Practice question
Question
A bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.3 \,
\text{m} \) along its axis. What is the magnetic field \( B \) at that point? (Take \( \mu_0 = 4\pi
\times 10^{-7} \, \text{T m A}^{-1} \)).
Explanation
**Permanent magnet requirement** is high retentivity to maintain field and high coercivity to resist demagnetization. Ability to retain magnetism after field removal is property of hard ferromagnets, related to domain wall pinning and anisotropy. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.5 A m² , r = 0.3 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.5/(0.3)³) = 10⁻⁷ × (3/0.027) = 1.11 × 10⁻⁵ T ≈ 1.1 × 10⁻⁵ T . Substituting values gives 1.1 × 10⁻⁵ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.