Practice question

**Magnetic field of bar magnet** follows inverse cube law B ∝ m/r³, unlike inverse square for electric dipole. Given B at distance r, moment m = B r³/(μ₀/4π) for equatorial, m = B r³/(2·μ₀/4π) for axial, enabling moment extraction from measured field. B = (μ₀/4π) (m/r³) , so m = (B r³/(μ₀/4π)) . Given: B = 6 × 10⁻⁶ T , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . m = (6 × 10⁻⁶ × (0.3)³/10⁻⁷) = (6 × 10⁻⁶ × 0.027/10⁻⁷) = 1.62 A m² . Substituting values gives 1.62 A m², which matches expected magnitude for this magnetic configuration, confirming dipole field