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. Torque is tau = m B sinθ . Given: m = 0.25 A m² , B = 0.5 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.25 × 0.5 × 0.5 = 0.0625 N m . Substituting values gives 0.0625 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on