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Magnetism and Matter

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Chapter Notes

Overview
This chapter explores magnetism as a subject in its own right, building on the idea that moving charges produce magnetic fields. It covers the behaviour of bar magnets as magnetic dipoles, the analogy between bar magnets and solenoids, Gauss's law for magnetism, and the classification of materials into dia-, para-, and ferromagnetic based on their response to external magnetic fields. The one big idea a student must walk away with is that isolated magnetic poles do not exist, and all magnetic phenomena arise from circulating currents or dipoles, with materials responding differently to magnetic fields due to their internal atomic structure.
Key Concepts (17)
Magnetic monopoles
Isolated magnetic north or south poles that do not exist; unlike electric charges, magnetic poles cannot be isolated.
e.g. If a bar magnet is broken into two halves, each half becomes a new magnet with both north and south poles, not isolated poles.
Magnetic field lines
Visual representation of the magnetic field; they form continuous closed loops, do not intersect, and the tangent at any point gives the direction of the net magnetic field B.
e.g. Iron filings sprinkled around a bar magnet arrange themselves along the field lines, showing closed loops from north to south outside the magnet.
Bar magnet as an equivalent solenoid
A bar magnet can be thought of as a large number of circulating currents in analogy with a solenoid, producing similar magnetic field patterns.
e.g. Cutting a bar magnet in half is like cutting a solenoid; we get two smaller solenoids with weaker magnetic properties.
Magnetic moment (m)
A vector quantity associated with a magnet or current loop; for a bar magnet, it is equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.
e.g. A current-carrying solenoid of 800 turns, area 2.5 × 10⁻⁴ m², and current 3.0 A has a magnetic moment m = NIA = 800 × 3.0 × 2.5 × 10⁻⁴ = 0.6 J T⁻¹.
Torque on a magnetic dipole in a uniform magnetic field
The torque experienced by a magnetic dipole of moment m placed in a uniform magnetic field B is given by τ = m × B, with magnitude τ = mB sinθ, where θ is the angle between m and B.
e.g. A compass needle of magnetic moment m placed at an angle of 30° to a uniform field B experiences a restoring torque that tends to align it with the field.
Magnetic potential energy (Um)
The energy of a magnetic dipole in a magnetic field, given by Um = –m·B = –mB cosθ, with zero potential energy chosen at θ = 90°.
e.g. When a magnet is aligned with the field (θ = 0°), potential energy is minimum (–mB), which is the most stable position.
Gauss's law for magnetism
The net magnetic flux through any closed surface is zero, reflecting the fact that isolated magnetic poles (monopoles) do not exist.
e.g. For any closed surface, the number of magnetic field lines entering equals the number leaving, so the net flux is always zero.
Magnetisation (M)
The net magnetic dipole moment per unit volume of a material, M = m_net / V, measured in A m⁻¹.
e.g. A paramagnetic material placed in an external field develops a net magnetisation M in the direction of the field.
Magnetic intensity (H)
A vector field defined by H = B/μ₀ – M, representing the contribution to the total magnetic field from external factors like currents, measured in A m⁻¹.
e.g. For a solenoid with n turns per metre carrying current I, H = nI, independent of the core material.
Magnetic susceptibility (χ)
A dimensionless quantity that measures how a magnetic material responds to an external field, defined by M = χH.
e.g. For paramagnetic materials, χ is small and positive (e.g., χ ≈ +10⁻⁵); for diamagnetic materials, χ is small and negative (e.g., χ ≈ –10⁻⁵).
Relative magnetic permeability (μr)
A dimensionless quantity defined as μr = 1 + χ, indicating how much the material enhances or reduces the magnetic field compared to vacuum.
e.g. For a ferromagnetic material like iron, μr >> 1 (e.g., 400), so the magnetic field inside is greatly enhanced.
Diamagnetism
A property of materials that have a tendency to move from stronger to weaker parts of an external magnetic field, developing a net magnetic moment opposite to the applied field.
e.g. Bismuth, copper, and water are diamagnetic; they are weakly repelled by a magnet.
Paramagnetism
A property of materials that get weakly magnetised when placed in an external magnetic field, with a tendency to move from weak to strong field regions.
e.g. Aluminium, sodium, and oxygen are paramagnetic; they are weakly attracted to a magnet.
Ferromagnetism
A property of materials that get strongly magnetised in an external magnetic field, exhibiting spontaneous alignment of atomic dipoles over macroscopic domains.
e.g. Iron, cobalt, and nickel are ferromagnetic; they are strongly attracted to a magnet and can retain magnetisation.
Domain
A macroscopic volume in a ferromagnetic material (typically ~1 mm, containing about 10¹¹ atoms) where atomic dipole moments spontaneously align in a common direction.
e.g. In an unmagnetised iron bar, domains are randomly oriented; when an external field is applied, domains align and grow in the direction of the field.
Meissner effect
The phenomenon of perfect diamagnetism in superconductors, where the external magnetic field is completely expelled from the material (χ = –1, μr = 0).
e.g. A superconductor repels a magnet, allowing magnetically levitated trains to float above the track.
Magnetising current (Im)
The additional current that would need to be passed through the windings of a solenoid in the absence of a core to produce the same magnetic field B as with the core.
e.g. For a solenoid with μr = 400 and I = 2 A, the magnetising current is Im = 794 A.
Formulas & Equations (10)
B = (μ₀ / 4π) × (2m / r³)
Reads as: B equals mu-zero over four pi times two m over r cubed
Variables: B = axial magnetic field (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), m = magnetic moment (A m²), r = distance from centre of magnet (m)
Use when: Calculating the magnetic field on the axis of a short bar magnet at a distance r >> l (length of magnet).
⚠ Forgetting that this formula applies only for r >> l, and using it for points close to the magnet where the dipole approximation fails.
B = –(μ₀ / 4π) × (m / r³)
Reads as: B equals minus mu-zero over four pi times m over r cubed
Variables: B = equatorial magnetic field (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), m = magnetic moment (A m²), r = distance from centre of magnet (m)
Use when: Calculating the magnetic field on the equatorial line (normal bisector) of a short bar magnet at a distance r >> l.
⚠ Omitting the negative sign, which indicates the direction is opposite to the magnetic moment vector.
τ = m × B
Reads as: tau equals m cross B
Variables: τ = torque (N m), m = magnetic moment (A m²), B = magnetic field (T)
Use when: Finding the torque on a magnetic dipole (e.g., bar magnet or current loop) placed in a uniform magnetic field.
⚠ Writing τ = mB instead of the vector cross product; the magnitude is mB sinθ, not mB.
U_m = –m·B = –mB cosθ
Reads as: U_m equals minus m dot B equals minus m B cos theta
Variables: U_m = magnetic potential energy (J), m = magnetic moment (A m²), B = magnetic field (T), θ = angle between m and B
Use when: Calculating the potential energy of a magnetic dipole in a uniform magnetic field, with zero energy at θ = 90°.
⚠ Forgetting that the zero of potential energy is fixed at θ = 90°, so the formula gives –mB at θ = 0° and +mB at θ = 180°.
B = μ₀ (H + M)
Reads as: B equals mu-zero times H plus M
Variables: B = total magnetic field inside material (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), H = magnetic intensity (A m⁻¹), M = magnetisation (A m⁻¹)
Use when: Relating the total magnetic field in a material to the applied field (H) and the material's response (M).
⚠ Confusing H with B; H is the magnetic intensity due to external sources, while B is the total field including the material's contribution.
M = χH
Reads as: M equals chi times H
Variables: M = magnetisation (A m⁻¹), χ = magnetic susceptibility (dimensionless), H = magnetic intensity (A m⁻¹)
Use when: Describing the linear response of a magnetic material to an applied magnetic intensity.
⚠ Assuming χ is constant for all materials; for ferromagnets, χ is not linear and depends on the history of magnetisation.
B = μ₀ μᵣ H = μH
Reads as: B equals mu-zero mu-r H equals mu H
Variables: B = magnetic field (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), μᵣ = relative magnetic permeability (dimensionless), μ = magnetic permeability of material (T m A⁻¹), H = magnetic intensity (A m⁻¹)
Use when: Calculating the magnetic field inside a material given its relative permeability and the applied magnetic intensity.
⚠ Using μ₀ instead of μ for the material; μ = μ₀μᵣ, not μ₀ alone.
B₀ = μ₀ nI
Reads as: B-zero equals mu-zero n I
Variables: B₀ = magnetic field inside an empty solenoid (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), n = number of turns per unit length (m⁻¹), I = current (A)
Use when: Finding the magnetic field inside a long solenoid in vacuum or air.
⚠ Forgetting that this formula applies only for an ideal infinite solenoid; for finite solenoids, the field at ends is weaker.
H = nI
Reads as: H equals n I
Variables: H = magnetic intensity (A m⁻¹), n = number of turns per unit length (m⁻¹), I = current (A)
Use when: Calculating the magnetic intensity inside a solenoid, independent of the core material.
⚠ Thinking H depends on the core; H is determined solely by the free current and geometry, not the material.
B = μ₀ n (I + I_m)
Reads as: B equals mu-zero n times I plus I_m
Variables: B = magnetic field (T), μ₀ = permeability of free space (4π × 10⁻⁷ T m A⁻¹), n = number of turns per unit length (m⁻¹), I = current (A), I_m = magnetising current (A)
Use when: Finding the equivalent current needed in an empty solenoid to produce the same B as with a magnetic core.
⚠ Confusing I_m with the actual current; I_m is a hypothetical current representing the core's contribution.
How It Works
Plotting magnetic field lines using iron filings
  1. Step 1: Sprinkle iron filings on a sheet of glass placed over a short bar magnet.
  2. Step 2: Gently tap the glass to allow the filings to align along the magnetic field lines.
  3. Step 3: Observe the pattern; the filings form continuous curves from the north pole to the south pole outside the magnet.
  4. Step 4: The pattern mimics the magnetic field lines, with denser filings indicating stronger field regions.
💡 Iron filings align along the field lines but do not show the direction; a compass needle is needed to determine direction.
Bar magnet as an equivalent solenoid
  1. Step 1: Consider a finite solenoid of n turns per unit length carrying current I.
  2. Step 2: Calculate the axial magnetic field at a large distance r (r >> l) using the formula B = (μ₀/4π) × (2m/r³), where m = NIA is the magnetic moment of the solenoid.
  3. Step 3: Compare this field with the experimentally observed far axial field of a bar magnet.
  4. Step 4: Conclude that a bar magnet produces the same magnetic field as an equivalent solenoid with the same magnetic moment.
💡 The analogy holds only at large distances; near the magnet, the field patterns differ slightly due to the finite size of the solenoid.
Behaviour of a magnetic dipole in a uniform magnetic field
  1. Step 1: Place a magnetic dipole of moment m in a uniform magnetic field B.
  2. Step 2: The dipole experiences a torque τ = m × B, which tends to align m with B.
  3. Step 3: The potential energy of the dipole is U_m = –m·B, with minimum at θ = 0° (stable equilibrium) and maximum at θ = 180° (unstable equilibrium).
  4. Step 4: No net force acts on the dipole in a uniform field; only torque is present.
💡 In a non-uniform field, a dipole experiences both a net force and a torque, as seen with an iron nail near a bar magnet.
Gauss's law for magnetism applied to closed surfaces
  1. Step 1: Consider a closed surface S (e.g., a Gaussian surface) in a magnetic field.
  2. Step 2: Divide S into small area elements ΔS and calculate the magnetic flux through each: Δφ_B = B·ΔS.
  3. Step 3: Sum the flux over all area elements to get the net flux φ_B = Σ B·ΔS.
  4. Step 4: Since magnetic field lines form closed loops, the number of lines entering equals the number leaving, so φ_B = 0 for any closed surface.
💡 This law is a direct consequence of the non-existence of magnetic monopoles; contrast with Gauss's law for electricity where net flux equals q/ε₀.
Magnetisation of a material in a solenoid
  1. Step 1: Place a material core inside a long solenoid carrying current I.
  2. Step 2: The magnetic intensity H = nI is set up by the solenoid, independent of the core.
  3. Step 3: The material develops a magnetisation M = χH, where χ is its magnetic susceptibility.
  4. Step 4: The total magnetic field inside the material is B = μ₀(H + M) = μ₀μᵣH.
  5. Step 5: For paramagnetic materials (χ > 0), B > B₀; for diamagnetic (χ < 0), B < B₀; for ferromagnetic (χ >> 1), B >> B₀.
💡 H is due to free currents only; M is the material's response; B is the total field. The magnetising current I_m is a hypothetical current that would produce the same B in an empty solenoid.
Domain alignment in ferromagnetic materials
  1. Step 1: In an unmagnetised ferromagnetic material, domains (macroscopic regions ~1 mm) have randomly oriented magnetisation, resulting in zero net magnetisation.
  2. Step 2: When an external magnetic field B₀ is applied, domains aligned with B₀ grow in size, and other domains rotate to align with the field.
  3. Step 3: This alignment and growth continue until a single 'giant' domain is formed, producing strong net magnetisation.
  4. Step 4: Upon removal of the external field, hard ferromagnets (e.g., alnico) retain magnetisation, while soft ferromagnets (e.g., soft iron) lose it.
💡 Domain formation is a cooperative effect requiring quantum mechanics; at high temperatures, thermal agitation destroys domain structure, and the material becomes paramagnetic.
Exam Traps — Don't Make These Mistakes
⚠ Confusing magnetic field lines with electric field lines
✗ Magnetic field lines start at north pole and end at south pole, similar to electric field lines starting at positive charge and ending at negative charge.
→ Magnetic field lines form continuous closed loops; they do not start or end at poles. Unlike electric field lines, there are no sources or sinks for magnetic field lines.
✓ Magnetic field lines are continuous closed loops; they emerge from the north pole and enter the south pole outside the magnet, but inside the magnet they go from south to north, forming closed loops.
⚠ Forgetting the negative sign in equatorial field formula
✗ The magnetic field on the equatorial line of a bar magnet is B = (μ₀/4π) × (m/r³).
→ The correct formula has a negative sign: B = –(μ₀/4π) × (m/r³). The negative sign indicates that the field direction is opposite to the magnetic moment vector.
✓ The equatorial magnetic field of a short bar magnet is B = –(μ₀/4π) × (m/r³), where the negative sign shows the field is antiparallel to m.
⚠ Misapplying Gauss's law for magnetism to open surfaces
✗ Gauss's law for magnetism states that the net magnetic flux through any surface is zero.
→ Gauss's law applies only to closed surfaces. For an open surface, the net flux can be non-zero.
✓ Gauss's law for magnetism states that the net magnetic flux through any closed surface is zero.
⚠ Confusing magnetic intensity H with magnetic field B
✗ The magnetic field inside a solenoid with a core is B = μ₀H.
→ B = μ₀(H + M), not μ₀H. The magnetisation M of the core contributes to the total field. Only in vacuum (M = 0) does B = μ₀H.
✓ The total magnetic field inside a material is B = μ₀(H + M) = μ₀μᵣH, where μᵣ = 1 + χ.
⚠ Assuming all ferromagnets retain magnetisation
✗ Ferromagnetic materials always retain their magnetisation after the external field is removed.
→ Only hard ferromagnets (e.g., alnico) retain magnetisation; soft ferromagnets (e.g., soft iron) lose magnetisation when the external field is removed.
✓ Ferromagnetic materials can be classified as hard (retain magnetisation, used for permanent magnets) or soft (lose magnetisation, used for electromagnets).
Compare & Contrast
Magnetic susceptibility (χ)
Diamagnetic materials
Paramagnetic materials
Diamagnetic materials have negative χ (e.g., –10⁻⁵), while paramagnetic materials have positive χ (e.g., +10⁻⁵).
Field lines and sources
Electric dipole
Magnetic dipole
Electric field lines begin at positive charges and end at negative charges (sources and sinks exist), while magnetic field lines form continuous closed loops with no sources or sinks (monopoles do not exist).
Retention of magnetisation after removal of external field
Hard ferromagnets
Soft ferromagnets
Hard ferromagnets (e.g., alnico) retain magnetisation and are used for permanent magnets, while soft ferromagnets (e.g., soft iron) lose magnetisation and are used for electromagnets.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks
Most asked: Bar magnet as a magnetic dipole and its field (axial and equatorial) · Torque and potential energy of a magnetic dipole in a uniform magnetic field · Gauss's law for magnetism and its implications · Classification of materials: diamagnetic, paramagnetic, ferromagnetic · Magnetic susceptibility, permeability, and their relation · Domain theory of ferromagnetism
Don't skip: Derivation or application of axial and equatorial magnetic field of a bar magnet; Numerical problems on torque and potential energy of a magnetic dipole; Comparison of dia-, para-, and ferromagnetic materials with examples and χ values; Gauss's law for magnetism and its contrast with Gauss's law for electricity
One-Line Revision (read aloud before the exam)
  1. Magnetic monopoles do not exist; cutting a bar magnet always yields two smaller magnets with both poles.
  2. Magnetic field lines form continuous closed loops and never intersect.
  3. A bar magnet is equivalent to a solenoid of the same magnetic moment at large distances.
  4. Torque on a magnetic dipole in a uniform field is τ = m × B, with magnitude mB sinθ.
  5. Magnetic potential energy is U_m = –m·B, minimum at θ = 0° (stable) and maximum at θ = 180° (unstable).
  6. Gauss's law for magnetism: net magnetic flux through any closed surface is zero.
  7. Magnetisation M is the net magnetic dipole moment per unit volume.
  8. Magnetic intensity H = nI for a solenoid, independent of the core material.
  9. Total magnetic field in a material is B = μ₀(H + M) = μ₀μᵣH.
  10. Magnetic susceptibility χ is negative for diamagnetic, positive small for paramagnetic, and large positive for ferromagnetic materials.
  11. Diamagnetic materials are weakly repelled by magnets; examples: bismuth, copper, water.
  12. Paramagnetic materials are weakly attracted; examples: aluminium, sodium, oxygen.
  13. Ferromagnetic materials are strongly attracted and can retain magnetisation; examples: iron, cobalt, nickel.
  14. Domains are macroscopic regions (~1 mm) in ferromagnets where atomic dipoles align spontaneously.
  15. Hard ferromagnets retain magnetisation (permanent magnets); soft ferromagnets lose it (electromagnets).
  16. The Meissner effect is perfect diamagnetism in superconductors, expelling all external magnetic field.
  17. At high temperatures, ferromagnets become paramagnetic due to loss of domain structure.

MCQs — 10 Questions with Answers

Q1. Which statement correctly describes the properties of magnetic field lines of a bar magnet?

  • A. They form continuous closed loops and the tangent at any point gives the direction of B ✓
  • B. They originate from the north pole and terminate at the south pole like electric field lines
  • C. They can intersect at a point where the magnetic field is stronger
  • D. They represent the direction of force on a stationary charge placed in the field

Answer: A — Magnetic field lines form closed loops (unlike electric dipole field lines) because monopoles don't exist, and the tangent gives B direction; options B, C, and D are incorrect properties of magnetic fields.

Q2. A magnetized compass needle with magnetic moment m = 0.4 A·m² is freely suspended in Earth's magnetic field B = 4 × 10⁻⁵ T. What is the torque when the needle makes an angle of 30° with the field direction?

  • A. 8 × 10⁻⁶ N·m ✓
  • B. 2 × 10⁻⁶ N·m
  • C. 4 × 10⁻⁶ N·m
  • D. 1.6 × 10⁻⁵ N·m

Answer: A — τ = mB sin θ = 0.4 × 4 × 10⁻⁵ × sin(30°) = 0.4 × 4 × 10⁻⁵ × 0.5 = 8 × 10⁻⁶ N·m.

Q3. A bar magnet is broken into two pieces: one piece is cut along its length (from N pole to S pole) and the other is cut transversely. Which statement is correct?

  • A. Transverse cut gives two monopoles; longitudinal cut gives two complete magnets
  • B. Both cuts produce two complete magnets, each with N and S poles ✓
  • C. Longitudinal cut gives monopoles; transverse cut gives complete magnets
  • D. Neither cut can produce isolated magnetic poles under any condition

Answer: B — Both transverse and longitudinal cuts produce two complete bar magnets with their own N and S poles; magnetic monopoles cannot exist because internal circulating currents maintain dipole structure in each piece.

Q4. The magnetic potential energy of a compass needle (m = 1 A·m²) in Earth's field (B = 2 × 10⁻⁵ T) is minimum when the needle is oriented at angle θ. What is θ and the minimum energy value?

  • A. θ = 90°, Um(min) = 0
  • B. θ = 0°, Um(min) = –2 × 10⁻⁵ J ✓
  • C. θ = 180°, Um(min) = +2 × 10⁻⁵ J
  • D. θ = 45°, Um(min) = –√2 × 10⁻⁵ J

Answer: B — Potential energy Um = –mB cos θ is minimum when cos θ = 1, i.e., θ = 0° (aligned with field), giving Um(min) = –1 × 2 × 10⁻⁵ = –2 × 10⁻⁵ J.

Q5. Which of the following is NOT a correct statement about paramagnetic, diamagnetic, and ferromagnetic materials?

  • A. Paramagnetic materials are weakly attracted to magnetic fields due to unpaired electrons
  • B. Diamagnetic materials are repelled by magnetic fields and have negative magnetic susceptibility
  • C. Ferromagnetic materials lose their permanent magnetism above the Curie temperature
  • D. Diamagnetic materials like copper are more strongly attracted to fields than paramagnetic materials ✓

Answer: D — Diamagnetic materials are repelled (χ < 0, very small), while paramagnetic materials are attracted (χ > 0); ferromagnetic materials (χ >> 1) are much more strongly attracted than both—statement D is false.

Q6. A solenoid with n = 1000 turns/m, current I = 2 A, and cross-sectional area A = 0.01 m² acts as a magnetic dipole. The magnetic moment is (μ₀ = 4π × 10⁻⁷ T·m/A):

  • A. m = 20 A·m² ✓
  • B. m = 0.02 A·m²
  • C. m = 200 A·m²
  • D. m = 2 A·m²

Answer: A — For a solenoid, magnetic moment m = nIA = 1000 × 2 × 0.01 = 20 A·m² (magnetic moment depends on turns density, current, and area, not μ₀ in the final formula).

Q7. Two identical bar magnets are placed with one aligned along the direction of the other's magnetic field. Which statement is true regarding their magnetic field patterns?

  • A. The field lines remain the same as a single magnet because forces cancel
  • B. The S pole of the second magnet attracts the N pole of the first; field pattern shows enhanced field between them ✓
  • C. The N poles repel each other; the field pattern shows cancellation between the magnets
  • D. The combined field pattern becomes identical to that of a non-magnetic material

Answer: B — When magnets are aligned N-S, the S pole of the second attracts the N pole of the first (opposite poles attract); the magnetic field lines flow from N to S of the system, and the field is enhanced in the region between them where lines are concentrated.

Q8. The axial magnetic field of a finite solenoid at large distance r from its center is B = (μ₀/4π) × (2m/r³). This formula demonstrates that:

  • A. A bar magnet and solenoid produce identical fields only at very large distances ✓
  • B. A solenoid cannot be equivalent to a bar magnet due to different geometries
  • C. The magnetic field depends only on the current and not on the solenoid's structure
  • D. Both solenoids and bar magnets violate the principle that magnetic monopoles don't exist

Answer: A — At large distances (far-field), both solenoid and bar magnet produce the same dipole field B ∝ m/r³; this justifies Ampere's hypothesis that a bar magnet is equivalent to circulating currents inside it.

Q9. Assertion: A magnetic dipole placed perpendicular to a uniform magnetic field experiences maximum torque. Reason: Torque on a magnetic dipole is given by τ = m × B, which is maximum when m ⊥ B.

  • A. Both assertion and reason are true, and reason is the correct explanation ✓
  • B. Both assertion and reason are true, but reason is not the correct explanation
  • C. Assertion is true but reason is false
  • D. Both assertion and reason are false

Answer: A — When m ⊥ B (θ = 90°), sin θ = 1 and τ = mB is maximum; the vector cross product τ = m × B is also maximum in magnitude when vectors are perpendicular, so both assertion and reason are correct and logically linked.

Q10. Iron becomes paramagnetic above its Curie temperature (TC ≈ 770°C). This occurs because: (A) magnetic domains break up and thermal energy overcomes atomic alignment, (B) electrons gain sufficient kinetic energy to unpair, (C) ferromagnetism is replaced by induced diamagnetism.

  • A. Only (A) is correct ✓
  • B. Only (B) is correct
  • C. Only (C) is correct
  • D. (A), (B), and (C) are all equally correct

Answer: A — Above TC, thermal motion randomizes atomic magnetic moments that were cooperatively aligned in ferromagnetic domains; this disorder, not electron unpairing or diamagnetic induction, causes the transition from ferromagnetic to paramagnetic behavior.

Flashcards

Why do magnetic field lines form closed loops unlike electric field lines?

Because magnetic monopoles do not exist; field lines must emerge from one pole and enter the other pole, forming continuous closed loops.

What is the magnetic moment of a bar magnet equivalent to?

The magnetic moment of an equivalent solenoid that produces the same magnetic field at large distances.

Write the formula for torque on a magnetic dipole in an external field.

τ = m × B, or in magnitude τ = mB sin θ where θ is angle between m and B.

What is the magnetic potential energy of a dipole in a uniform field?

Um = –m · B = –mB cos θ, where energy is minimum at θ = 0° and maximum at θ = 180°.

What happens when a bar magnet is cut transversely into two pieces?

Each piece becomes a complete magnet with its own north and south poles; magnetic monopoles are never isolated.

Define paramagnetic substances and give one example.

Paramagnetic substances have unpaired electrons and are weakly attracted to external magnetic field; examples are aluminum, oxygen, and iron oxide.

What is diamagnetism and name one diamagnetic material.

Diamagnetism is the property of all materials to be repelled by a magnetic field due to paired electrons; bismuth is strongly diamagnetic.

Define ferromagnetism and what happens above the Curie temperature.

Ferromagnetism is strong permanent magnetism due to aligned atomic magnetic moments; above Curie temperature TC, thermal motion destroys alignment and it becomes paramagnetic.

How is a bar magnet equivalent to a current-carrying solenoid?

A bar magnet can be thought of as a solenoid with circulating currents; both produce identical magnetic field patterns at large distances.

At what angle θ is the torque on a magnetic dipole maximum in an external field?

Torque is maximum at θ = 90° where τ = mB (dipole is perpendicular to field).

Important Board Questions

Define magnetic dipole moment and write the expression for magnetic potential energy of a magnetic dipole in a uniform magnetic field. What is the condition for maximum stability? [2 marks]

Magnetic moment m = nIA for solenoid or Ampere's definition; potential energy Um = –m·B = –mB cos θ; maximum stability when θ = 0° (m parallel to B, Um = –mB).

A bar magnet of magnetic moment m = 0.5 A·m² is suspended freely in Earth's magnetic field B = 3 × 10⁻⁵ T. (a) Calculate the torque when the magnet makes an angle of 60° with the field. (b) At what angle is the potential energy zero? (c) Show that the potential energy is minimum when the magnet is aligned with the field. [5 marks]

Use τ = mB sin θ for part (a); potential energy Um = –mB cos θ is zero when cos θ = 0, so θ = 90°; for part (c), differentiate Um with respect to θ and show dUm/dθ = 0 at θ = 0° with d²Um/dθ² < 0 (confirming minimum).

State and derive that a bar magnet is equivalent to a solenoid carrying current. Show that the axial magnetic field of a finite solenoid at large distance r is B = (μ₀/4π) × (2m/r³), where m is the magnetic moment. Why do magnetic field lines form closed loops unlike electric field lines? [6 marks]

Use Ampere's hypothesis: atomic circulating currents create magnetic dipole; derive solenoid field by integrating Biot-Savart for circular loops at different positions; far-field (r >> length) gives dipole form ∝ m/r³; explain: magnetic monopoles don't exist, so flux must close on itself; compare with electric dipole where field lines terminate on charges.

Assertion-Reason Questions for Magnetism and Matter

Q1. Assertion (A):

When a bar magnet is broken into two pieces by cutting perpendicular to its length, each piece behaves like a complete bar magnet with two poles.

Reason (R):

According to Ampere's hypothesis, a bar magnet consists of circulating atomic currents distributed throughout its volume, similar to a finite solenoid.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true, and the reason correctly explains the assertion. Because the magnet is modelled as distributed circulating currents (like a solenoid), breaking it creates new solenoid-like structures with poles at each end. This is why magnetic monopoles don't exist—the dipole nature is preserved.

Q2. Assertion (A):

The magnetic field lines of a bar magnet form continuous closed loops, unlike electric field lines which begin and end on charges.

Reason (R):

Magnetic monopoles do not exist in nature, so magnetic field lines cannot have sources or sinks like electric field lines do.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both are true and causally related. The non-existence of magnetic monopoles (Gauss's law for magnetism: ∮B·dA = 0) ensures field lines form closed loops. Electric field lines have sources (positive charges) and sinks (negative charges), but magnets always have both poles together.

Q3. Assertion (A):

A magnetised needle suspended in Earth's magnetic field experiences zero torque when it is perpendicular to the field direction.

Reason (R):

The restoring torque on a magnetic dipole in a uniform field is given by τ = mB sinθ, which is zero when θ = 90°.

Answer: D — A is false but R is true — The assertion is false. When perpendicular (θ = 90°), sin(90°) = 1, so τ = mB (maximum), not zero. The torque is zero when aligned (θ = 0°) or anti-aligned (θ = 180°). The reason is correctly stated but leads to a false assertion—a common student mistake.

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Case-Based Questions (CBQ) for Magnetism and Matter

Case Study:

A geophysics student investigates Earth's magnetic field using a magnetometer. The device contains a small magnetised needle with magnetic moment m = 0.5 A·m² suspended in a non-uniform region where the field strength varies from 30 μT in one location to 50 μT in another location nearby (distance ~10 cm). The needle aligns itself with the field direction and experiences a slight restoring torque when displaced from equilibrium. Earth's average magnetic field strength is approximately 40 μT. The student uses this apparatus to create a simple compass at two different latitudes to study the variation of Earth's magnetism.

Q1. At the first location where B = 30 μT, if the needle is aligned with Earth's field (θ = 0°), the magnetic potential energy is approximately

  • A. –1.5 × 10⁻⁸ J ✓
  • B. –6.0 × 10⁻⁸ J
  • C. 0 J
  • D. +1.5 × 10⁻⁸ J

Answer: A

Q2. If the needle is displaced to θ = 60° at the location where B = 40 μT, the restoring torque is approximately

  • A. 1.73 × 10⁻⁸ N·m ✓
  • B. 1.0 × 10⁻⁸ N·m
  • C. 0.87 × 10⁻⁸ N·m
  • D. 2.0 × 10⁻⁸ N·m

Answer: A

Q3. The needle experiences a net force in the non-uniform field region (30 μT to 50 μT gradient). This force arises from

  • A. The torque acting on the dipole
  • B. The gradient in the magnetic field acting on the magnetic dipole moment ✓
  • C. The gravitational pull on the magnetised needle
  • D. Repulsion between Earth's field and the needle's poles

Answer: B

Q4. At higher magnetic latitudes, Earth's field strength is greater than at the equator. A compass needle at higher latitudes will oscillate with a

  • A. Variable period depending on the needle's initial displacement
  • B. Longer period because the restoring torque is weaker
  • C. Period independent of field strength
  • D. Shorter period because the restoring torque is stronger ✓

Answer: D

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Previous Year Questions (PYQ) for Magnetism and Matter

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 55/1/1 · 1 mark

Assertion (A): All atoms have a net magnetic moment. Reason (R): A current loop does not always behave as a magnetic dipole.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Both Assertion (A) and Reason (R) are false.

CBSE 2026 · Set 55/2/1 · 1 mark

Which of the following materials has positive and small value of magnetic susceptibility?

  • Cu
  • Al
  • Bi
  • Ni

CBSE 2026 · Set 55/2/2 · 1 mark

Which of the following substance has relative magnetic permeability μ_r >> 1?

  • Aluminium
  • Copper
  • Lead
  • Nickel

CBSE 2026 · Set 55/2/3 · 1 mark

Which one of the following materials has magnetic susceptibility χ in the range −1 ≤ χ < 0 ?

  • Aluminium
  • Copper chloride
  • Copper
  • Nickel

CBSE 2026 · Set 55/4/1 · 1 mark

Assertion (A): When a ferromagnetic substance is heated to high temperature it becomes paramagnetic in nature. Reason (R): The disappearance of magnetisation of a ferromagnet is abrupt and not gradual.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A)
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A)
  • Assertion (A) is true, but Reason (R) is false
  • Both Assertion (A) and Reason (R) are false

CBSE 2026 · Set 55/4/2 · 1 mark

A coil having 100 closely wound turns and area of cross-section 300 mm² carries a current of 5 mA. The magnitude of the magnetic moment associated with the coil is:

  • 6 × 10⁻⁴ Am²
  • 3 × 10⁻⁵ Am²
  • 1·5 × 10⁻⁴ Am²
  • 5 × 10⁻⁵ Am²

CBSE 2025 · Set 55/4/1 · 1 mark

A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is :

  • 30°
  • 45°
  • 60°
  • 75°

CBSE 2025 · Set 55/4/1 · 1 mark

Which one out of the following materials is not paramagnetic ?

  • Aluminium
  • Sodium Chloride
  • Calcium
  • Copper Chloride

CBSE 2025 · Set 55/4/2 · 1 mark

The materials having negative magnetic susceptibility are :

  • diamagnetic
  • paramagnetic
  • ferromagnetic
  • non-magnetic

CBSE 2025 · Set 55/4/3 · 1 mark

A bar magnet is initially at right angles to a uniform magnetic field. The magnet is rotated till the torque acting on it becomes one-half of its initial value. The angle through which the bar magnet is rotated is:

  • 30°
  • 45°
  • 60°
  • 75°

CBSE 2025 · Set 55/4/3 · 1 mark

Which of the following substances has magnetic permeability less than that of free space?

  • Sodium
  • Iron
  • Aluminium
  • Copper

CBSE 2025 · Set 55/5/1 · 1 mark

A current of (10/π) A is maintained in a circular loop of radius 14 cm. The value of dipole moment associated with the loop is:

  • 0·019 Am²
  • 0·14 Am²
  • 0·196 Am²
  • 0·615 Am²

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Class 12 Physics Formula Sheet

Every formula from all 14 chapters on one page, checked against the NCERT textbook and the CBSE 2026-27 syllabus. Free PDF download.

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Class 12 Physics Previous Year Questions, Chapter-wise

All 962 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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