Magnetism and Matter — Exam Practice

Answer Because isolated magnetic monopoles don't exist — every field line leaving a north pole must eventually return to a south pole, with no starting or ending point in open space, giving Gauss's Law for magnetism: net magnetic flux through any closed surface is always exactly zero.

Answer Treat the bar magnet as two poles ±q_m, separated by 2a, with magnetic moment m = q_m(2a). At an axial point distance r from the centre, the fields due to each pole (using the pole model, analogous to Coulomb's Law) partially cancel due to opposite pole signs but add along the axis.

Working through the geometry and taking r ≫ a: B_axial = (μ₀/4π) × 2m/r³.

Answer (a) Magnetic susceptibility χ = M/H, measuring how strongly a material magnetizes in response to an applied field H.

Diamagnetic: χ small and negative, roughly temperature-independent. Paramagnetic: χ small and positive, following Curie's Law (χ ∝ 1/T). Ferromagnetic: χ large and positive, strongly temperature- and history-dependent (hysteresis), vanishing above the Curie temperature.

(b) B = μH = μ₀μ_rH → μ_r = B/(μ₀H) = 0.6 / (4π×10⁻⁷ × 1500) ≈ 318.3.

Solution Diamagnetic materials are weakly repelled by an external field, corresponding to a small, negative χ — present (weakly) in every material, though usually masked by stronger para- or ferromagnetic effects if present.

Solution Above the Curie temperature, thermal agitation overcomes the internal alignment that gives ferromagnetism its strength, and the material's response weakens to ordinary paramagnetic behaviour.

Solution At the magnetic equator, Earth's field is entirely horizontal — the field makes no angle with the horizontal plane, so dip = 0°. Dip reaches 90° at the magnetic poles, where the field points straight down (or up).

Solution Since magnetic monopoles don't exist, every field line entering a closed surface must also leave it — net flux is always exactly zero, regardless of the surface's shape or position.

Solution B_eq = (μ₀/4π)(m/r³) = 10⁻⁷ × 5 / (0.1)³ = 10⁻⁷ × 5 / 0.001 = 5×10⁻⁴ T.

Solution A compass needle aligns with the horizontal component of the field. Where that component is zero (near the magnetic poles, where dip ≈ 90°), the needle has nothing to align it horizontally, and can settle in any horizontal direction — a genuine practical limitation of compasses near the poles.

Solution B_H = B cos(dip) → B = B_H/cos(dip) = 0.3/cos45° = 0.3/0.707 ≈ 0.424 G.