Magnetism and Matter – Class 12 Physics Chapter 5 Notes & NCERT PDF

Class 12PhysicsChapter 5NCERT book: Physics Part I - Class 12

Notes and a simple summary of Chapter 5 of the NCERT Class 12 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 5: Magnetism and Matter

Introduction

Magnetism is far older than electricity as a science. Lodestone was known thousands of years ago, and the magnetic compass guided sailors long before anyone understood why it pointed north. In Chapter 4 you learnt that moving charges and currents produce magnetic fields. This chapter turns that idea around and asks: how do ordinary materials such as iron, copper, aluminium or water respond to a magnetic field, and why does a bar magnet behave like a current loop? The chapter builds a clear bridge between the bar magnet, the current-carrying solenoid and the electric dipole, then classifies all materials as diamagnetic, paramagnetic or ferromagnetic.

Key Concepts

5.2 The Bar Magnet

Magnetic field lines form continuous closed loops: outside the magnet they run from N to S, inside from S to N. They never intersect, and their density shows field strength.

A bar magnet behaves like a solenoid. Each turn of a solenoid is a tiny current loop (a magnetic dipole); together they give the same far field as a bar magnet. Magnetic moment of a solenoid: m = NIA, directed along the axis by the right-hand rule.

Cutting a bar magnet never isolates a pole. Each piece is a complete magnet; magnetic monopoles are not known to exist.

Dipole in a uniform field: the net force is zero, but there is a torque τ = m × B, magnitude τ = mB sin θ. Potential energy U = −m·B = −mB cos θ. U is minimum (−mB, stable) at θ = 0° and maximum (+mB, unstable) at θ = 180°.

Electrostatic analogue: replace p by m and 1/ε₀ by µ₀. At a far point on the axis, B = µ₀·2m/(4πr³); on the equatorial line, B = −µ₀m/(4πr³) (half the axial value, opposite to m).

5.3 Magnetism and Gauss's Law

The net magnetic flux through any closed surface is zero: ∮B·dS = 0. Every field line that enters a closed surface also leaves it. This is the mathematical statement that isolated magnetic poles do not exist, unlike the electric case where flux equals q/ε₀.

5.4 Magnetisation and Magnetic Intensity

Magnetisation M = net magnetic moment per unit volume, M = m_net/V. Unit: A m⁻¹.

Inside a long solenoid with a core, B = µ₀(H + M), where H = nI is the magnetic intensity (A m⁻¹), set by the external current, and M is the material's own contribution.

For most materials, M = χH, where χ (susceptibility) is dimensionless.

Then B = µ₀(1 + χ)H = µ₀µᵣH = µH, with µᵣ = 1 + χ (relative permeability) and µ = µ₀µᵣ (permeability).

5.5 Magnetic Properties of Materials

Diamagnetic (−1 ≤ χ < 0, typically about −10⁻⁵): weakly repelled, move from stronger to weaker field; field lines are pushed out of the sample. Examples: bismuth, copper, lead, silicon, nitrogen gas (at STP), water, sodium chloride. Diamagnetism is present in all substances but is masked in others. A superconductor is a perfect diamagnet (χ = −1, µᵣ = 0): it expels the field completely (Meissner effect).

Paramagnetic (small positive χ, about +10⁻⁵): weakly attracted, move from weaker to stronger field. Atoms have permanent dipole moments that are randomly oriented; an external field aligns them partly. Examples: aluminium, sodium, calcium, oxygen (at STP), copper chloride.

Ferromagnetic (χ ≫ 1, µᵣ > 1000): strongly attracted. Atomic dipoles align spontaneously in regions called domains, each about 1 mm across and containing about 10¹¹ atoms. An external field grows the aligned domains. Examples: iron, cobalt, nickel, gadolinium.

Hard ferromagnets (Alnico, lodestone) keep their magnetisation after the field is removed and make permanent magnets. Soft ferromagnets (soft iron) lose it quickly and are used in electromagnets and transformer cores.

Worked Examples

Example 1 (Book Example 5.5). A solenoid with n = 1000 turns per metre carries I = 2 A and has a core with µᵣ = 400. Find H, B, M and the magnetising current I_M.

H = nI = 1000 × 2 = 2 × 10³ A/m. B = µᵣµ₀H = 400 × 4π × 10⁻⁷ × 2 × 10³ ≈ 1.0 T. M = (µᵣ − 1)H = 399 × 2 × 10³ ≈ 8 × 10⁵ A/m. Writing B = µ₀n(I + I_M) gives I_M = χI = 399 × 2 ≈ 800 A (the book quotes 794 A because it uses the rounded B = 1.0 T). Note: the book prints this relation with µᵣ; it should read µ₀.

Example 2 (Exercise 5.1). A bar magnet at 30° to a 0.25 T field feels a torque of 4.5 × 10⁻² J. m = τ/(B sin θ) = 0.045/(0.25 × 0.5) = 0.36 J/T.

Example 3 (Exercise 5.3 and 5.4). A solenoid of 800 turns, cross-section 2.5 × 10⁻⁴ m², current 3.0 A: m = NIA = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 A m². In a 0.25 T field at 30°: τ = 0.6 × 0.25 × 0.5 = 0.075 N m.

Example 4 (Exercise 5.5). m = 1.5 J/T in B = 0.22 T, initially aligned. Work to turn it normal: W = mB(cos 0° − cos 90°) = 1.5 × 0.22 = 0.33 J. Work to turn it opposite: W = 2mB = 0.66 J.

Example 5 (Exercise 5.7). m = 0.48 J/T, r = 10 cm. Axial field B = (µ₀/4π)(2m/r³) = 10⁻⁷ × 0.96/10⁻³ = 9.6 × 10⁻⁵ T = 0.96 G. Equatorial field = 0.48 G, directed opposite to m.

Formula and Key-Facts Sheet

Solenoid moment: m = NIA

Torque: τ = m × B, τ = mB sin θ

Energy: U = −mB cos θ; work to rotate from θ₁ to θ₂: W = mB(cos θ₁ − cos θ₂)

Axial field: B = (µ₀/4π)(2m/r³); equatorial field: B = (µ₀/4π)(m/r³)

Gauss's law: ∮B·dS = 0

M = m_net/V; H = nI; B = µ₀(H + M)

M = χH; µᵣ = 1 + χ; µ = µ₀µᵣ

µ₀ = 4π × 10⁻⁷ T m A⁻¹; 1 G = 10⁻⁴ T

Diamagnetic χ < 0; paramagnetic χ small and positive; ferromagnetic χ ≫ 1

Board Exam Focus

Board papers regularly ask: compare the three classes of materials (3 marks); derive or state the torque and potential energy of a dipole; state Gauss's law for magnetism; numericals on m = NIA, τ = mB sin θ and work done in rotating a magnet.

Answer outline: "Distinguish between dia-, para- and ferromagnetic substances." Give susceptibility sign and size (negative; small positive; very large positive), µᵣ (< 1; slightly > 1; ≫ 1), behaviour in a non-uniform field (move to weaker field; move to stronger field; strongly to stronger field), effect on field lines (expelled; slightly concentrated; strongly concentrated) and two examples each.

Answer outline: "Write the potential energy of a dipole in a field and state stable and unstable positions." U = −mB cos θ; stable equilibrium at θ = 0° (U = −mB), unstable at θ = 180° (U = +mB); work to turn it from 0° to 180° is 2mB.

Answer outline: "State Gauss's law in magnetism and its significance." ∮B·dS = 0 for any closed surface; field lines are closed loops; no isolated magnetic monopoles.

Key Terms

Magnetic dipole moment (चुंबकीय द्विध्रुव आघूर्ण)

Magnetic field lines (चुंबकीय क्षेत्र रेखाएँ)

Torque (बल आघूर्ण)

Magnetisation (चुंबकन)

Magnetic intensity (चुंबकीय तीव्रता)

Magnetic susceptibility (चुंबकीय प्रवृत्ति)

Relative permeability (आपेक्षिक चुंबकशीलता)

Diamagnetism (प्रतिचुंबकत्व)

Paramagnetism (अनुचुंबकत्व)

Ferromagnetism (लौहचुंबकत्व)

Domain (डोमेन / प्रभाव क्षेत्र)

Superconductor (अतिचालक)

Permanent magnet (स्थायी चुंबक)

Electromagnet (विद्युत चुंबक)

Common Mistakes

Writing U = −mB sin θ; the energy uses cos θ, the torque uses sin θ.

Forgetting that inside a magnet field lines run S to N, so the lines are closed loops.

Using µᵣ where µ₀ is needed in B = µ₀n(I + I_M), or treating χ and µᵣ as equal (µᵣ = 1 + χ).

Saying diamagnetic materials have negative permeability. Their χ is negative; µᵣ is slightly less than 1 but still positive.

Forgetting the factor 2 on the axis (axial field is twice the equatorial field at the same distance).

Mixing gauss and tesla: 1 G = 10⁻⁴ T.

💡 Key Learning Points

✓A bar magnet is equivalent to a solenoid; its magnetic moment is m = NIA, and cutting it never isolates a pole.
✓A dipole in a uniform field feels no net force but a torque τ = m × B; its energy is U = −mB cos θ, least at θ = 0°.
✓Gauss's law for magnetism, ∮B·dS = 0, expresses the absence of magnetic monopoles.
✓B = µ₀(H + M), M = χH, µᵣ = 1 + χ and µ = µ₀µᵣ link the applied field, the material's response and the total field.
✓Materials are diamagnetic (χ < 0), paramagnetic (small χ > 0) or ferromagnetic (χ ≫ 1, domains); hard ferromagnets make permanent magnets and soft ones make electromagnet cores.

👨‍🏫 Teaching Tips

→Sprinkle iron filings around a bar magnet and around a solenoid side by side so students see the same field pattern.
→Build a table on the board comparing electric dipole and magnetic dipole formulas (p ↔ m, 1/ε₀ ↔ µ₀).
→Break a magnetised needle into pieces and test each with a compass to show there are no monopoles.
→Draw the U versus θ graph from 0° to 180° to show stable and unstable equilibrium visually.
→Use a three-column comparison chart for dia-, para- and ferromagnetism and make students add examples.
→Explain domains with a class activity: students first face random directions, then all turn to face one wall when a 'field' is applied.
→Show a video of magnetic levitation over a superconductor to make the Meissner effect memorable.
→Solve one numerical each on m = NIA, τ = mB sin θ and W = mB(cos θ₁ − cos θ₂) before assigning exercises.

📋 Assessment Questions

Calculate the torque on a bar magnet or solenoid in a uniform field and the work done to rotate it.
State Gauss's law for magnetism and explain what it implies about magnetic monopoles.
Define magnetisation, magnetic intensity, susceptibility and relative permeability, and use the relations between them.
Classify given substances as diamagnetic, paramagnetic or ferromagnetic with reasons based on χ and µᵣ.
Distinguish hard and soft ferromagnetic materials by their uses.

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