Practice question
Question
A bar magnet with \( m = 3.5 \, \text{A m}^2 \) is at \( 0.6 \, \text{m} \) along its axis. What is \(
B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).
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. B = (μ₀/4π) (2m/r³) . Given: m = 3.5 A m² , r = 0.6 m , (μ₀/4π) = 10⁻⁷ . B = 10⁻⁷ × (2 × 3.5/(0.6)³) = 10⁻⁷ × (7.0/0.216) ≈ 3.24 × 10⁻⁶ T . Substituting values gives 3.24 × 10⁻⁶ 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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