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 = (μ₀/4π) (m/r³) . Given: m = 1.2 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (1.2/(0.3)³) = 10⁻⁷ × (1.2/0.027) ≈ 4.44 × 10⁻⁶ T . Substituting values gives 4.44 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.