A bar magnet of length 6 cm is placed in the magnetic meridian with N pole pointing towards the geographical north. Two neutral points, separated by a distance of 8 cm are obtained on the equatorial axis of the magnet. If B_H = 1.2 10⁻⁵ T , then the pole strength of the magnet is:

Step-by-step solution

Given: 2l = 6 , cm l = 0.03 , m , d = 8 2 = 4 , cm = 0.04 , m , B_H = 1.2 10⁻⁵ , T At neutral point on equatorial axis: B = B_H ₀ 4 m 2l (d^2 + l^2)^ 3/2 = B_H d^2 + l^2 = (0.04)^2 + (0.03)^2 = 0.0016 + 0.0009 = 0.0025 (0.0025)^ 3/2 = (0.05)^3 = 1.25 10⁻⁴ 10⁻⁷ m 0.06 1.25 10⁻⁴ = 1.2 10⁻⁵ m 0.06 = 1.2 10⁻⁵ 1.25 10⁻⁴ 10⁻⁷ = 1.5 10⁻² m = 0.015 0.06 = 0.25 , A·m Correct option: (2)