In magnetic dipole
$${\text{Force}} \propto \frac{1}{{{r^4}}}$$
In the given question, $${\text{Force}} \propto {x^{ - n}}$$
Hence, $$n = 4$$

Given that : $${B_1} = 1.2 \times {10^{ - 2}}T,$$    orientation of dipole with the field $${B_1},{\theta _1} = {15^ \circ }$$
Hence, orientation of dipole with $${B_2},{\theta _2} = {60^ \circ } - {15^ \circ } = {45^ \circ }\left( {{\text{figure}}} \right)$$

As the dipole is in equilibrium, therefore, the torque on the dipole due to the two fields must be equal and opposite.
If $$M$$ be the magnetic dipole moment of the dipole, then
$$\eqalign{ & {\tau _1} = {\tau _2}\,{\text{or}}\,M{B_1}\sin {\theta _1} = M{B_2}\sin {\theta _2} \cr & {\text{or,}}\,\,{B_2} = \frac{{{B_1}\sin {\theta _1}}}{{\sin {\theta _2}}} = \frac{{1.2 \times {{10}^{ - 2}}\sin {{15}^ \circ }}}{{\sin {{45}^ \circ }}} \cr & = \frac{{1.2 \times {{10}^{ - 2}} \times 0.2588}}{{0.7071}} = 4.4 \times {10^{ - 3}}\,Tesla \cr} $$

$$B = {\mu _0}{\mu _r}H \Rightarrow {\mu _r} \propto \frac{B}{H} = {\text{slope}}\,{\text{of}}\,B - H\,{\text{curve}}$$
According to the given graph, slope of the graph is highest at point $$Q.$$

Work done in rotating the dipole from $${\theta _1}$$ to $${\theta _2}$$ is
\[W = - MB\left( {\cos {\theta _2} - \cos {\theta _1}} \right)\,\,\,\left( {\begin{array}{*{20}{r}} {M = {\rm{Magnetic\,moment}}}\\ {B = {\rm{Magnetic\,induction}}} \end{array}} \right)\]
Case I
$${W_1} = - MB\left( {\cos {{90}^ \circ } - \cos {0^ \circ }} \right) = MB\,......\left( {\text{i}} \right)$$
Case II
$$\eqalign{ & {W_2} = - MB\left( {\cos {{60}^ \circ } - \cos {0^ \circ }} \right) \cr & = - MB\left( {\frac{1}{2} - 1} \right) = \frac{1}{2}MB\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii), $${W_2} = \frac{1}{2}{W_1}$$
As $${W_1} = n{W_2}$$
$$\therefore n = 2$$

Magnetic field in solenoid $$B = {\mu _0}ni$$
$$\eqalign{ & \Rightarrow \frac{B}{{{\mu _0}}} = ni\,\,\,\left( {{\text{where}}\,n = {\text{number of turns per unit length}}} \right) \cr & \Rightarrow \frac{B}{{{\mu _0}}} = \frac{{Ni}}{L} \cr & \Rightarrow 3 \times {10^3} = \frac{{100i}}{{10 \times {{10}^{ - 2}}}} \cr & \Rightarrow i = 3A \cr} $$

$$\eqalign{ & {B_1} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2M}}{{{r^3}}} = {10^{ - 7}} \times \frac{{2 \times 1}}{{{1^3}}} = 2 \times {10^{ - 7}}T \cr & {\text{and}}\,\,{B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{M}{{{r^3}}} = {10^{ - 7}} \times \frac{1}{{{1^3}}} = {10^{ - 7}}T \cr & {\text{Now,}}\,B = \sqrt {B_1^2 + B_2^2} = \sqrt 5 \times {10^{ - 7}}T. \cr} $$

As we know the relation between the magnetic permeability and susceptibility of material, i.e
$${\mu _r} = 1 + {\chi _m}\,......\left( {\text{i}} \right)$$
$$\because $$ For diamagnetic substances, $${\mu _r} < 1$$
So, according to equation (i), the magnetic susceptibility $$\left( {{\chi _m}} \right)$$  of diamagnetic substance will be negative.
While in the case of para and ferromagnetic substances, diamagnetic susceptibility is positive.

The strength of the earths magnetic field is not constant. It varies from one place to other place on the surface of earth. Its value being of the order of $${10^{ - 5}}T.$$

$$\eqalign{ & {\text{Loss in }}P.E. = {\text{gain in }}K.E. \cr & \therefore {E_k} = {U_i} - {U_j} = - MB\cos {90^ \circ } - \left( { - MB\cos {0^ \circ }} \right) \cr & = 4 \times 25 \times {10^{ - 6}} = {10^{ - 4}}\,J \cr} $$

The magnetic moment, $$M = ml\,\,\left( {_{l\, = {\text{ length of magnet}}}^{m\,\, = \,\,{\text{pole}}\,{\text{strength}}}} \right)$$
According to question,
$$l = \frac{\pi }{3} \times r$$
So, $$r = \frac{{3l}}{\pi }$$
New magnetic moment, $$M' = m \times r$$
$$ = m \times \frac{{3l}}{\pi } = \frac{3}{\pi } \cdot ml = \frac{{3M}}{\pi }$$