Practice question

**Field due to bar magnet** on axial line is B_axial = (μ₀/4π)·2m/r³, equatorial B_eq = (μ₀/4π)·m/r³, where μ₀/4π = 10⁻⁷ T·m/A, m magnetic moment (A·m²), r distance (m). Axial field twice equatorial at same distance and parallel to moment, equatorial opposite to moment direction. B = μ₀ (H + M) , so H = (B/μ₀) - M . Given: B = 0.25 T , M = 1.8 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . (B/μ₀) = (0.25/4π × 10⁻⁷) ≈ 1.989 × 10⁵ A m⁻¹ . H = 1.989 × 10⁵ - 1.8 × 10⁵ = 1.89 × 10⁴ A m⁻¹