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 μ 0 4 π = 10 - 7 in SI units and B H = Horizontal component of earth's magnetic field = 3.6 × 10 - 5 Tesla.

Step-by-step solution

Horizontal component of earth's magnetic field is along geographical south its North. M → = Magnetic dipole moment. At neutral point on the equatorial line the magnetic field B ′ cancels B H ∴ B ′ = B H ⇒ μ 0 4 π M x 3 = B H   ⇒ 10 - 7 × M 30 100 3 = 3.6 × 10 - 5 ⇒ 10 - 7 × 10 3 × M 27 = 3.6 × 10 - 5 M = 27 × 3.6 × 10 - 5 10 - 4 M = 27 × 3.6 × 10 - 1 = 9.72 M ≈ 9.7