Practice question
Question
The magnetic field contribution \( B_m \) due to a material with \( M = 1.6 \times 10^5 \, \text{A
m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).
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_m = μ₀ M . Given: M = 1.6 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 1.6 × 10⁵ = 0.20096 T ≈ 0.20 T . Substituting values gives 0.20 T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.
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