Problem Solving Strategy
Find out the relation of time period with the earth’s horizontal magnetic field and then compare it for the two given cases.
Time period in vibration magnetometer is given by
$$\eqalign{ & T = 2\pi \sqrt {\frac{I}{{M \times {B_H}}}} \cr & \Rightarrow T \propto \frac{1}{{\sqrt {{B_H}} }} \cr & {\text{So, for two different cases}} \cr & \Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{{\left( {{B_H}} \right)}_2}}}{{{{\left( {{B_H}} \right)}_1}}}} \cr & \Rightarrow \frac{2}{{{T_2}}} = \sqrt {\frac{{18}}{{24}}} \cr & \therefore T = 2.3\,s \approx 2\,s \cr} $$

$$B = \frac{{{\mu _0}}}{{4\pi }}\frac{M}{{{{\left( {{r^2} + {\ell ^2}} \right)}^{\frac{3}{2}}}}} = {B_H}$$
Given that $${\mu _0} = 4\pi \times {10^{ - 7}}T\,m{A^{ - 1}},$$
$$\eqalign{ & M = 1.34\,A{m^2}, \cr & r = 15\,cm = 0.15\,m\,\,{\text{and}}\,\,\ell = 5.0\,cm = 0.05\,m \cr & \therefore {B_H} = {10^{ - 7}} \times \frac{{1.34}}{{{{\left[ {{{\left( {0.15} \right)}^2} + {{\left( {0.5} \right)}^2}} \right]}^{\frac{3}{2}}}}} \cr & = {10^{ - 7}} \times \frac{{1.34}}{{0.025\sqrt {0.025} }} = 0.34 \times {10^{ - 4}}\,T \cr} $$

$$\eqalign{ & {\text{As}}\,\,BI = {\mu _0}M{I_M} = {\mu _0}\left( {I + {I_M}} \right) \cr & {\text{Here,}}\,\,I = 0 \cr & {\text{Then}}\,\,{\mu _0}MI = {\mu _0}\left( {{I_M}} \right) \cr & \Rightarrow {I_M} = MI = {10^6} \times 0.1 = {10^{ + 5}}A \cr} $$

Magnetic field inside the superconductor is zero. Diamagnetic substances are repelled in external magnetic field.

When a bar magnet cut perpendicular to its axis into two equal parts then $$m' = m$$   and length $$\ell ' = \frac{\ell }{2}$$
$$\therefore M' = m\ell ' = m\frac{\ell }{2} = \frac{M}{2}$$

$$\eqalign{ & W = MB\cos {60^ \circ } \cr & {\text{and}}\,\tau = MB\sin {60^ \circ } \cr & \therefore \tau = \sqrt 3 W. \cr} $$

Ferromagnetic substance has magnetic domains whereas paramagnetic substances have magnetic dipoles which get attracted to a magnetic field.
Diamagnetic substances do not have magnetic dipole but in the presence of external magnetic field due to their orbital motion of electrons these substances are repelled.

Iron is ferromagnetic.

$$\frac{{{M_1}}}{{{M_2}}} = \frac{{T_2^2 + T_1^2}}{{T_2^2 - T_1^2}} = \frac{{{6^2} + {4^2}}}{{{6^2} - {4^2}}} = \frac{{52}}{{20}} = \left( {2.6} \right):1$$

The magnetic moment of the coil is
$$M = NIA = 16 \times 0.75 \times \pi \times {\left( {0.1} \right)^2} = 0.377\,A{m^2}$$
If $$K$$ be the moment of inertia of the coil about its axis of rotation, then its period of oscillation in a magnetic field $$B$$ is given by
$$T = 2\pi \sqrt {\frac{K}{{MB}}} $$
or its frequency $$v$$ is $$ = \frac{1}{T} = \frac{1}{{2\pi }}\sqrt {\frac{{MB}}{K}} $$
This gives $$K = \frac{{MB}}{{4{\pi ^2}{v^2}}}$$
Given that $$B = 5.0 \times {10^{ - 2}}T,M = 0.377\,A{\text{ - }}{m^2}\,{\text{and}}\,v = 2{s^{ - 1}}$$
$$\therefore K = \frac{{0.377 \times 5.0 \times {{10}^{ - 2}}}}{{4 \times {{\left( {3.14} \right)}^2} \times {{\left( 2 \right)}^2}}} = 1.2 \times {10^{ - 4}}kg{m^2}$$