Practice question
Question
A magnetic dipole of moment \( 0.8 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at
\( 45^\circ \). What is the torque on it?
Explanation
**Axial versus equatorial** field comparison shows B_axial = 2 B_eq for same r. Using μ₀ = 4π×10⁻⁷ T·m/A, calculation involves r³ = (0.3)³ = 0.027 m³, so B = 10⁻⁷·m/r³ yields moment estimation. Torque is tau = m B sinθ . Given: m = 0.8 A m² , B = 0.5 T , θ = 45° , sin 45° = (1/√(2)) ≈ 0.707 . Substitute: tau = 0.8 × 0.5 × 0.707 ≈ 0.2828 N m ≈ 0.28 N m . Substituting values gives 0.28 N m, 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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