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. U_m = -m B cosθ . Given: m = 0.6 A m² , B = 0.8 T , θ = 0° , cos 0° = 1 . U_m = -0.6 × 0.8 × 1 = -0.48 J . Substituting values gives -0.48 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.