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Question

A bar magnet with \( m = 1.6 \, \text{A m}^2 \) is at \( 0.8 \, \text{m} \) along its axis. What is \(
B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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Explanation

**Hysteresis loop** plots B versus H for ferromagnetic material, retentivity (remanence) is residual B at H=0 after saturation, coercivity is reverse H needed to reduce B to zero. Hard ferromagnets have high retentivity and coercivity, retaining strong magnetism after external field removed, suitable for permanent magnets. B = (μ₀/4π) (2m/r³) . Given: m = 1.6 A m² , r = 0.8 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 1.6/(0.8)³) = 10⁻⁷ × (3.2/0.512) ≈ 6.25 × 10⁻⁷ T . Substituting values gives 6.25 × 10⁻⁷ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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