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

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

(a) $$\begin{aligned} B_V & =\frac{\alpha_0}{4 \pi} \frac{2 m \cos \theta}{r^3} \quad\text{.... (i)}\\ B_H & =\frac{\alpha_0}{4 \pi} \frac{\sin \theta m}{r^3}\quad\text{.... (ii)} \end{aligned}$$

Squaring both the equations and adding, we get

$$\begin{aligned} B_V^2+B_H^2 & =\left(\frac{\alpha_0}{4 \pi}\right) \frac{m^2}{r^6}\left[4 \cos ^2 \theta+\sin ^2 \theta\right] \\ B & =\sqrt{B_V^2+B_H^2}=\frac{\alpha_0}{4 \pi} \frac{m}{r^3}\left[3 \cos ^2 \theta+1\right]^{1 / 2}\quad\text{... (iii)} \end{aligned}$$

From Eq. (iii), the value of $B$ is minimum, if $\cos \theta=\frac{\pi}{2}$

$\theta=\frac{\pi}{2}$. Thus, the magnetic equator is the locus.

(b) Angle of dip,

$$\begin{aligned} & \tan \delta=\frac{B_V}{B_H}=\frac{\frac{\alpha_0}{4 \pi} \cdot \frac{2 m \cos \theta}{r^3}}{\frac{\alpha_0}{4 \pi} \cdot \frac{\sin \theta \cdot m}{r^3}}=2 \cot \theta \quad\text{.... (iv)}\\ & \tan \delta=2 \cot \theta \end{aligned}$$

For dip angle is zero i.e., $\delta=0$

$$\begin{aligned} \cot \theta & =0 \\ \theta & =\frac{\pi}{2} \end{aligned}$$

It means that locus is again magnetic equator.

(c) $\tan \delta=\frac{B_V}{B_H}$

Angle of dip i.e., $\delta= \pm 45$

$$\begin{aligned} \frac{B_V}{B_H} & =\tan ( \pm 45\Upsilon) \\ \frac{B_V}{B_H} & =1 \\ 2 \cot \theta & =1 \quad\text{[From Eq. (iv)]}\\ \cot \theta & =\frac{1}{2} \\ \tan \theta & =2 \\ \Rightarrow\quad\theta & =\tan ^{-1}(2) \end{aligned}$$

Thus, $\theta=\tan ^{-1}(2)$ is the locus.

$P$ is in the plane $S$, needle is in north, so the declination is zero.

$P$ is also on the magnetic equator, so the angle of dip $=0$, because the value of angle of dip at equator is zero. $Q$ is also on the magnetic equator, thus the angle of dip is zero. As earth tilted on its axis by $11.3^{\circ}$, thus the declination at $Q$ is $11.3^{\circ}$.