A bar magnet of length 12 cm is placed such that its north pole points towards the geographic north. Two neutral points which are separated by 16 cm are obtained on the equatorial axis of the bar magnet. What is the pole strength of the bar magnet if the horizontal component of the earth's field is 1.25 10⁻⁵ T ?

Step-by-step solution

Length of the bar magnet, 2l = 12 cm = 0.12 m l = 0.06 m The neutral points are obtained on the equatorial axis. The distance between the two neutral points is 16 cm . Distance of each neutral point from the centre of the magnet, d = 16 2 = 8 cm = 0.08 m At the neutral point on the equatorial line, the magnetic field due to the magnet is equal in magnitude and opposite in direction to the horizontal component of the Earth's magnetic field B_H . B_ eq = B_H ₀ 4 M (d^2 + l^2)^ 3/2 = B_H Since magnetic moment M = m 2l