Solution :

(At null points , field due to a magnet is equal and opposite to the horizontal component of earth magnetic field )

Given $ d=14\;cm$

$\qquad= 0.14\;m$

$H= 0.6 \;G$

$B_1= \large\frac{\mu_0}{4 \pi} . \frac{2m}{d^3}$

But the magnetic field is equal to the horizontal component of earth's magnetic field .

(ie) $B_1=\large\frac{\mu_0}{4 \pi} . \frac{2m}{d^3}$

on the equatorial line of magnet at same distance (d) magnetic field due to the magnet.

$B_2 =\large\frac{\mu_0}{4 \pi} \frac{m}{d^3}$

$\qquad= \large\frac{B_1}{2}$

$\qquad= \large\frac{H}{2}$

The total magnetic field on equatorial line at this point is

$B= B_1+B_2$

$\quad= H+ \large\frac{H}{2}$

$\quad= \large\frac{3}{2}$$H$

$\therefore B=\large\frac{3}{2} $$ \times 0.36 $

$\qquad= 0.54\;G$

Hence the magnetic of earth's magnetic field is $0.54 \;G$

Answer : $\text{(i)magnitude : 0.54 G (ii) direction : along the direction of earth's field}$