Practice question
Question
A magnetic dipole of moment \( 0.15 \, \text{A m}^2 \) is in a uniform field of \( 0.8 \, \text{T} \)
at \( 60^\circ \). What is the torque on it?
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. Torque is tau = m B sinθ . Given: m = 0.15 A m² , B = 0.8 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.15 × 0.8 × 0.866 ≈ 0.1039 N m ≈ 0.104 N m . Substituting values gives 0.104 N m, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque
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