Complete Summary and Solutions for Magnetism and Matter – NCERT Class XII Physics Part I, Chapter 5 – Magnetic Properties of Materials, Magnetic Field, and Magnetic Effects
Detailed summary and explanation of Chapter 5 'Magnetism and Matter' from the NCERT Class XII Physics Part I textbook, covering magnetic field intensity, magnetic field lines, magnetic properties of materials (diamagnetism, paramagnetism, ferromagnetism), hysteresis, and practical applications—along with all NCERT questions and answers.
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Categories: NCERT, Class XII, Physics Part I, Chapter 5, Magnetism, Magnetic Properties, Magnetic Field, Hysteresis, Summary, Questions, Answers
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Magnetism and Matter - Class 12 Physics Chapter 5 Ultimate Study Guide 2025
Magnetism and Matter
Chapter 5: Physics - Ultimate Study Guide | NCERT Class 12 Notes, Questions, Derivations & Quiz 2025
Full Chapter Summary & Detailed Notes - Magnetism and Matter Class 12 NCERT
Overview & Key Concepts
Chapter Goal: Understand magnetism in matter, bar magnets, Gauss's law, magnetic properties like dia-, para-, ferromagnetism. Exam Focus: Field lines, dipole, susceptibility; 2025 Updates: Applications in MRI, superconductors. Fun Fact: Earth's magnetism known since ancient times. Core Idea: Magnetic fields from currents, matter classification. Real-World: Compasses, magnets in devices. Expanded: All subtopics point-wise with evidence (e.g., Fig 5.1 filings), examples (e.g., solenoid), debates (monopoles).
Wider Scope: From basics to properties; sources: Text, figures (5.1-5.8), examples.
Expanded Content: Include calculations, analogies; links (e.g., to electrostatics); point-wise breakdown.
5.1 Introduction
Summary in Points: Magnetic phenomena universal. Galaxies to atoms permeated by fields. Earth's magnetism predates humans. Word 'magnet' from Magnesia island, 600 BC ore. Previous: Currents produce fields (Oersted, Ampere, Biot-Savart). Now: Magnetism as subject. Ideas: (i) Earth magnet south to north. (ii) Bar magnet aligns north-south, north pole to geographic north. (iii) Like repel, unlike attract. (iv) No isolated poles; breaking gives two magnets. (v) Magnets from iron alloys.
Begin with: Bar magnet behavior, Gauss's law, material classification: para-, dia-, ferromagnetism.
Summary in Points: Iron filings on glass over bar magnet show pattern (Fig 5.1). Suggests two poles like electric dipole. North to geographic north, south to south. Similar around current solenoid.
Summary in Points: Pattern allows plotting lines (Fig 5.2). Properties: (i) Closed loops (unlike electric). (ii) Tangent direction B. (iii) Density proportional magnitude. (iv) No intersection. Plot with compass needle.
Bar, solenoid, electric dipole; Gaussian surfaces.
5.2.2 Bar Magnet as an Equivalent Solenoid
Summary in Points: Resemblance lines suggest bar as circulating currents like solenoid. Cutting bar like solenoid gives smaller ones. Compass deflections similar. Calculate axial field finite solenoid (Fig 5.3a): B = μ0 m / (4π r^3). Equals bar far field. Magnetic moment equivalent.
Axial calculation; needle in uniform B (Fig 5.3b).
5.2.3 The Dipole in a Uniform Magnetic Field
Summary in Points: Compass needle m oscillates in B. Torque τ = m × B = m B sinθ. Potential energy U_m = -m · B = -m B cosθ. Min at θ=0° (stable), max 180° (unstable).
Expanded: Evidence: Eq (5.2-5.3); debates: Zero choice; real: Compass alignment.
Summary in Points: Lines continuous closed loops. Net flux zero any closed surface (unlike electric). φ_B = ∫ B · dS = 0. Reflection no monopoles; fields from dipoles/loops.
Expanded: Evidence: Fig 5.5; debates: Monopoles search; real: No net flux.
Diagram: Gauss Surfaces
Vector area ΔS; flux calculation.
5.4 Magnetisation and Magnetic Intensity
Summary in Points: Atomic moments add to net M = m_net / V. Solenoid B_0 = μ0 n I; with core B = B_0 + B_m, B_m = μ0 M. H = B/μ0 - M. B = μ0 (H + M). M = χ H, χ susceptibility. μ_r = 1 + χ, μ = μ0 μ_r.
Expanded: Evidence: Eqs (5.7-5.15); debates: H external; real: Core enhancement.
Diagram: Filled Solenoid
Interior field increase.
5.5 Magnetic Properties of Materials
Summary in Points: Classify by χ: Dia- (negative small), Para+ (positive small), Ferro+ (large). Table 5.2. Dia-: Repel, field reduced (Fig 5.7a). Para-: Attract weak, field enhanced (Fig 5.7b). Ferro-: Attract strong, domains align (Fig 5.8).
Expanded: Evidence: Figs 5.7-5.8; debates: Temperature dependence; real: Superconductors.
Diagram: Properties
Field lines near dia-/para- substances; domains.
5.5.1 Diamagnetism
Summary in Points: Move strong to weak field, repel. Orbiting e- equivalent loops; induced opposite field (Lenz). Materials: Bismuth, copper, water. Universal but weak; superconductors perfect (χ=-1, Meissner).
Expanded: Evidence: Fig 5.7a; debates: In all; real: Levitated trains.
5.5.2 Paramagnetism
Summary in Points: Weakly magnetised, move weak to strong. Atomic dipoles align in B_0 at low T. Field concentrated. Materials: Aluminium, oxygen. χ, μ_r depend T; saturation all aligned.
Expanded: Evidence: Fig 5.7b; debates: Random motion; real: Oxygen paramag.
5.5.3 Ferromagnetism
Summary in Points: Strongly magnetised, attract strong. Dipoles align in domains (~1mm, 10^11 atoms). B_0 aligns/grows domains. Hard (permanent like alnico), soft (disappear like iron). Elements: Fe, Co, Ni. μ_r >1000; Curie T to para-.
All terms from chapter; detailed with examples, relevance. Expanded: 25+ terms grouped by subtopic; added advanced like "susceptibility", "permeability".
Magnetic Pole
North/south ends bar magnet. Ex: North to geographic north. Relevance: No isolation.
Magnetic Dipole
Bar magnet two poles. Ex: Moment m. Relevance: Basic element.
Field Lines
Visualize B; closed loops. Ex: Density strength. Relevance: Direction/tangent.
Magnetic Moment
m = pole strength × distance. Ex: Solenoid equivalent. Relevance: Torque/energy.
Magnetisation
M = m_net / V. Ex: A m^{-1}. Relevance: Material response.
Magnetic Intensity
H = B/μ0 - M. Ex: External field. Relevance: Core independent.
χ large positive. Ex: Iron. Relevance: Strong attract, domains.
Domain
Aligned region ~1mm. Ex: 10^11 atoms. Relevance: Ferro- basis.
Tip: Group by type (fields/properties/materials); examples for recall. Depth: Debates (e.g., monopoles). Errors: Confuse χ and μ_r. Interlinks: To Ch4 currents. Advanced: Vector forms. Real-Life: MRI. Graphs: Hysteresis (though next ch). Coherent: Evidence → Interpretation. For easy learning: Flashcard per term with example.
Key Formulas - All Important Equations
List of all formulas from chapter; grouped, with units/explanations.
Formula
Description
Units/Example
B = μ0 m / (4π r^3)
Axial field solenoid/bar
T; far field
τ = m B sinθ
Torque on dipole
N m
U_m = - m B cosθ
Potential energy
J
B_E = - μ0 m / (4π r^3)
Equatorial field
T
B_A = μ0 2 m / (4π r^3)
Axial field
T
∫ B · dS = 0
Gauss's law magnetism
Wb
M = m_net / V
Magnetisation
A/m
B = μ0 (H + M)
Total field
T
M = χ H
Susceptibility
Dimensionless
μ_r = 1 + χ
Relative permeability
Dimensionless
B = μ H
With permeability
T
Tip: Memorize with units; practice analogies to electric.
Derivations - Detailed Guide
Key derivations with steps; from PDF (e.g., solenoid field, torque, energy).
All solved examples from the PDF with detailed explanations.
Example 5.1: (a) What happens if a bar magnet is cut into two pieces: (i) transverse to its length, (ii) along its length? (b) A magnetised needle in a uniform magnetic field experiences a torque but no net force. An iron nail near a bar magnet, however, experiences a force of attraction in addition to a torque. Why? (c) Must every magnetic configuration have a north pole and a south pole? What about the field due to a toroid? (d) Two identical looking iron bars A and B are given, one of which is definitely known to be magnetised. (We do not know which one.) How would one ascertain whether or not both are magnetised? If only one is magnetised, how does one ascertain which one? [Use nothing else but the bars A and B.]
Simple Explanation: Effects of cutting, force vs torque, poles, identify magnets.
Solution (a): In either case, one gets two magnets, each with a north and south pole.
Solution (b): No force if the field is uniform. The iron nail experiences a non-uniform field due to the bar magnet. There is induced magnetic moment in the nail, therefore, it experiences both force and torque. The net force is attractive because the induced south pole (say) in the nail is closer to the north pole of magnet than induced north pole.
Solution (c): Not necessarily. True only if the source of the field has a net non-zero magnetic moment. This is not so for a toroid or even for a straight infinite conductor.
Solution (d): Try to bring different ends of the bars closer. A repulsive force in some situation establishes that both are magnetised. If it is always attractive, then one of them is not magnetised. In a bar magnet the intensity of the magnetic field is the strongest at the two ends (poles) and weakest at the central region. This fact may be used to determine whether A or B is the magnet. In this case, to see which one of the two bars is a magnet, pick up one, (say, A) and lower one of its ends; first on one of the ends of the other (say, B), and then on the middle of B. If you notice that in the middle of B, A experiences no force, then B is magnetised. If you do not notice any change from the end to the middle of B, then A is magnetised.
Simple Way: Poles inseparable; induced moments.
Example 5.2: Figure 5.4 shows a small magnetised needle P placed at a point O. The arrow shows the direction of its magnetic moment. The other arrows show different positions (and orientations of the magnetic moment) of another identical magnetised needle Q. (a) In which configuration the system is not in equilibrium? (b) In which configuration is the system in (i) stable, and (ii) unstable equilibrium? (c) Which configuration corresponds to the lowest potential energy among all the configurations shown?
Simple Explanation: Equilibrium, stability, energy for configurations.
Solution: Potential energy of the configuration arises due to the potential energy of one dipole (say, Q) in the magnetic field due to other (P). Use the result that the field due to P is given by the expression [Eqs. (5.4) and (5.5)]: B_p = - μ0 m_p / (4π r^3) (on the normal bisector); B_p = μ0 2 m_p / (4π r^3) (on the axis) where m_p is the magnetic moment of the dipole P. Equilibrium is stable when m_Q is parallel to B_P, and unstable when it is anti-parallel to B_P. For instance for the configuration Q3 for which Q is along the perpendicular bisector of the dipole P, the magnetic moment of Q is parallel to the magnetic field at the position 3. Hence Q3 is stable. Thus, (a) PQ1 and PQ2 (b) (i) PQ3, PQ6 (stable); (ii) PQ5, PQ4 (unstable) (c) PQ6
Simple Way: Stable parallel, unstable anti; lowest energy aligned.
Example 5.3: Many of the diagrams given in Fig. 5.6 show magnetic field lines (thick lines in the figure) wrongly. Point out what is wrong with them. Some of them may describe electrostatic field lines correctly. Point out which ones.
Simple Explanation: Identify incorrect field lines.
Solution (a): Wrong. Magnetic field lines can never emanate from a point, as shown in figure. Over any closed surface, the net flux of B must always be zero, i.e., pictorially as many field lines should seem to enter the surface as the number of lines leaving it. The field lines shown, in fact, represent electric field of a long positively charged wire. The correct magnetic field lines are circling the straight conductor, as described in Chapter 4.
Solution (b): Wrong. Magnetic field lines (like electric field lines) can never cross each other, because otherwise the direction of field at the point of intersection is ambiguous. There is further error in the figure. Magnetostatic field lines can never form closed loops around empty space. A closed loop of static magnetic field line must enclose a region across which a current is passing. By contrast, electrostatic field lines can never form closed loops, neither in empty space, nor when the loop encloses charges.
Solution (c): Right. Magnetic lines are completely confined within a toroid. Nothing wrong here in field lines forming closed loops, since each loop encloses a region across which a current passes. Note, for clarity of figure, only a few field lines within the toroid have been shown. Actually, the entire region enclosed by the windings contains magnetic field.
Solution (d): Wrong. Field lines due to a solenoid at its ends and outside cannot be so completely straight and confined; such a thing violates Ampere’s law. The lines should curve out at both ends, and meet eventually to form closed loops.
Solution (e): Right. These are field lines outside and inside a bar magnet. Note carefully the direction of field lines inside. Not all field lines emanate out of a north pole (or converge into a south pole). Around both the N-pole, and the S-pole, the net flux of the field is zero.
Solution (f): Wrong. These field lines cannot possibly represent a magnetic field. Look at the upper region. All the field lines seem to emanate out of the shaded plate. The net flux through a surface surrounding the shaded plate is not zero. This is impossible for a magnetic field. The given field lines, in fact, show the electrostatic field lines around a positively charged upper plate and a negatively charged lower plate. The difference between Fig. [5.6(e) and (f)] should be carefully grasped.
Solution (g): Wrong. Magnetic field lines between two pole pieces cannot be precisely straight at the ends. Some fringing of lines is inevitable. Otherwise, Ampere’s law is violated. This is also true for electric field lines.
Simple Way: Closed loops, no emanation, no crossing.
Example 5.4: (a) Magnetic field lines show the direction (at every point) along which a small magnetised needle aligns (at the point). Do the magnetic field lines also represent the lines of force on a moving charged particle at every point? (b) Magnetic monopoles do not exist. If magnetic monopoles existed, how would the Gauss’s law of magnetism be modified? (c) Does a bar magnet exert a torque on itself due to its own field? Does one element of a current-carrying wire exert a force on another element of the same wire? (d) Magnetic field arises due to charges in motion. Can a system have magnetic moments even though its net charge is zero?
Simple Explanation: Lines of force? Monopoles? Self-torque? Net zero charge moments?
Solution (a): No. The magnetic force is always normal to B (remember magnetic force = q v × B). It is misleading to call magnetic field lines as lines of force.
Solution (b): Gauss’s law of magnetism states that the flux of B through any closed surface is always zero ∫_s B · dS = 0. If monopoles existed, the right hand side would be equal to the monopole (magnetic charge) q_m enclosed by S. [Analogous to Gauss’s law of electrostatics, ∫_s B · dS = μ0 q_m where q_m is the (monopole) magnetic charge enclosed by S.]
Solution (c): No. There is no force or torque on an element due to the field produced by that element itself. But there is a force (or torque) on an element of the same wire. (For the special case of a straight wire, this force is zero.)
Solution (d): Yes. The average of the charge in the system may be zero. Yet, the mean of the magnetic moments due to various current loops may not be zero. We will come across such examples in connection with paramagnetic material where atoms have net dipole moment through their net charge is zero.
Example 5.5: A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current I_m.
Simple Explanation: Calculate fields in cored solenoid.
Solution (a): The field H is dependent of the material of the core, and is H = n I = 1000 × 2.0 = 2 × 10^3 A/m.
Solution (b): The magnetic field B is given by B = μ_r μ0 H = 400 × 4π × 10^{-7} (N/A^2) × 2 × 10^3 (A/m) = 1.0 T
Solution (c): Magnetisation is given by M = (B – μ0 H)/ μ0 = (μ_r μ0 H – μ0 H)/μ0 = (μ_r – 1) H = 399 × H ≅ 8 × 10^5 A/m
Solution (d): The magnetising current I_M is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. Thus B = μ0 n (I + I_M). Using I = 2A, B = 1 T, we get I_M = 794 A.
Simple Way: H from windings, M from core, B total.
Tip: All textbook examples covered with full details from PDF.
NCERT Textbook Exercise Questions & Solutions
All NCERT exercise questions with detailed solutions (5.1 to 5.7 from PDF).
5.1 A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^{-2} J. What is the magnitude of magnetic moment of the magnet?
5.2 A short bar magnet of magnetic moment m = 0.32 J T^{-1} is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution:
(a) Stable: θ=0°, U = -m B = -0.32×0.15 = -0.048 J.
(b) Unstable: θ=180°, U = m B = 0.048 J.
Long Note: Energy min stable.
5.3 A closely wound solenoid of 800 turns and area of cross section 2.5 × 10^{-4} m^2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Solution:
Acts as bar: Field lines similar. m = n I A = 800 × 3 × 2.5×10^{-4} = 0.6 J/T.
Long Note: Direction by right-hand rule.
5.4 If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Solution:
τ = m B sinθ = 0.6 × 0.25 × sin30° = 0.075 N m.
Long Note: Tends align.
5.5 A bar magnet of magnetic moment 1.5 J T^{-1} lies aligned with the direction of a uniform magnetic field of 0.22 T. (a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction? (b) What is the torque on the magnet in cases (i) and (ii)?
Solution:
(a) (i) W = m B (1 - cos90°) = 1.5×0.22×1 = 0.33 J. (ii) W = m B (1 - cos180°) = 0.66 J.
(b) (i) τ = m B sin90° = 0.33 N m. (ii) τ = m B sin180° = 0.
Long Note: Work against potential.
5.6 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^{-4} m^2, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane. (a) What is the magnetic moment associated with the solenoid? (b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^{-2} T is set up at an angle of 30° with the axis of the solenoid?
Solution:
(a) m = n I A = 2000 × 4 × 1.6×10^{-4} = 1.28 J/T.
(b) Force zero uniform; τ = m B sinθ = 1.28 × 0.075 × sin30° = 0.048 N m.
Long Note: No net force dipole uniform.
5.7 A short bar magnet has a magnetic moment of 0.48 J T^{-1}. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Solution:
(a) Axial: B = μ0 2 m /(4π r^3) = 4π×10^{-7} × 2×0.48 / (4π × 0.1^3) = 9.6×10^{-5} T along S to N.
(b) Equatorial: B = - μ0 m /(4π r^3) = 4.8×10^{-5} T along N to S.
Long Note: Directions opposite.
Tip: At least 7 exercise questions covered with detailed point-wise solutions.
Lab Activities - Step-by-Step Guide
From PDF (e.g., plot field lines, dipole oscillation); explain how to do.
Activity 1: Plot Magnetic Field Lines of Bar Magnet
Step-by-Step:
Step 1: Place bar on paper.
Step 2: Sprinkle iron filings.
Step 3: Tap gently.
Step 4: Observe pattern.
Observation: Lines from N to S, closed.
Precaution: Uniform sprinkle.
Activity 2: Measure Dipole Moment
Step-by-Step:
Step 1: Suspend needle in B.
Step 2: Measure period T.
Step 3: m = (4π² I / T²) / B_h.
Step 4: Compare values.
Observation: Torque restores.
Precaution: Horizontal component.
Note: PDF implies experiments like filings; general for verification.
Key Concepts - In-Depth Exploration
Core ideas with examples, pitfalls, interlinks. Expanded: All concepts with steps/examples/pitfalls.
Magnetic Field Lines
Steps: 1. Closed loops, 2. Tangent B, 3. Density magnitude, 4. No cross. Ex: Bar patterns. Pitfall: Not force lines. Interlink: Gauss law. Depth: Compass plotting.
Bar as Solenoid
Steps: 1. Currents analog, 2. Field calculation, 3. Moment equal. Ex: Cutting similar. Pitfall: Atomic level. Interlink: Ampere. Depth: Far field.
Dipole in Field
Steps: 1. Torque aligns, 2. Energy min parallel. Ex: Compass. Pitfall: Zero choice. Interlink: Electric analog. Depth: Stable/unstable.
