Practice question
Question
The magnetic field contribution \( B_m \) due to a material with \( M = 1.8 \times 10^5 \, \text{A
m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).
Explanation
**Ferromagnetism** shows large positive χ ≈ 10³ to 10⁵, strong attraction, domain structure with spontaneous magnetization, hysteresis, retentivity. Distinction based on sign and magnitude of χ and behaviour in non-uniform field, explaining attraction versus repulsion. B_m = μ₀ M . Given: M = 1.8 × 10⁵ A m⁻¹ , μ₀ = 4π × 10⁻⁷ . B_m = 4π × 10⁻⁷ × 1.8 × 10⁵ = 0.22608 T ≈ 0.23 T . Substituting values gives 0.23 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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