Force between two short bar magnets is given by
$$\eqalign{ & F = \frac{{{\mu _0}}}{{4\pi }}\frac{{6{M_1}{M_2}}}{{{r^4}}} \cr & \therefore \frac{{{F_1}}}{{{F_2}}} = \frac{{r_2^4}}{{r_1^4}} = {\left( {\frac{{2r}}{r}} \right)^4} \cr & {\text{or}}\,\,{F_2} = \frac{{{F_1}}}{{16}} = \frac{{4.8}}{{16}} = 0.3\,N. \cr} $$

$$\eqalign{ & {I_1} = \frac{{m{\ell ^2}}}{{12}}\,{\text{and}}\,{I_2} = \frac{{\left( {2m} \right){\ell ^2}}}{{12}} = 2{I_1}. \cr & {\text{Now,}}\,\,\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}}} \,or\,\frac{{{T_0}}}{{{T_2}}} = \sqrt {\frac{1}{2}} \cr & \therefore {T_2} = \sqrt 2 {T_0}. \cr} $$

At geomagnetic poles, there is no horizontal component of earth field and so compass needle may stay at any position.

Magnetic intensity on end side -on position is twice than broad side on position.

When a bar magnet is placed in an external magnetic field $$B,$$ a magnetic torque $$\tau $$ acts on it, which is given by $$\tau = M \times B = \left| M \right| \times \left| B \right|\sin \theta $$      ($$\theta =$$  angle between $$M$$ and $$B$$ )

$$\because $$ Work done in rotating the magnet
$$W = MB\left( {\cos {\theta _0} - \cos \theta } \right)$$
Where, $$M =$$  magnetic moment of the magnet
$$B =$$  magnetic field
$$W = MB\left( {\cos {0^ \circ } - \cos {{60}^ \circ }} \right)$$
$$ = MB\left( {1 - \frac{1}{2}} \right) = \frac{{MB}}{2}$$
$$\therefore MB = 2W\,.......\left( {\text{i}} \right)$$
Torque on a magnet in this position is given by,
$$\eqalign{ & \tau = M \times B \cr & = MB \cdot \sin \theta = 2W \cdot \sin {60^ \circ }\,\,\left[ {{\text{from}}\,{\text{Eq}}{\text{.}}\left( {\text{i}} \right)} \right] \cr & = 2W\frac{{\sqrt 3 }}{2} = W\sqrt 3 \cr} $$

Magnetic dipole moment
$$\eqalign{ & M = m \times \ell \cr & M' = m \times r \cr} $$

From figure
$$\eqalign{ & \ell = \frac{{\pi r}}{3}\,\,{\text{or}}\,\,r = \frac{{3\ell }}{\pi } \cr & {\text{so,}}\,M' = m \times r \cr & = \frac{{m \times 3\ell }}{\pi } = \frac{3}{\pi }M \cr} $$

Diamagnetic substances are weakly magnetised in a direction opposite to that of applied magnetic field. These are repelled in an external magnetic field i.e. have a tendency to move from high to low field region, i.e. it is repelled by both North and South poles of a bar magnet.

The earth’s core is hot and molten. Hence, convective current in earth’s core is responsible for it’s magnetic field.

work done $$ = MB\left( {\cos {\theta _1} - \cos {\theta _2}} \right)$$
$$\eqalign{ & = MB\left( {\cos {0^ \circ } - \cos {{60}^ \circ }} \right) \cr & = MB\left( {1 - \frac{1}{2}} \right) = \frac{{2 \times {{10}^4} \times 6 \times {{10}^{ - 4}}}}{2} = 6\,J \cr} $$