so if $\alpha_0$ and $e$ together should have the value $\alpha_0 e^2$, as e has the dimensions of charge.

Let $$\chi=\alpha_0 e^2 m^a v^b R^c\quad\text{.... (i)}$$

where $a, b, c$ are the power of $m, v$ and $R$ respectively, such that relation (i) is satisfied.

Dimensional equation of (i) is

$$\begin{aligned} {\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0 \mathrm{Q}^0\right] } & =\left[\mathrm{MLQ}^{-2}\right] \times\left[\mathrm{Q}^2\right]\left[\mathrm{M}^2\right] \times\left(\mathrm{LT}^{-1}\right)^b \times[\mathrm{L}]^c \\ & =\left[\mathrm{M}^{1+a}+\mathrm{L}^{1+b+c} \mathrm{~T}^{-b} \mathrm{Q}^0\right] \end{aligned}$$

Equating the powers of $M, L$ and $T$, we get

$$\begin{aligned} & 0=1+a \Rightarrow a=-1,0=1+b+c \quad\text{.... (ii)}\\ & 0=-b \Rightarrow b=0,0=1+0+c \text { or } c=-1 \end{aligned}$$

Putting values in Eq. (i), we get

$$\begin{aligned} \chi & =\propto_0 e^2 m^{-1} V^2 R^{-1}=\frac{\alpha_0 e^2}{m R} \quad\text{.... (iii)}\\ \text{Here,}\quad\propto_0 & =4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{~A} A^{-1} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m & =9.1 \times 10^{-31} \mathrm{~kg}, R=10^{-10} \mathrm{~m} \\ \chi & =\frac{\left(4 \pi \times 10^{-7}\right) \times\left(1.6 \times 10^{-19}\right)^2}{\left(9.1 \times 10^{-31}\right) \times 10^{-10}} \approx 10^{-4} \\ \therefore\quad\frac{\chi \quad}{\chi_{\text {(given solid) }}} & =\frac{10^{-4}}{10^{-5}}=10 \end{aligned}$$