Practice question
Question
A bar magnet with magnetic moment \( 1.8 \, \text{A m}^2 \) is placed at a distance of \( 0.5 \,
\text{m} \) 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
**Bar magnet properties** include dipole moment m = pole strength × separation, unit A·m², field lines emerge from north and enter south outside. Lines never cross, ensuring single valued B at any point, and pattern reflects dipole nature with symmetric loops around magnet. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.8 A m² , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.8/(0.5)³) = 10⁻⁷ × (3.6/0.125) = 2.88 × 10⁻⁶ T . Substituting values gives 2.88 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming
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