devarshi-dt-logo

Question:

Two short bar magnets of length 1 cm each have magnetic moments 1.20 Am² and 1.00 Am² respectively. They are placed on a horizontal table parallel to each other with their N poles pointing towards the South. They have a common magnetic equator and are separated by a distance of 20.0 cm. The value of the resultant horizontal magnetic induction at the mid-point O of the line joining their centres is close to (Horizontal component of earth's magnetic induction is 3.6 × 10⁻⁵ Wb/m²)

3.50 × 10⁻⁸ Wb/m²

5.80 × 10⁻⁸ Wb/m²

3.6 × 10⁻⁵ Wb/m²

2.56 × 10⁻⁸ Wb/m²

Solution:

Given that, m1 = 1.20 A.m²
m2 = 1.00 A.m³
B along equator : B = μ₀/4π * M/d³
Here B1 = μ₀/4π × 1.20/10⁷
B1 = 10⁻⁷ × 1.20 Wb/m²
B2 = μ₀/4π × 1/10⁷ = 10⁻⁷ Wb/m²
θ = 0° ∴ B resultant = 10⁻⁷ B = √B₁² + B₂² + 2B₁B₂
we know resultant will be directed from N→S thus θ = 0° = (√1.44 + 1 + 1.4) × 10⁻⁷ = √4.94 × 10⁻⁷
B = 2.226 × 10⁻⁸ Wb/m² ⇒ a
Thus B = 2.56 × 10⁻⁸ Wb/m²