Class 12 Physics Chapter 5: Magnetism and Matter worksheet with answer key

Answer: (c) They start at the north pole and end at the south pole.

The lines of a magnet run outside from N to S and continue inside the magnet from S to N, forming closed loops; they do not begin or end anywhere. The other three are properties stated for field lines.

Answer: (d) B/16

Axial: B = (μ₀/4π)(2m/r³). Equatorial at 2r: (μ₀/4π) m/(2r)³ = (μ₀/4π) m/8r³ = B/16. Forgetting the factor 2 of the axial field gives B/8; counting it twice gives B/32; ignoring the change of distance gives B/2.

Answer: (c) 2.5 × 10⁻⁵ T

Halving for the equatorial position and dividing by 2³ = 8 for the doubled distance: B = 4.0 × 10⁻⁴ × (1/2) × (1/8) = 2.5 × 10⁻⁵ T. Dropping the 1/2 gives 5.0 × 10⁻⁵ T.

Answer: (c) is a magnet with its own north and south pole

Magnetic poles cannot be isolated: breaking a magnet gives smaller magnets, each with a north and a south pole. Isolated poles (monopoles) do not exist.

Answer: (b) 0.30 J

U = −mB cos θ = −2.0 × 0.30 × cos 120° = −0.60 × (−0.5) = +0.30 J. The sign is positive because the moment points more against the field than along it.

On the axis, B = (μ₀/4π)(2m/r³), so m = Br³/(2 × 10⁻⁷) = 3.2 × 10⁻⁶ × (0.50)³/(2 × 10⁻⁷) = 2.0 A m². On the equatorial line at the same distance the field is half the axial value, (μ₀/4π)(m/r³) = 1.6 µT, directed opposite to m.

• m from the axial field: m = 2.0 A m² (1)
• equatorial field 1.6 µT, opposite to m (1)

H = nI is the same in both cases, and B = μ₀(1 + χ)H. Ratio = (1 + 1.0 × 10⁻⁵)/(1 − 1.0 × 10⁻⁵) ≈ 1 + 2.0 × 10⁻⁵ = 1.00002. The fields differ by only 2 parts in 10⁵: a minuscule difference in χ, yet one liquid is pushed out of strong-field regions and the other is drawn in.

• H = nI unchanged; B = μ₀(1 + χ)H (1)
• ratio ≈ 1.00002, a difference of 2 parts in 10⁵, though the behaviours are opposite (1)

(a) τ = mB sin θ is largest at θ = 90°: τ_max = 2.5 × 0.040 = 0.10 N m, with the moment perpendicular to the field. (b) sin θ = 1/2, so θ = 30° or 150°. U = −mB cos θ: at 30°, U = −0.10 × 0.866 = −0.087 J; at 150°, U = +0.087 J.

• τ_max = mB = 0.10 N m at θ = 90° (1)
• θ = 30° or 150°, U = −0.087 J and +0.087 J (1)

(a) τ_max = mB, so m = 0.060/0.30 = 0.20 A m². (b) U = −mB cos θ = −0.060 cos θ = −0.030 J gives cos θ = 0.5, θ = 60°; τ = mB sin 60° = 0.060 × 0.866 = 0.052 N m. (c) The unstable equilibrium is θ = 180°: W = mB(cos 60° − cos 180°) = 0.060 × 1.5 = 0.090 J.

• m = τ_max/B = 0.20 A m² (1)
• θ = 60°, τ = 0.052 N m (1)
• W = mB(cos 60° − cos 180°) = 0.090 J (1)

(a) 1/ε₀ → μ₀ (equivalently 1/4πε₀ → μ₀/4π), p → m. (b) Torque m × B; energy −m·B. (c) Axial field of a short bar magnet: B_A = (μ₀/4π)(2m/r³), along m.

• 1/ε₀ → μ₀ and p → m (1)
• torque m × B and energy −m·B (1)
• B_A = μ₀2m/4πr³ (1)

(a) Within each domain the atomic dipoles are aligned, but the magnetisation varies randomly from domain to domain, so the moments cancel and there is no bulk magnetisation. (b) In the magnet's field the domains turn towards the field and those already along it grow, so the field lines are highly concentrated; in the non-uniform field it moves towards high field, that is, it is attracted. (c) The nail is soft iron, a soft ferromagnet whose magnetisation disappears when the field is removed; lodestone is a hard ferromagnet whose magnetisation persists.

• domains randomly oriented, so moments cancel (1)
• domains align and grow in the field; attracted towards high field (1)
• soft ferromagnet loses magnetisation; lodestone is a hard ferromagnet (1)