Q.
A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of $30 \,cm$ from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $Am^2$ is close to :
(Given $\frac{\mu_{0}}{4\pi}=10^{-7}$ in SI units and $B_H$ = Horizontal component of earth's magnetic field = $3.6 \times 10^{-5}$ Tesla.)
Solution: