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. The magnetic field on the equatorial line is B = (μ₀/4π) (m/r³) . Given: m = 2 A m² , r = 0.1 m , (μ₀/4π) = 10⁻⁷ T m A⁻¹ . Substitute: B = 10⁻⁷ × (2/(0.1)³) = 10⁻⁷ × (2/0.001) = 2 × 10⁻⁴ T . Substituting values gives 2 × 10⁻⁴ T, which matches expected magnitude for this