Magnetism and Matter Class 12 Notes (2026-27) — CBSE
Class 12 Physics Chapter 5 notes: the bar magnet as a magnetic dipole, field lines, dipole moment, and the torque and potential energy of a dipole in a field.
Magnetism and Matter — Class 12 Physics Notes
Chapter Snapshot
This chapter treats a bar magnet as a magnetic dipole. Because a magnet's field pattern is identical to that of an electric dipole (and of a current-carrying solenoid), all the dipole results carry over: magnetic moment, field lines, the axial and equatorial fields, and the torque and potential energy of the dipole in a uniform field.
Board relevance: a short chapter; usually a torque/potential-energy or dipole-field question. The electric-dipole ↔ magnetic-dipole analogy makes the formulas easy to remember.
Syllabus note (rationalised): the sections on the Earth's magnetism, magnetisation and magnetic intensity, and magnetic properties of materials (dia-, para-, ferromagnetism) have been removed. This note covers the retained bar-magnet dipole content.
Key Concepts & Definitions
Bar magnet as a magnetic dipole — every magnet has two poles, north and south, of equal strength. The poles cannot be isolated: breaking a magnet gives two smaller magnets, each with both poles (there are no magnetic monopoles).
Magnetic field lines of a bar magnet:
- Run from north to south outside the magnet and south to north inside (forming closed loops).
- Never intersect; are denser where the field is stronger.
Magnetic dipole moment (m) — a vector directed from S to N of the magnet; magnitude m = qₘ × 2l (pole strength × magnetic length). Unit: A·m². For a current loop/solenoid, m = NIA.
Bar magnet ↔ solenoid: a solenoid of N turns carrying current I behaves like a bar magnet of moment m = NIA, one end north and the other south. This is why the two produce identical field patterns.
Formulas
By analogy with the electric dipole (replace 1/4πε₀ → μ₀/4π and p → m):
Position (r ≫ l) Magnetic field of a bar magnet
Axial (end-on) B = (μ₀/4π)(2m/r³)
Equatorial (broadside-on) B = (μ₀/4π)(m/r³)
- The axial field is twice the equatorial field at the same distance, and points along m; the equatorial field points opposite to m.
- Both fall off as 1/r³ (dipole field).
Torque on a dipole in a uniform field:
τ = m × B, magnitude τ = mB sin θ
Potential energy of the dipole:
U = −m·B = −mB cos θ
- θ = 0° (aligned): U = −mB, minimum, stable equilibrium.
- θ = 90°: U = 0, torque maximum.
- θ = 180° (anti-aligned): U = +mB, maximum, unstable.
Work to rotate the dipole from θ₁ to θ₂: W = mB(cos θ₁ − cos θ₂).
Worked Examples
Example 1 — Torque: A bar magnet of moment 0.4 A·m² is placed at 30° to a uniform field of 0.2 T. Find the torque.
τ = mB sin θ = 0.4 × 0.2 × sin 30° = 0.4 × 0.2 × 0.5 = 0.04 N·m.
Example 2 — Potential energy: For the same magnet, find the PE when aligned and when anti-aligned.
Aligned (θ = 0°): U = −mB = −0.4 × 0.2 = −0.08 J (stable). Anti-aligned (θ = 180°): U = +0.08 J (unstable).
Example 3 — Work done: Find the work to rotate the magnet from θ = 0° to θ = 90° in the 0.2 T field.
W = mB(cos 0° − cos 90°) = 0.08(1 − 0) = 0.08 J.
Example 4 — Equivalent solenoid: A solenoid of 400 turns, area 3 × 10⁻⁴ m², carrying 2 A has what magnetic moment?
m = NIA = 400 × 2 × 3 × 10⁻⁴ = 0.24 A·m² — it behaves like a bar magnet of this moment.
Important Question Patterns
1. Torque/PE (2–3 marks): τ = mB sin θ and U = −mB cos θ; identify stable/unstable equilibrium; work to rotate.
2. Dipole field (2–3 marks): axial vs equatorial field of a bar magnet; the 2:1 ratio; direction relative to m.
3. Bar magnet ↔ solenoid (2 marks): m = NIA; why they give the same field.
4. Field lines / monopoles (1–2 marks): properties of magnetic field lines; why isolated poles don't exist.
5. Analogy (2 marks): map electric-dipole results to the magnetic dipole.
⚡ Quick Revision
- Bar magnet = magnetic dipole (N and S of equal strength; no monopoles). Field lines: N→S outside, closed loops.
- Magnetic moment m (S→N), unit A·m²; for a loop/solenoid m = NIA.
- Dipole field (1/r³): axial B = (μ₀/4π)(2m/r³), equatorial B = (μ₀/4π)(m/r³) — axial is twice equatorial.
- Torque τ = m × B = mB sin θ; PE U = −m·B = −mB cos θ (min at θ = 0, stable).
- Work to rotate: W = mB(cos θ₁ − cos θ₂).
- A bar magnet is equivalent to a solenoid of the same magnetic moment.
- Removed from syllabus: Earth's magnetism, magnetisation/intensity, dia/para/ferromagnetism.
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