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 Am² is close to (Given μ₀/4π = 10⁻⁷ in SI units and BH = Horizontal component of earth's magnetic field = 3.6 × 10⁻⁵ Tesla).

Magnetic field due to bar magnet of pole strength of m is given as at equatorial position is given as
B = μ₀/4π * (m/r³)
where r is the distance from the center of the magnet.
At neutral points, the magnetic field due to the bar magnet is equal and opposite to the horizontal component of the Earth's magnetic field.
Therefore,
B = BH
μ₀/4π * (m/r³) = BH
10⁻⁷ * (m/(0.3)³) = 3.6 × 10⁻⁵
m = 3.6 × 10⁻⁵ * (0.3)³ / 10⁻⁷
m = 3.6 × 10⁻⁵ * 0.027 / 10⁻⁷
m = 9.72 × 10⁻³ / 10⁻⁷
m = 9.72 × 10⁴ Am²
m ≈ 9.7 Am²
Therefore, the magnetic moment of the magnet is approximately 9.7 Am².