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Question

A bar magnet with magnetic moment \( 1.0 \, \text{A m}^2 \) is placed at a distance of \( 0.5 \,
\text{m} \) along its axis. What is the magnetic field \( B \) at that point? (Take \( \mu_0 = 4\pi
\times 10^{-7} \, \text{T m A}^{-1} \)).

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Explanation

**Elements of Earth's field** include declination D, inclination I, horizontal component B_H, total field B = √(B_H² + B_V²). B_H provides compass direction, declination varies with location, important for navigation, inclination 0° at magnetic equator, 90° at poles. The magnetic field along the axis is B = (μ₀/4π) (2m/r³) . Given: m = 1.0 A m² , r = 0.5 m , (μ₀/4π) = 10⁻⁷ . Substitute: B = 10⁻⁷ × (2 × 1.0/(0.5)³) = 10⁻⁷ × (2.0/0.125) = 1.6 × 10⁻⁶ T . Substituting values gives 1.6 × 10⁻⁶ T, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³

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