Question

A bar magnet is 10 cm long and is kept with its North (N) pole pointing North. A neutral point is formed at a distance of 15 cm from each pole. Given the horizontal component of earth's field to be 0.4 Gauss. The pole strength of the magnet is __________ A.m. 

(a) 9 

(b) 6.75 

(c) 27 

(d) 13.5

Answer 1

The correct option (d) 13.5

Explanation:

2ℓ = 10 cm = 10^–1 m

r = 15 cm = 15 × 10^–2 m

OP = d = √(15² – 5²) = 10√2 cm = √(200) cm

From tangent law,

B_H = (μ₀/4π)[{(2ℓ)(m)}/(OP² + AO²)^3/2]

∴ 0.4 ×10^–4 = [(4π × 10^–7 × 10^–1 × m)/{4π [(200 + 25) × 10^–4]^3/2}]

∴  0.4 × 10⁴ = [m/(15 × 10^–2)³]

Hence

m = 13.5 A.m.