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Magnetism and Matter: Class 12 Physics Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A geologist observes that a rock sample repels a compass needle. Laboratory tests show the sample moves from stronger to weaker field regions between electromagnet poles, has no residual magnetism after field removal, and has susceptibility χ = −0.000085. A solenoid with 500 turns, length 25 cm, and current 2 A uses this material as its core.

A geologist is surveying a region and uses a magnetic compass to navigate. She notices that near a certain rocky outcrop, the compass needle is slightly repelled from the rock surface instead of being attracted to it. She collects a small sample of the rock and takes it to a laboratory. In the lab, the following observations are recorded:
(i) When placed between the poles of a strong electromagnet, the sample moves from the stronger field region to the weaker field region.
(ii) When the external magnetic field is removed, the sample shows no residual magnetism.
(iii) The susceptibility (χ) of the material is measured to be −0.000085.
(iv) A cylindrical bar of this material, wound with N = 500 turns over a length of 25 cm and carrying a current of 2 A, is used as a solenoid core.

Based on this scenario, answer the following:
(a) Identify the magnetic class of the rock sample. Name ONE common example of such a material.
(b) What is the relative permeability (μ<sub>r</sub>) of this material? Is it greater than, equal to, or less than 1?
(c) Calculate the magnetisation (M) of the material inside the solenoid. (Given: χ = −0.000085)
(d) The geologist's compass needle is deflected away from the rock. Justify this behaviour using the concept of magnetic susceptibility.

Show answer
(a) Identification of magnetic class:
The sample moves from stronger to weaker field regions, shows no residual magnetism, and has a small negative susceptibility (χ < 0). These are the characteristic properties of a DIAMAGNETIC material.

∴ The rock sample is DIAMAGNETIC.

Common example: Bismuth (Bi) [also acceptable: Copper (Cu), Gold (Au), Silver (Ag)].

(b) Relative permeability:
For any magnetic material, the relative permeability is given by:

μ<sub>r</sub> = 1 + χ

Substituting χ = −0.000085:

μ<sub>r</sub> = 1 + (−0.000085)

∴ μ<sub>r</sub> = 0.999915

Since χ < 0, μ<sub>r</sub> is slightly less than 1.

(c) Magnetisation (M) inside the solenoid:
The magnetising field H inside a solenoid is given by:

H = nI

where n = N/L = number of turns per unit length, and I = current.

n = 500 / 0.25 = 2000 turns m<sup>−1</sup>

→ H = 2000 × 2 = 4000 A m<sup>−1</sup>

Using the relation between magnetisation and susceptibility:

M = χ H

M = (−0.000085) × 4000

∴ M = −0.34 A m<sup>−1</sup>

(The negative sign confirms magnetisation is opposite to the applied field, consistent with diamagnetic behaviour.)

(d) Justification of compass deflection:
Magnetic susceptibility χ is defined as the ratio of magnetisation M to the magnetising field H, and it measures how a material responds to an external magnetic field.

For the diamagnetic rock sample, χ < 0 (χ = −0.000085). This means when an external magnetic field (from the Earth or nearby magnet) is applied, the sample acquires a small magnetisation in the DIRECTION OPPOSITE to the applied field.

As a result, the rock sample creates a weak field that opposes the external field near its surface. The compass needle, which aligns with the net local field, is therefore slightly pushed away (repelled) from the rock instead of being attracted.

Because diamagnets are repelled by both poles of a bar magnet and move toward regions of weaker field, the compass needle deflects away from the rock surface.
Q2Case-based4 marks

A geologist studies two rock samples near a strong bar magnet. Sample P causes field lines to crowd inside it and aligns parallel to the field (strongly attracted). Sample Q causes field lines to become sparser inside it and aligns perpendicular to the field (weakly repelled).

A geologist is studying rock samples from two different volcanic sites. She places both samples near a strong bar magnet and observes the following:

• Sample P: The field lines around it become denser inside the sample and sparser outside it. When suspended freely on a thread, it aligns itself parallel to the applied field and is attracted towards the stronger region of the field.

• Sample Q: The field lines around it become slightly sparser inside and denser just outside. When suspended freely, it aligns itself perpendicular to the applied field and is weakly repelled from the stronger region.

Based on these observations, answer the following:

(i) Identify the magnetic nature of Sample P and Sample Q. Give one identifying property of each. [1 mark]

(ii) For Sample P, the applied magnetic field intensity H = 1200 A/m and the intensity of magnetisation M = 2.4 × 10⁵ A/m. Calculate the relative permeability μᵣ of Sample P. [1 mark]

(iii) The geologist heats Sample P above a certain critical temperature and repeats the experiment. She finds that it no longer shows strong attraction; instead, it behaves like Sample P at room temperature had very weak alignment. Name this critical temperature and explain what happens to the internal structure of the material at this temperature. [1 mark]

(iv) A uniform external magnetic field B⃗ = 0.4 T is applied. Compare the magnetic field inside Sample P (χ = 200) with that inside Sample Q (χ = −0.00015). Which has a greater magnetic field inside it, and by approximately how much? [1 mark]

Show answer
(i) Sample P is a ferromagnetic substance.
Property: It has very high positive susceptibility (χ >> 1) and aligns strongly parallel to the applied field.
Sample Q is a diamagnetic substance.
Property: It has small negative susceptibility (χ < 0) and is weakly repelled by the field, aligning perpendicular to it.

(ii) By definition, magnetic susceptibility χ = M/H.
→ χ = (2.4 × 10⁵) / 1200 = 200
Relative permeability: μᵣ = 1 + χ
→ μᵣ = 1 + 200 = 201
∴ μᵣ of Sample P = 201

(iii) The critical temperature is called the Curie Temperature.
Below the Curie temperature, a ferromagnetic material has small regions called magnetic domains, each with a net magnetic moment aligned in the same direction. These domains align with an external field, giving strong magnetisation.
Above the Curie temperature, thermal agitation disrupts the alignment of magnetic dipoles within domains — the domain structure breaks down completely and the material becomes paramagnetic (very weak, temperature-dependent alignment with the field).

(iv) The magnetic field inside a material is related to the applied field by:
B_inside = μᵣ × B₀ = (1 + χ) × B₀

For Sample P (ferromagnetic, χ = 200):
B_P = (1 + 200) × 0.4 = 201 × 0.4 = 80.4 T

For Sample Q (diamagnetic, χ = −0.00015):
B_Q = (1 + (−0.00015)) × 0.4 = 0.99985 × 0.4 ≈ 0.39994 T

Difference ≈ 80.4 − 0.39994 ≈ 80.0 T
∴ Sample P has a far greater magnetic field inside it, approximately 80 T greater than inside Sample Q. The ferromagnetic sample amplifies the field enormously due to its very large relative permeability, while the diamagnetic sample marginally reduces it.
Q3Case-based4 marks

A geologist surveys a rocky terrain and observes unusual compass deflections near certain rock samples. By carefully studying the field-line patterns and magnetic response of the samples in controlled external fields, she classifies the samples into different magnetic categories. Her measurements reveal that one sample crowds the magnetic field lines slightly into itself, while another sample actually repels the field lines faintly. These contrasting behaviours form the basis for the questions below.

A geologist is surveying a region and notices that a compass needle placed near a rock sample gets slightly attracted towards the sample and the field lines around the sample appear to converge (crowd slightly) into it. She records the following observations:
(i) The sample weakly attracts both poles of a bar magnet.
(ii) When placed in an external magnetic field B⃗, the sample develops a weak magnetisation in the direction of B⃗.
(iii) The relative permeability μᵣ of the sample is slightly greater than 1.
(iv) On heating the sample strongly, its magnetic behaviour practically vanishes above a certain temperature.

Based on these observations, answer the following:
(a) Identify the magnetic class (type) of the material. Give ONE characteristic property that distinguishes it from diamagnetic materials. [1]
(b) The sample has a volume of 2×10⁻⁵ m³ and, when placed in an external field of 0.4 T, acquires a uniform magnetisation M = 3.2 A m⁻¹. Calculate the magnetic moment of the sample. [1]
(c) A small bar magnet of moment m = 0.5 A m² is placed at the centre of a circular coil of 50 turns, radius 0.2 m, carrying a current of 2 A, with its axis along the axis of the coil. Calculate the magnitude of the torque on the bar magnet if it is tilted so that its axis makes an angle of 30° with the magnetic field of the coil. [1]
(d) The geologist also finds a second sample whose magnetic susceptibility χ is small and negative. Identify its magnetic class and state whether its relative permeability μᵣ is greater than, equal to, or less than 1. [1]

Show answer
(a) The material is PARAMAGNETIC.

Distinguishing property from diamagnetic materials:
A paramagnetic material has a positive (though small) magnetic susceptibility (χ > 0) and is feebly attracted towards the stronger region of an external magnetic field, whereas a diamagnetic material has a negative susceptibility (χ < 0) and is feebly repelled. (Any one valid point accepted.)

(b) By definition, the magnetic moment of a uniformly magnetised sample is:

m = M × V

where M is the magnetisation and V is the volume.

Substituting:
m = 3.2 × 2×10⁻⁵

∴ Magnetic moment of sample = 6.4×10⁻⁵ A m²

(c) The magnetic field at the centre of a circular coil of N turns, radius R, carrying current I is given by:

B = μ₀NI / 2R

Substituting (μ₀ = 4π×10⁻⁷ T m A⁻¹, N = 50, I = 2 A, R = 0.2 m):

B = (4π×10⁻⁷ × 50 × 2) / (2 × 0.2)

B = (4π×10⁻⁷ × 100) / 0.4

B = (4π×10⁻⁵) / 0.4

B = π×10⁻⁴ T ≈ 3.14×10⁻⁴ T

The torque on a magnetic dipole in a uniform field is:

τ = mB sinθ

Substituting (m = 0.5 A m², B = π×10⁻⁴ T, θ = 30°, sin 30° = 0.5):

τ = 0.5 × π×10⁻⁴ × 0.5

τ = 0.25π×10⁻⁴

∴ Torque on bar magnet = 7.85×10⁻⁵ N m (≈ 0.25π×10⁻⁴ N m)

(d) The second sample is DIAMAGNETIC.

For a diamagnetic material, χ < 0, and since μᵣ = 1 + χ, its relative permeability μᵣ is LESS THAN 1.
Q4Case-based4 marks

A geophysicist studies two rock samples: Sample P (strongly attracted to a magnet, retains magnetisation) and Sample Q (weakly repelled, no retained magnetisation). Sample P has magnetic dipole moment M = 0.48 A m²; Earth's horizontal component B_H = 0.30 × 10⁻⁴ T; distance on equatorial line r = 0.20 m.

A geophysicist is analysing the magnetic properties of two rock samples collected from different geological sites. Sample P is found to be strongly attracted to a bar magnet and retains its magnetisation even after the magnet is removed. Sample Q is weakly repelled by the same bar magnet and loses any induced magnetisation the moment the external field is removed.

(i) Identify the magnetic class of Sample P and Sample Q. Give one distinguishing property of each.

(ii) Sample P, when analysed, is found to have a magnetic dipole moment M = 0.48 A m² and is placed with its axis along the direction of Earth's horizontal component of magnetic field B_H = 0.30 × 10⁻⁴ T. Calculate the torque acting on it when its axis makes an angle of 30° with B_H.

(iii) The geophysicist now places Sample P (treated as a small bar magnet of moment M = 0.48 A m²) at a point on the equatorial line at a distance r = 0.20 m from its centre. Calculate the magnitude of the magnetic field at that point.

(iv) If the temperature of Sample P is raised well above its Curie temperature, predict what happens to its magnetic behaviour. Justify your answer.

Show answer
(i) Identification and distinguishing properties: [1 mark]

Sample P → Ferromagnetic material.
Distinguishing property: Ferromagnetic materials have very large positive susceptibility (χ >> 1) and possess spontaneous domain alignment; they retain magnetisation (hysteresis) after the external field is removed.

Sample Q → Diamagnetic material.
Distinguishing property: Diamagnetic materials have small negative susceptibility (χ < 0, typically −10⁻⁵ to −10⁻⁶); they are feebly repelled by external magnetic fields and do not retain any magnetisation.

──────────────────────────────────────
(ii) Torque on Sample P: [1 mark]

The torque on a magnetic dipole in a uniform field is given by:
τ = M B sinθ

Substituting values (M = 0.48 A m², B = 0.30 × 10⁻⁴ T, θ = 30°):
τ = 0.48 × 0.30 × 10⁻⁴ × sin 30°
τ = 0.48 × 0.30 × 10⁻⁴ × 0.5
τ = 0.48 × 0.15 × 10⁻⁴

∴ τ = 7.2 × 10⁻⁶ N m

──────────────────────────────────────
(iii) Magnetic field at the equatorial point: [1 mark]

The magnetic field at a point on the equatorial line of a short magnetic dipole is given by:
B_eq = (μ₀ / 4π) × (M / r³)

where μ₀ / 4π = 10⁻⁷ T m A⁻¹, M = 0.48 A m², r = 0.20 m.

Substituting:
B_eq = 10⁻⁷ × (0.48 / (0.20)³)
B_eq = 10⁻⁷ × (0.48 / 8 × 10⁻³)
B_eq = 10⁻⁷ × (0.48 / 0.008)
B_eq = 10⁻⁷ × 60

∴ B_eq = 6.0 × 10⁻⁶ T (direction antiparallel to M⃗)

──────────────────────────────────────
(iv) Effect of raising temperature above Curie temperature: [1 mark]

When the temperature of Sample P is raised well above its Curie temperature (T_C), the ferromagnetic material undergoes a phase transition and becomes paramagnetic.

Justification: Below T_C, thermal energy is insufficient to disrupt the aligned magnetic domains responsible for ferromagnetism. Above T_C, thermal agitation destroys the long-range domain alignment completely. The susceptibility then follows the Curie–Weiss law: χ = C / (T − T_C), and the material shows only weak, random alignment with an external field — characteristic of paramagnetism. The material will no longer retain magnetisation on removal of the external field.
Q5Case-based4 marks

A geophysics student studies three volcanic rock samples X, Y, and Z placed in a strong external magnetic field. Sample X is feebly magnetised along the field and loses magnetism on removal. Sample Y is strongly magnetised along the field and retains magnetism on removal. Sample Z is feebly magnetised opposite to the field and loses magnetism on removal.

A geophysics student is studying the magnetic properties of three rock samples collected from different volcanic sites. She places each sample inside a strong external magnetic field and observes the following:

• Sample X: Gets feebly magnetised in the direction of the external field; loses magnetism completely when the field is removed.
• Sample Y: Gets strongly magnetised in the direction of the external field; retains a large amount of magnetism even after the field is removed.
• Sample Z: Gets feebly magnetised in the direction OPPOSITE to the external field; loses magnetism when the field is removed.

Based on these observations, answer the following questions:

(i) Identify the magnetic class (paramagnetic, diamagnetic, or ferromagnetic) of each sample X, Y, and Z.
(ii) The student measures the magnetic susceptibility (χ) of Sample X. State the sign and approximate magnitude (small/large) of χ for Sample X.
(iii) Sample Y is made into a toroid of mean radius 10 cm, wound with 500 turns, and a current of 2 A is passed through it. Given that the relative permeability μᵣ of Sample Y is 800, calculate the magnetic field B inside the toroid.
(iv) Explain briefly why Sample Z cannot be used as a permanent magnet.

Show answer
(i) Identification of magnetic class of each sample:

Sample X → Paramagnetic
Reasoning: Feebly magnetised along the field direction; magnetism disappears on removal of field. Susceptibility χ is small and positive.

Sample Y → Ferromagnetic
Reasoning: Strongly magnetised along the field; retains magnetism after removal (shows remanence). Susceptibility χ is large and positive.

Sample Z → Diamagnetic
Reasoning: Feebly magnetised opposite to the applied field; no residual magnetism. Susceptibility χ is small and negative.

∴ X — Paramagnetic, Y — Ferromagnetic, Z — Diamagnetic.

(ii) Magnetic susceptibility of Sample X (Paramagnetic):

For a paramagnetic material, the susceptibility χ is positive (magnetisation is along the field) and small in magnitude.

∴ Sign of χ: Positive (+); Magnitude: Small (typically 10⁻⁵ to 10⁻³).

(iii) Calculation of magnetic field B inside the toroid of Sample Y:

The magnetic field inside a toroid is given by:

B = μ₀ μᵣ n I

where:
• μ₀ = 4π × 10⁻⁷ T m A⁻¹
• μᵣ = 800 (relative permeability of Sample Y)
• n = number of turns per unit length = N / (2πr)
• N = 500 turns, r = 10 cm = 0.10 m
• I = 2 A

Calculating n:
n = 500 / (2π × 0.10) = 500 / (0.2π) = 2500/π turns m⁻¹

Substituting:
B = (4π × 10⁻⁷) × 800 × (2500/π) × 2

B = (4π × 10⁻⁷) × 800 × (2500/π) × 2

= 4π × 10⁻⁷ × (800 × 2500 × 2) / π

= 4 × 10⁻⁷ × 800 × 2500 × 2

= 4 × 10⁻⁷ × 4,000,000

= 4 × 10⁻⁷ × 4 × 10⁶

= 16 × 10⁻¹ = 1.6 T

∴ Magnetic field inside the toroid, B = 1.6 T

(iv) Why Sample Z (Diamagnetic) cannot be used as a permanent magnet:

A permanent magnet requires a material that can be strongly magnetised along an applied field and retains a large residual magnetism (high remanence) after the field is removed. Diamagnetic materials like Sample Z are feebly magnetised in the direction opposite to the applied field and lose all magnetism as soon as the external field is removed (no remanence). Therefore, Sample Z cannot store any magnetic moment permanently and is completely unsuitable for making a permanent magnet.
Q6Case-based4 marks

A geologist on a field expedition uses a bar magnet (magnetic dipole moment M = 2.0 A m²) for navigation. She observes compass needle behaviour at various positions around the magnet. The equatorial line is the perpendicular bisector of the magnet's axis. (μ₀/4π = 10⁻⁷ T m A⁻¹)

A geologist on a field expedition carries a small compass needle and a bar magnet for navigation. She notices that when she places the bar magnet with its north pole pointing geographically north, the compass needle at a point on the equatorial line of the bar magnet (at a distance of 10 cm from its centre) deflects and aligns itself parallel to the bar magnet's axis. The bar magnet has a magnetic dipole moment of 2.0 A m².

(i) What is the direction of the magnetic field at a point on the equatorial line of a bar magnet?
(ii) Write the formula for the magnetic field on the equatorial line of a bar magnet and calculate its magnitude at a distance of 10 cm from its centre. (Use μ₀/4π = 10⁻⁷ T m A⁻¹)
(iii) If the geologist replaces the bar magnet with a circular current-carrying loop of radius 5 cm carrying a current of 2 A, what is the magnetic dipole moment of the loop?
(iv) Name the class of magnetic materials that the geologist's compass needle is most likely made of, and state ONE property that makes it suitable for this purpose.

Diagram for question 6: Magnetism and Matter
Show answer
(i) Direction of magnetic field on the equatorial line:

The magnetic field at a point on the equatorial line of a bar magnet is directed opposite to the direction of the magnetic dipole moment (M⃗), i.e., it points from the north pole to the south pole of the bar magnet (antiparallel to M⃗).

∴ The field at the equatorial point is directed from N-pole to S-pole of the bar magnet. [1 mark]

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(ii) Magnetic field on the equatorial line — formula and calculation:

The magnetic field at an equatorial point at distance r from the centre of a bar magnet is given by:

B_eq = (μ₀/4π) × M / (r² + l²)^(3/2)

For a short bar magnet (l << r), this simplifies to:

B_eq = (μ₀/4π) × M / r³

Given: M = 2.0 A m², r = 10 cm = 0.10 m, μ₀/4π = 10⁻⁷ T m A⁻¹

Substituting:

B_eq = 10⁻⁷ × 2.0 / (0.10)³
= 10⁻⁷ × 2.0 / 10⁻³
= 2.0 × 10⁻⁴ T

∴ B_eq = 2.0 × 10⁻⁴ T [1 mark]

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(iii) Magnetic dipole moment of the circular current loop:

The magnetic dipole moment of a current-carrying loop is given by:

m = I × A = I × π r²

Given: I = 2 A, r = 5 cm = 0.05 m

Substituting:

m = 2 × π × (0.05)²
= 2 × π × 2.5 × 10⁻³
= 2 × 3.14 × 2.5 × 10⁻³
= 1.57 × 10⁻² A m²

∴ m ≈ 1.57 × 10⁻² A m² (or π/200 A m²) [1 mark]

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(iv) Class of magnetic material for compass needle:

The compass needle is most likely made of a ferromagnetic material (e.g., hardened steel or alnico alloy).

Property that makes it suitable: Ferromagnetic materials have high retentivity (they retain strong magnetisation even after the external magnetising field is removed), so the needle maintains a permanent magnetic moment and reliably aligns with the Earth's magnetic field.

∴ Material: Ferromagnetic; Key property: High retentivity. [1 mark]
Q7Case-based4 marks

Earth's magnetic field plays a crucial role in navigation and geophysical studies. A compass needle — essentially a magnetic dipole — aligns itself along the horizontal component of Earth's field (B_H). The oscillation of such a needle about equilibrium, when disturbed, provides a method to measure magnetic moments. Magnetic materials are classified based on their behaviour in an external field: diamagnetic, paramagnetic, and ferromagnetic. Every ferromagnetic material has a characteristic Curie temperature above which it loses its ferromagnetic property. The Earth's total field B_E is related to its horizontal component B_H and the angle of dip δ by: B_H = B_E cos δ.

A geophysics research team is studying the Earth's magnetic field at a remote location. They set up a small compass needle (a magnetic dipole of magnetic moment m = 0.12 A m²) at a point where the Earth's horizontal component of magnetic field is B_H = 3.6 × 10⁻⁵ T. The needle is displaced by a small angle from its equilibrium position and released.

(i) Name the type of magnetism exhibited by the material used in a permanent bar magnet. State ONE property that distinguishes it from paramagnetism.

(ii) When the compass needle oscillates about its equilibrium position, the time period of oscillation is given by T = 2π√(I/mB_H), where I is the moment of inertia of the needle. If the moment of inertia of this needle about its centre is I = 4.8 × 10⁻⁷ kg m², calculate the time period of oscillation.

(iii) The team heats the bar magnet to a temperature well above its Curie temperature. State and explain what happens to its magnetic properties.

(iv) At the same location, if the angle of dip is 60°, calculate the magnitude of the Earth's total magnetic field B_E at that location.

Show answer
(i) The material used in a permanent bar magnet exhibits FERROMAGNETISM.

Distinguishing property:

| Property | Ferromagnetism | Paramagnetism |
|---|---|---|
| Susceptibility (χ) | Very large positive (χ >> 1) | Small positive (χ << 1, typically 10⁻⁵ to 10⁻³) |

∴ Ferromagnetic materials have domains of spontaneously aligned dipoles giving very high susceptibility, whereas paramagnetic materials have randomly oriented dipoles giving only weak magnetisation.
(1 mark: correct identification + one valid distinguishing point)

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(ii) By the given formula, the time period of oscillation of a magnetic dipole in a uniform field is:

T = 2π √(I / mB_H)

Substituting the values:

I = 4.8 × 10⁻⁷ kg m², m = 0.12 A m², B_H = 3.6 × 10⁻⁵ T

mB_H = 0.12 × 3.6 × 10⁻⁵ = 4.32 × 10⁻⁶ N m

I / (mB_H) = (4.8 × 10⁻⁷) / (4.32 × 10⁻⁶) = 0.1111 s²

√(I / mB_H) = √0.1111 ≈ 0.3333 s

T = 2π × 0.3333

∴ T = 2 × 3.14 × 0.3333 ≈ 2.09 s ≈ 2.1 s
(1 mark: correct substitution and calculation with unit)

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(iii) When the bar magnet is heated well above its Curie temperature:

The ferromagnetic material LOSES its ferromagnetism and becomes PARAMAGNETIC.

Reason: Below the Curie temperature, ferromagnetic materials have small regions called domains in which magnetic dipoles are spontaneously aligned parallel to each other, giving the material a large net magnetisation. When heated above the Curie temperature, thermal agitation becomes strong enough to DESTROY the domain structure. The dipoles become randomly oriented (as in a paramagnetic substance), and the material loses its strong magnetisation. On cooling back below the Curie temperature, domains reform and ferromagnetism is restored.
(1 mark: correct statement + correct reason in terms of domain structure)

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(iv) The relationship between the total magnetic field B_E, horizontal component B_H, and angle of dip δ is:

B_H = B_E cos δ

⟹ B_E = B_H / cos δ

Given: B_H = 3.6 × 10⁻⁵ T, δ = 60°, cos 60° = 0.5

B_E = (3.6 × 10⁻⁵) / 0.5

∴ B_E = 7.2 × 10⁻⁵ T
(1 mark: correct formula, substitution, and answer with unit)
Q8Case-based4 marks

A bar magnet of magnetic dipole moment M = 0.5 A m² is placed horizontally on a table. A compass needle is used to explore the magnetic field at various points around the magnet. Point P lies on the axial line and point Q lies on the equatorial line, both at a distance r = 20 cm = 0.20 m from the centre of the magnet. (Use: μ₀/4π = 10⁻⁷ T m A⁻¹)

A bar magnet is placed on a table. A student observes that a compass needle placed at point P (on the axial line, 20 cm from the centre of the magnet) shows a deflection, and the needle aligns along the field of the magnet. The magnetic dipole moment of the bar magnet is 0.5 A m².

(i) Name the physical quantity that describes the strength of a magnetic dipole. Write its SI unit.
(ii) Calculate the magnitude of the magnetic field at point P on the axial line.
(iii) If the compass needle is now moved to point Q on the equatorial line at the same distance (20 cm) from the centre, how does the magnitude of the field at Q compare with that at P? Give the ratio B_axial : B_equatorial.
(iv) A student claims that the field lines of a bar magnet form closed loops, unlike electric field lines of a dipole which begin and end on charges. Is the student correct? Justify briefly.

Diagram for question 8: Magnetism and Matter
Show answer
(i) The physical quantity that describes the strength of a magnetic dipole is the Magnetic Dipole Moment (m⃗ or M).
It is defined as m⃗ = M = NIA (for a current loop), or M = pole strength × magnetic length for a bar magnet.
SI unit: A m² (ampere metre squared).
[1 mark]

(ii) The magnetic field on the axial line of a magnetic dipole is given by:
B_axial = (μ₀/4π) × (2M/r³)

Substituting values:
B_axial = 10⁻⁷ × (2 × 0.5) / (0.20)³
B_axial = 10⁻⁷ × 1.0 / (8 × 10⁻³)
B_axial = 10⁻⁷ / (8 × 10⁻³)
B_axial = (1/8) × 10⁻⁴

∴ B_axial = 1.25 × 10⁻⁵ T
[1 mark]

(iii) The magnetic field on the equatorial line of a magnetic dipole is:
B_equatorial = (μ₀/4π) × (M/r³)

Comparing with B_axial = (μ₀/4π) × (2M/r³), at the same distance r:

B_axial / B_equatorial = 2M / M = 2

∴ B_axial : B_equatorial = 2 : 1

The field at P (axial) is twice the field at Q (equatorial) at the same distance.
[1 mark]

(iv) Yes, the student is correct.
Because magnetic monopoles do not exist in nature — every magnet always has a north pole and a south pole together. Therefore, magnetic field lines always form closed continuous loops: outside the magnet they go from the N-pole to the S-pole, and inside the magnet they continue from S-pole back to N-pole.
In contrast, electric field lines of a dipole originate on the positive charge and terminate on the negative charge — they are open lines.
∴ The field lines of a bar magnet always form closed loops, which is consistent with Gauss's law for magnetism: ∮ B⃗ · dA⃗ = 0.
[1 mark]
Q9MCQ1 mark

The magnetic permeability of a ferromagnetic material is

Show answer
Option (C) is correct.

Explanation: For a ferromagnetic material, the magnetic susceptibility χ is very large and positive (χ >> 1). Since relative permeability μr = 1 + χ, the permeability of a ferromagnetic material is also very large (μr >> 1). Examples: iron, nickel, cobalt.
Q10Short Answer1 mark

Assertion (A) : A diamagnetic substance, when placed in a non-uniform magnetic field, moves from the stronger region to the weaker region of the field.
Reason (R) : The magnetic susceptibility of a diamagnetic material is small and negative.

Show answer
Option (a) is correct.

Explanation: A diamagnetic material has magnetic susceptibility χ that is small and negative (e.g., χ ≈ −10⁻⁵ for bismuth). Because the induced magnetisation opposes the applied field (μᵣ < 1, i.e., permeability less than μ₀), the net force on the material is directed towards the region of weaker field — hence a diamagnetic sample is repelled from the stronger part of a non-uniform field. Reason (R) — that susceptibility is small and negative — is precisely the property that causes this behaviour, so R correctly explains A.
Q11MCQ1 mark

A bar magnet of magnetic dipole moment M is placed along the x-axis. The ratio of the magnetic field at a point on its axial line to the magnetic field at a point on its equatorial line, both at the same distance r from the centre, is:

Show answer
Option (b) is correct.

Explanation: The magnetic field due to a short magnetic dipole of moment M at distance r:
• Axial point: B<sub>axial</sub> = (μ<sub>0</sub>/4π)(2M/r<sup>3</sup>)
• Equatorial point: B<sub>eq</sub> = (μ<sub>0</sub>/4π)(M/r<sup>3</sup>)
∴ B<sub>axial</sub> / B<sub>eq</sub> = 2 : 1
Q12Short Answer1 mark

Assertion (A) : A paramagnetic substance is weakly attracted towards the stronger region of an external magnetic field.

Reason (R) : The magnetic susceptibility of a paramagnetic substance is small and negative.

Show answer
Option (C) is correct.

Explanation:
Assertion (A) is TRUE. A paramagnetic substance has a small positive magnetic susceptibility (χ > 0, small). When placed in a non-uniform external magnetic field, the induced magnetisation aligns with the field, and the substance experiences a net force directed towards the stronger region of the field — hence it is weakly attracted towards the stronger field region.

Reason (R) is FALSE. The magnetic susceptibility of a paramagnetic substance is small and POSITIVE (not negative). It is a diamagnetic substance that has small negative susceptibility (χ < 0).

∴ Assertion (A) is true but Reason (R) is false — Option (C).
Q13MCQ1 mark

A bar magnet is placed inside a solenoid and the solenoid is filled with a paramagnetic material. Which one of the following correctly describes the magnetic susceptibility (χ) of the paramagnetic material?

Show answer
Option (C) is correct.

Explanation: For a paramagnetic material, the magnetic susceptibility χ is small and positive (0 < χ < ε). Paramagnetic materials are weakly attracted towards a magnetic field, so χ > 0 but remains very small (typically 10⁻⁵ to 10⁻³). A large positive χ (>> 1) corresponds to ferromagnetic materials, χ = 0 to non-magnetic materials, and χ < −1 is not physically meaningful (diamagnetic materials have χ small and negative, e.g. −10⁻⁵).
Q14Short Answer2 marks

A bar magnet of magnetic moment 0.5 A m² is placed in a uniform magnetic field of strength 0.2 T such that its axis makes an angle of 30° with the field direction. Find (i) the torque acting on the magnet and (ii) the potential energy of the magnet in this position.

Show answer
Given: m = 0.5 A m², B = 0.2 T, θ = 30°.

(i) The torque on a magnetic dipole in a uniform field is given by:

τ = mB sinθ

→ τ = 0.5 × 0.2 × sin 30°
→ τ = 0.5 × 0.2 × 0.5

∴ τ = 0.05 N m

(ii) The potential energy of a magnetic dipole in a uniform field is given by:

U = −mB cosθ

→ U = −0.5 × 0.2 × cos 30°
→ U = −0.1 × (√3/2)
→ U = −0.1 × 0.866

∴ U = −0.0866 J ≈ −8.66 × 10⁻² J
Q15Short Answer2 marks

A bar magnet of magnetic moment 2.5 J T⁻¹ is placed with its axis along the direction of a uniform magnetic field of magnitude 0.4 T. Calculate the torque acting on the magnet and the potential energy of the magnet in this position.

Show answer
The torque on a magnetic dipole in a uniform field is given by:
τ = MB sinθ

Here, m = 2.5 J T⁻¹, B = 0.4 T, θ = 0° (axis along field).

→ τ = 2.5 × 0.4 × sin 0° = 2.5 × 0.4 × 0 = 0

∴ Torque τ = 0 N m

The potential energy of a magnetic dipole in a uniform field is given by:
U = −MB cosθ

→ U = −2.5 × 0.4 × cos 0° = −2.5 × 0.4 × 1 = −1.0

∴ Potential energy U = −1.0 J

(The negative sign indicates this is the position of stable equilibrium.)
Q16Short Answer2 marks

A bar magnet of magnetic moment 2.5 A m² is placed with its axis along the direction of a uniform magnetic field of magnitude 0.4 T. Calculate: (a) the torque acting on the magnet, and (b) the potential energy of the magnet in this position.

Show answer
The torque on a magnetic dipole in a uniform field is given by:
τ = mB sinθ
and the potential energy is given by:
U = −m⃗ · B⃗ = −mB cosθ

Here, m = 2.5 A m², B = 0.4 T, and θ = 0° (axis of magnet along B⃗).

(a) τ = mB sinθ = 2.5 × 0.4 × sin 0° = 2.5 × 0.4 × 0
∴ τ = 0 N m

(b) U = −mB cosθ = −2.5 × 0.4 × cos 0° = −2.5 × 0.4 × 1
∴ U = −1.0 J

The negative potential energy indicates this is the position of minimum (stable) equilibrium.
Q17Short Answer2 marks

A short bar magnet of magnetic moment m = 0.6 J T⁻¹ is placed with its axis making an angle of 60° with a uniform external magnetic field of magnitude B = 0.2 T. Calculate the torque acting on the magnet and state whether this torque tends to align or dis-align the magnet with the field.

Show answer
The torque acting on a magnetic dipole in a uniform magnetic field is given by:
τ = m B sin θ

Substituting the given values (m = 0.6 J T⁻¹, B = 0.2 T, θ = 60°):
τ = 0.6 × 0.2 × sin 60°
τ = 0.12 × (√3/2)
τ = 0.12 × 0.866

∴ τ = 0.104 N m ≈ 0.10 N m

This torque acts in such a direction as to align the magnetic moment m⃗ along the direction of the external magnetic field B⃗ (i.e., it tends to reduce θ toward zero).
Q18Short Answer3 marks

A bar magnet of magnetic moment 0.6 J T⁻¹ is placed in a uniform external magnetic field of 0.4 T. (a) What is the work done in rotating the magnet from its equilibrium position (θ = 0°) to a position where it makes an angle of 60° with the field direction? (b) What is the torque on the magnet when it makes an angle of 30° with the field?

Show answer
Given: Magnetic moment M = 0.6 J T⁻¹, External field B = 0.4 T.

(a) Work done in rotating a magnetic dipole in a uniform magnetic field:

W = MB(cos θ₁ − cos θ₂)

Here, θ₁ = 0° (equilibrium position) and θ₂ = 60°.

W = MB(cos 0° − cos 60°)

W = 0.6 × 0.4 × (1 − 0.5)

W = 0.24 × 0.5

∴ W = 0.12 J

(b) Torque on the magnet at angle θ with the field:

τ = MB sin θ

Here, θ = 30°.

τ = 0.6 × 0.4 × sin 30°

τ = 0.24 × 0.5

∴ τ = 0.12 N m
Q19Short Answer3 marks

A research team is studying the magnetic behaviour of three unknown materials X, Y, and Z. They record the following observations:
• Material X: When placed in a non-uniform external magnetic field, it moves from the stronger region to the weaker region. Its susceptibility χ is a small negative constant, independent of temperature.
• Material Y: It shows a very strong attraction towards a magnetic field. When the external field is removed, it retains magnetisation. Its B–H curve forms a wide hysteresis loop.
• Material Z: Its susceptibility follows Curie's law χ = C/T, where C is the Curie constant and T is the absolute temperature. At temperature T_C it undergoes a phase transition.

(i) Identify the magnetic class (diamagnetic / paramagnetic / ferromagnetic) of each material X, Y, and Z. Give one distinguishing property for each. [1 mark]
(ii) The hysteresis loop of material Y has a large area. What does this imply about energy loss per cycle? State one practical application for which this property makes Y suitable. [1 mark]
(iii) Material Z is initially ferromagnetic and is heated above the temperature T_C mentioned above. Name the temperature T_C and explain, at the atomic level, why Z becomes paramagnetic above T_C. [1 mark]
(iv) A small sample of material X (volume 2 × 10⁻⁶ m³, χ = −8 × 10⁻⁶) is placed in a uniform magnetic field of intensity H = 10⁵ A m⁻¹. Calculate the magnetisation M of the sample and hence find the magnetic moment m of the sample. [1 mark]

Show answer
(i) Identification of magnetic classes:

Material X → Diamagnetic.
Distinguishing property: It is repelled by a magnetic field (moves from stronger to weaker field region); susceptibility χ is small and negative (−1 ≤ χ < 0).

Material Y → Ferromagnetic.
Distinguishing property: It shows spontaneous magnetisation, retains magnetisation after the field is removed (permanent magnet behaviour), and has a large positive susceptibility (χ >> 1).

Material Z → Paramagnetic (at temperatures above T<sub>C</sub>; ferromagnetic below T<sub>C</sub>).
Distinguishing property: Susceptibility obeys Curie's law χ = C/T — it is small, positive, and inversely proportional to absolute temperature.

(ii) Large area of hysteresis loop — energy loss:

The area of the B–H hysteresis loop represents the energy dissipated (lost as heat) per unit volume per cycle of magnetisation and demagnetisation.
A large loop area → large energy loss per cycle (high hysteretic loss).

Practical application: Material Y is suitable for use as a permanent magnet (e.g., in loudspeakers, electric motors, or generators) because its high retentivity and high coercivity mean it retains strong magnetisation and resists demagnetisation — the large hysteresis is acceptable since it is not being repeatedly cycled.

(iii) Curie temperature and atomic-level explanation:

The temperature T<sub>C</sub> is called the Curie Temperature (Curie Point).

Below T<sub>C</sub>: In a ferromagnetic material, atoms (or ions) have permanent magnetic dipole moments. The exchange interaction between neighbouring atoms causes their dipole moments to align spontaneously in small regions called magnetic domains, giving rise to strong net magnetisation.

Above T<sub>C</sub>: Thermal agitation (thermal energy ~ k<sub>B</sub>T) becomes large enough to overcome the exchange interaction between neighbouring atomic dipoles. The long-range order within domains is destroyed; the domains disappear. The atomic dipoles are now randomly oriented on average, so the material behaves like a paramagnet — showing only weak, field-induced alignment described by Curie's law.
∴ Material Z becomes paramagnetic above T<sub>C</sub>.

(iv) Calculation of magnetisation M and magnetic moment m:

Given:
Volume V = 2 × 10<sup>−6</sup> m<sup>3</sup>
Susceptibility χ = −8 × 10<sup>−6</sup>
Magnetic field intensity H = 10<sup>5</sup> A m<sup>−1</sup>

Formula for magnetisation:
M = χ H

Substituting:
M = (−8 × 10<sup>−6</sup>) × (10<sup>5</sup> A m<sup>−1</sup>)
M = −8 × 10<sup>−1</sup> A m<sup>−1</sup>

∴ M = −0.8 A m<sup>−1</sup>

(The negative sign confirms diamagnetic behaviour — magnetisation is opposite to H.)

Formula for magnetic moment:
m = M × V

Substituting:
m = (−0.8 A m<sup>−1</sup>) × (2 × 10<sup>−6</sup> m<sup>3</sup>)
m = −1.6 × 10<sup>−6</sup> A m<sup>2</sup>

∴ Magnetic moment m = −1.6 × 10<sup>−6</sup> A m<sup>2</sup>

(Magnitude |m| = 1.6 × 10<sup>−6</sup> A m<sup>2</sup>, directed opposite to the applied field.)
Q20Short Answer3 marks

A geologist on a field expedition carries a sensitive magnetometer to study magnetic anomalies in the Earth's crust. At a certain location, she places a small bar magnet of magnetic moment M = 0.60 A m² on a horizontal surface. She then measures the magnetic field at two points:

(i) Point P, situated on the axial line of the magnet at a distance of 20 cm from its centre.
(ii) Point Q, situated on the equatorial line of the same magnet at the same distance of 20 cm from its centre.

She also notices that when she places a small sample of the rock (collected from the site) near the magnet, the rock gets weakly repelled by both poles of the magnet.

(a) Calculate the magnetic field at point P (axial point). [1 mark]
(b) Calculate the magnetic field at point Q (equatorial point). [1 mark]
(c) Compare the directions of the magnetic fields at P and Q relative to the magnetic moment M of the bar magnet. [1 mark]
(d) Identify the magnetic nature of the rock sample. Give ONE characteristic property that justifies this identification. [1 mark]

Diagram for question 20: Magnetism and Matter
Show answer
(a) Magnetic field at axial point P:

By the formula for the magnetic field on the axial line of a short magnetic dipole:

B_axial = (μ₀ / 4π) × (2M / r³)

Substituting values: μ₀/4π = 10⁻⁷ T m A⁻¹, M = 0.60 A m², r = 20 cm = 0.20 m

B_P = 10⁻⁷ × (2 × 0.60) / (0.20)³

B_P = 10⁻⁷ × 1.20 / (8 × 10⁻³)

B_P = 10⁻⁷ × 150

∴ B_P = 1.5 × 10⁻⁵ T

(b) Magnetic field at equatorial point Q:

By the formula for the magnetic field on the equatorial line of a short magnetic dipole:

B_equatorial = (μ₀ / 4π) × (M / r³)

Substituting values: M = 0.60 A m², r = 0.20 m

B_Q = 10⁻⁷ × 0.60 / (0.20)³

B_Q = 10⁻⁷ × 0.60 / (8 × 10⁻³)

B_Q = 10⁻⁷ × 75

∴ B_Q = 0.75 × 10⁻⁵ T

(Also note: B_axial = 2 × B_equatorial, confirming B_P = 2B_Q ✓)

(c) Direction comparison:

The magnetic field at the axial point P is directed along the direction of the magnetic moment M⃗ (i.e., from the S-pole to the N-pole of the bar magnet, along the axis).

The magnetic field at the equatorial point Q is directed opposite to the direction of the magnetic moment M⃗ (i.e., from the N-pole to the S-pole, antiparallel to M⃗).

∴ B⃗_axial is parallel to M⃗, whereas B⃗_equatorial is antiparallel to M⃗.

(d) Identification of the rock sample:

Since the rock is weakly repelled by both poles of the bar magnet, it is a diamagnetic material.

Characteristic property: Diamagnetic materials have negative (small) magnetic susceptibility (χ < 0). When placed in an external magnetic field, they develop a magnetisation opposite to the applied field, causing weak repulsion. Examples: bismuth, copper, water.
Q21Short Answer3 marks

A geophysics research team is surveying a region near an active volcanic zone. They use a sensitive magnetometer and record the following observations at three test sites:

• Site P: A cylindrical rock sample is found to strongly attract the magnetometer probe even when the external field is removed.
• Site Q: A mineral sample placed in a non-uniform external magnetic field B⃗ moves weakly toward the stronger field region. Its susceptibility χ is measured as +3.2 × 10⁻⁵.
• Site R: A crystal sample, when placed inside a solenoid, produces an internal field slightly less than the applied field B₀. Its relative permeability μᵣ = 0.99982.

The team also finds a uniformly magnetised cylindrical bar (from Site P) of length 12 cm and cross-sectional area 2.0 cm², with a magnetic moment of 1.44 A m².

(i) Identify the magnetic material class at each site (P, Q, R) and give one distinguishing property of each.
(ii) For the sample at Site Q, calculate the magnetisation M (intensity of magnetisation) if the applied field H = 8.0 × 10⁴ A m⁻¹.
(iii) Calculate the pole strength of the bar from Site P.
(iv) The bar from Site P is heated above its Curie temperature and then brought near a compass needle. Predict and justify the observation.

Show answer
MARKING SCHEME (4 marks — each sub-part 1 mark, ECF applies)

(i) Identification of magnetic material at each site:

• Site P — Ferromagnetic material.
Distinguishing property: Retains magnetisation even after the external field is removed (has non-zero remanence / permanent magnetism); very high susceptibility (χ >> 1).

• Site Q — Paramagnetic material.
Distinguishing property: χ is small and positive (χ = +3.2 × 10⁻⁵ here); magnetised weakly in the direction of the applied field; moves toward stronger field region.

• Site R — Diamagnetic material.
Distinguishing property: μᵣ < 1 (here μᵣ = 0.99982 < 1); χ is small and negative; internal field is slightly less than applied field (material weakly opposes the field).

∴ P → Ferromagnetic; Q → Paramagnetic; R → Diamagnetic.
[1 mark for all three correctly identified with one property each]

─────────────────────────────────────
(ii) Magnetisation M for Site Q sample:

By definition, magnetisation M is related to magnetic susceptibility χ and applied magnetic field intensity H by:

M = χ · H

Given: χ = +3.2 × 10⁻⁵, H = 8.0 × 10⁴ A m⁻¹

Substituting:
M = (3.2 × 10⁻⁵) × (8.0 × 10⁴) A m⁻¹
M = 3.2 × 8.0 × 10⁻⁵⁺⁴ A m⁻¹
M = 25.6 × 10⁻¹ A m⁻¹

∴ M = 2.56 A m⁻¹
[1 mark: formula + correct substitution + answer with unit]

─────────────────────────────────────
(iii) Pole strength of the bar from Site P:

The magnetic moment of a bar magnet is given by:

m = q_m × 2l

where q_m is the pole strength and 2l is the magnetic length (= geometrical length for a uniformly magnetised bar).

Given: m = 1.44 A m², 2l = 12 cm = 0.12 m

Rearranging:
q_m = m / (2l) = 1.44 / 0.12

∴ q_m = 12 A m
[1 mark: formula + substitution + correct answer with unit]

─────────────────────────────────────
(iv) Effect of heating the Site P bar above its Curie temperature:

Below the Curie temperature, ferromagnetic materials possess strong spontaneous magnetisation due to alignment of magnetic domains. Above the Curie temperature, thermal agitation destroys the domain structure completely.

When the bar is heated above its Curie temperature, it loses its ferromagnetic property and behaves like an ordinary paramagnetic material (very weak magnetisation).

Observation: When the demagnetised (now paramagnetic) bar is brought near the compass needle, the needle shows negligible or no deflection (no attraction or repulsion), unlike the strong deflection it would have caused before heating.

Justification: Above the Curie temperature, the domain structure collapses → net magnetic moment becomes negligible → the bar no longer acts as a magnet → it cannot exert a significant torque on the compass needle.

∴ The compass needle shows negligible deflection, confirming the bar has lost its permanent magnetism above the Curie temperature.
[1 mark: correct prediction + valid justification using Curie temperature / domain theory]
Q22Short Answer3 marks

A geophysicist is analysing magnetic data from two different locations on Earth's surface. At Location P, the angle of dip is 30° and the horizontal component of Earth's magnetic field is measured as 4×10⁻⁵ T. At Location Q, the horizontal component is the same (4×10⁻⁵ T) but the vertical component is 4√3×10⁻⁵ T.

(i) Calculate the total magnetic field intensity (B) at Location P.
(ii) Calculate the angle of dip at Location Q.
(iii) A small compass needle (magnetic moment m = 0.06 A m²) is placed at Location P with its axis initially aligned along the total Earth's field. It is then rotated in the vertical plane until the torque on it equals half its maximum possible torque at that location. What is the angle of rotation from its initial position?
(iv) A sample of bismuth (a diamagnetic material) is brought near the north pole of a strong bar magnet at Location Q. Describe the nature of force experienced by the bismuth sample and explain the physical reason using the concept of magnetisation.

Diagram for question 22: Magnetism and Matter
Show answer
PART (i): Total magnetic field at Location P

At Location P, angle of dip δ = 30°, horizontal component BH = 4×10⁻⁵ T.

By definition of angle of dip:

cos δ = BH / B

→ B = BH / cos δ

→ B = (4×10⁻⁵) / cos 30°

→ B = (4×10⁻⁵) / (√3/2)

→ B = (8×10⁻⁵) / √3

∴ B = (8/√3) × 10⁻⁵ ≈ 4.62×10⁻⁵ T

─────────────────────────────────────

PART (ii): Angle of dip at Location Q

At Location Q: BH = 4×10⁻⁵ T, BV = 4√3×10⁻⁵ T.

By definition:

tan δQ = BV / BH

→ tan δQ = (4√3×10⁻⁵) / (4×10⁻⁵)

→ tan δQ = √3

∴ δQ = 60°

─────────────────────────────────────

PART (iii): Angle of rotation for half-maximum torque

At Location P, total field B = (8/√3)×10⁻⁵ T.

The torque on a magnetic dipole in a uniform field is:

τ = mB sinθ

where θ is the angle between the magnetic moment and B⃗.

Initially, the needle is aligned along B⃗, so θ₀ = 0° and τ₀ = 0.

Maximum torque occurs at θ = 90°:

τmax = mB sin 90° = mB

For torque to equal half the maximum value:

mB sinθ = (1/2) mB

→ sinθ = 1/2

→ θ = 30°

The needle is rotated from θ₀ = 0° to θ = 30°.

∴ Angle of rotation = 30°

─────────────────────────────────────

PART (iv): Force on bismuth (diamagnetic) near the north pole

Bismuth is a diamagnetic material (susceptibility χm < 0, small negative value).

When a diamagnetic material is placed in a non-uniform magnetic field (such as near the pole of a bar magnet), it develops a magnetisation M⃗ directed opposite to the applied field B⃗ (because χm < 0).

Physical reason: The external field slightly modifies the orbital motion of electrons, inducing a net magnetic moment that opposes B⃗ (Lenz's law analogy at the atomic level). This gives the material a polarity such that its nearer face to the north pole also becomes a north pole.

As a result, the bismuth sample experiences a repulsive force and is pushed away from the north pole, towards the region of weaker field.

∴ The bismuth sample is repelled from the north pole of the bar magnet at Location Q.
Q23Short Answer3 marks

An electron of mass m and charge e revolves in a circular orbit of radius r around the nucleus of a hydrogen atom with speed v.
(a) Derive an expression for the magnetic moment (μ) associated with the orbital motion of the electron.
(b) Write the expression relating the magnetic moment vector (μ⃗) and the orbital angular momentum vector (L⃗) of the electron, and state the significance of the negative sign.
(c) A paramagnetic sample is placed in an external magnetic field. State what happens to its magnetisation when the external field is (i) increased at constant temperature, and (ii) temperature is raised at constant external field.

Show answer
(a) Magnetic Moment of the Orbiting Electron

An electron moving in a circular orbit of radius r with speed v constitutes a current loop.

Time period of revolution:
T = 2πr / v

Equivalent current due to the orbiting electron (charge e, taking magnitude):
I = e / T = ev / 2πr

The magnetic moment of a current loop is μ = I × A, where A = πr² is the area of the orbit.

∴ μ = (ev / 2πr) × πr²

∴ μ = evr / 2

(b) Relation between μ⃗ and L⃗

The orbital angular momentum of the electron is:
L = mvr

Dividing μ by L:
μ / L = (evr/2) / (mvr) = e / 2m

∴ μ = (e / 2m) L

In vector form, since the electron carries negative charge (−e), the magnetic moment vector is antiparallel to the angular momentum vector:

∴ μ⃗ = −(e / 2m) L⃗

Significance of the negative sign: The negative sign indicates that the magnetic moment vector (μ⃗) is directed opposite to the orbital angular momentum vector (L⃗). This arises because the electron (negatively charged) revolving anticlockwise produces a conventional current in the clockwise direction, so μ⃗ and L⃗ point in opposite directions.

(c) Behaviour of a Paramagnetic Sample

Curie's law states: M = C B / T, where M is magnetisation, B is the applied field, T is the absolute temperature, and C is the Curie constant.

(i) When the external field is increased at constant temperature:
Magnetisation M increases, because more atomic magnetic dipoles align along the field direction (M ∝ B at constant T).

(ii) When the temperature is raised at constant external field:
Magnetisation M decreases, because increased thermal agitation randomly disturbs the alignment of magnetic dipoles (M ∝ 1/T at constant B).
Q24Short Answer3 marks

A geophysicist is studying two remote locations X and Y on Earth's surface. At location X, the observed dip angle is 0° and the horizontal component of Earth's magnetic field is measured as 4×10⁻⁵ T. At location Y, the dip angle is 60° and the horizontal component is 2×10⁻⁵ T.

(a) Identify the geographic location of X. Calculate the total (resultant) magnetic field at X. (1 mark)

(b) Calculate the total magnetic field at Y and determine the vertical component of Earth's field at Y. (2 marks)

(c) A small bar magnet of magnetic moment m = 0.4 A m² is placed at location Y, oriented so that its axis aligns with Earth's total field at Y. Calculate the torque experienced by this magnet when it is then rotated by 90° from this equilibrium position. (1 mark)

Show answer
MARKING SCHEME — Magnetism and Matter [4 marks]

─────────────────────────────────────────
Part (a): Location of X and total field at X [1 mark]
─────────────────────────────────────────

At location X, dip angle δ = 0°.

By definition of magnetic dip (angle of inclination): the angle δ is the angle made by Earth's total magnetic field B⃗ with the horizontal plane at a given location.

Since δ = 0°, the total field is purely horizontal → Location X is on the magnetic equator.

Relation between components and total field:
H = B cos δ

At X: δ = 0°, so cos 0° = 1
∴ B = H / cos δ = (4×10⁻⁵) / 1

∴ Total field at X, B<sub>X</sub> = 4×10⁻⁵ T

(The vertical component V = B sin 0° = 0, confirming the field is entirely horizontal at the equator.)

─────────────────────────────────────────
Part (b): Total field and vertical component at Y [2 marks]
─────────────────────────────────────────

Given at Y: dip δ = 60°, horizontal component H<sub>Y</sub> = 2×10⁻⁵ T.

Using: H = B cos δ
→ B = H / cos δ = (2×10⁻⁵) / cos 60° = (2×10⁻⁵) / (0.5)

∴ Total field at Y, B<sub>Y</sub> = 4×10⁻⁵ T … [1 mark — substitution + answer with unit]

Vertical component:
V = B sin δ = (4×10⁻⁵) × sin 60° = (4×10⁻⁵) × (√3 / 2)
V = (4×10⁻⁵) × 0.866

∴ Vertical component at Y, V<sub>Y</sub> = 3.46×10⁻⁵ T ≈ 3.5×10⁻⁵ T … [1 mark — substitution + answer with unit]

[Alternatively: using V² = B² − H²
V = √((4×10⁻⁵)² − (2×10⁻⁵)²) = √(16×10⁻¹⁰ − 4×10⁻¹⁰) = √(12×10⁻¹⁰) = 2√3 × 10⁻⁵ ≈ 3.46×10⁻⁵ T ✓]

─────────────────────────────────────────
Part (c): Torque on bar magnet at Y [1 mark]
─────────────────────────────────────────

The magnet initially aligns along Earth's total field B⃗<sub>Y</sub> (equilibrium position, θ = 0°). It is then rotated by 90° from this position, so θ = 90°.

Formula for torque on a magnetic dipole:
τ = mB sin θ

Substituting values:
m = 0.4 A m², B<sub>Y</sub> = 4×10⁻⁵ T, θ = 90°, sin 90° = 1

τ = 0.4 × 4×10⁻⁵ × 1

∴ Torque, τ = 1.6×10⁻⁵ N m … [1 mark — formula + substitution + ∴ answer + unit]
Q25Short Answer3 marks

A geologist working in a remote field notices that her compass needle is behaving erratically near a large rock outcrop. On testing samples in a laboratory, two rock specimens are identified:
• Specimen P: When suspended freely near a strong magnet, it is weakly attracted and aligns along the field direction.
• Specimen Q: When brought near a magnet, it is weakly repelled and sets itself at right angles to the external field.
Based on this, answer the following questions:
(i) Identify the magnetic nature of Specimen P and Specimen Q.
(ii) Which specimen, P or Q, has a small positive value of magnetic susceptibility (χ)? Give one reason.
(iii) When Specimen P is placed in an external magnetic field, will the magnetic field lines inside the specimen be more concentrated or less concentrated than outside? Justify.
(iv) A third specimen R is strongly attracted to a magnet and retains its magnetism even after the magnet is removed. Name the type of magnetic material R belongs to. Give one example.

Show answer
(i) Specimen P — weakly attracted and aligns along the field → Paramagnetic material.
Specimen Q — weakly repelled and aligns perpendicular to the field → Diamagnetic material.

(ii) Specimen P (paramagnetic) has a small positive value of magnetic susceptibility (χ > 0).
Reason: Paramagnetic materials have permanent atomic magnetic dipoles that align partially along the external field, so the induced magnetisation is in the direction of the applied field, giving a small positive χ.

(iii) For Specimen P (paramagnetic): The magnetic field lines inside the specimen will be more concentrated (denser) than outside.
Justification: A paramagnetic material has χ > 0, so its relative permeability μᵣ = 1 + χ > 1. Since B⃗ = μᵣμ₀H⃗, the magnetic flux density inside is greater than in free space, causing field lines to crowd (concentrate) inside the specimen.

(iv) Specimen R belongs to a Ferromagnetic material.
Example: Iron (Fe) [Acceptable alternatives: Nickel (Ni), Cobalt (Co)]
Q26Short Answer3 marks

A bar magnet of magnetic dipole moment M is placed in a uniform external magnetic field B⃗. The axis of the magnet makes an angle of 60° with the direction of B⃗.

(a) Write the expression for the torque acting on the magnet.
(b) Calculate the torque if M = 2 A m², B = 0.5 T.
(c) In which orientation is the magnet in (i) stable equilibrium and (ii) unstable equilibrium? Give the value of torque in each case.

Show answer
(a) Expression for Torque:

When a magnetic dipole of moment M⃗ is placed in a uniform external magnetic field B⃗, the torque acting on it is given by:

τ⃗ = M⃗ × B⃗

∴ Magnitude: τ = MB sinθ

where θ is the angle between the magnetic moment M⃗ and the field B⃗. [1 mark]

(b) Calculation of Torque:

Given: M = 2 A m², B = 0.5 T, θ = 60°

Using τ = MB sinθ:

τ = 2 × 0.5 × sin 60°

τ = 1 × (√3/2)

∴ τ = 0.866 N m ≈ 0.87 N m [1 mark]

(c) Equilibrium Positions:

(i) Stable Equilibrium — when the magnetic moment M⃗ is aligned parallel to B⃗, i.e., θ = 0°.

τ = MB sin 0° = 0

The magnet returns to this position when slightly displaced; potential energy U = −MB (minimum).

(ii) Unstable Equilibrium — when the magnetic moment M⃗ is aligned antiparallel to B⃗, i.e., θ = 180°.

τ = MB sin 180° = 0

The magnet does not return to this position when displaced; potential energy U = +MB (maximum).

∴ In both equilibrium orientations the torque τ = 0, but only θ = 0° gives stable equilibrium. [1 mark]
Q27Short Answer3 marks

A geophysics research team is studying the magnetic properties of three rock samples collected from different geological sites. Sample P is strongly attracted to a bar magnet and retains its magnetisation even after the magnet is removed. Sample Q is weakly attracted to the bar magnet and loses its magnetisation as soon as the magnet is removed. Sample R is weakly repelled by both poles of the bar magnet.

(i) Identify the magnetic category (ferromagnetic / paramagnetic / diamagnetic) of each sample P, Q, and R.

(ii) The team places Sample Q in a non-uniform magnetic field. In which direction will it move — towards the stronger region or weaker region of the field? Justify using the sign of its susceptibility.

(iii) A bar magnet made from a material similar to Sample P has magnetic moment m = 2.4 A m² and is placed with its axis making an angle of 30° with a uniform external magnetic field B = 0.08 T. Calculate the torque acting on it.

(iv) The team observes that when Sample P is heated above a certain temperature, it starts behaving like Sample Q. Name this temperature and state what happens to the magnetic domains of Sample P at this temperature.

Show answer
(i) Identification of magnetic categories:

Sample P — Ferromagnetic: strongly attracted and retains magnetisation (has permanent magnetic domains).
Sample Q — Paramagnetic: weakly attracted and loses magnetisation when field is removed (χ is small and positive).
Sample R — Diamagnetic: weakly repelled by both poles (χ is small and negative).

[1 mark — all three correct; ½ if only two correct]

(ii) Direction of motion of Sample Q (paramagnetic) in a non-uniform field:

For a paramagnetic material, magnetic susceptibility χ > 0 (small positive value).
Because χ > 0, the material is feebly attracted towards the region of stronger magnetic field.

∴ Sample Q will move towards the stronger region of the non-uniform magnetic field.

[1 mark]

(iii) Torque on the bar magnet:

The torque on a magnetic dipole in a uniform external field is given by:

τ = m B sin θ

where m = magnetic moment, B = external field, θ = angle between m⃗ and B⃗.

Substituting values:

τ = 2.4 × 0.08 × sin 30°

τ = 2.4 × 0.08 × 0.5

τ = 0.192 × 0.5

∴ τ = 0.096 N m

[1 mark]

(iv) Curie Temperature and domain behaviour:

The temperature above which a ferromagnetic material loses its ferromagnetic property and becomes paramagnetic is called the Curie Temperature (T<sub>C</sub>).

At and above the Curie temperature, the thermal energy becomes large enough to disrupt the alignment of magnetic dipoles within the domains. The large-scale magnetic domains of Sample P break down and the material no longer has spontaneous bulk magnetisation; it instead behaves as a paramagnetic substance (small, positive χ, random dipole orientation).

∴ Above T<sub>C</sub>, the ordered domain structure of Sample P is destroyed and the material becomes paramagnetic — consistent with the behaviour of Sample Q.

[1 mark]
Q28Short Answer3 marks

A geophysics research team is mapping the Earth's magnetic field at a remote location. They place a small compass needle (modelled as a magnetic dipole of magnetic moment m = 0.08 A m²) at a point where the Earth's horizontal component of magnetic field is B_H = 0.4 × 10⁻⁴ T and the angle of dip is 60°.

(i) Calculate the magnitude of the Earth's total magnetic field B_E at that location.

(ii) The compass needle is displaced by a small angle from its equilibrium position and released. If the moment of inertia of the needle about its pivot is I = 2 × 10⁻⁶ kg m², calculate the time period of its oscillation.

(iii) The team then carries the same compass to a place where the angle of dip is 90°. State what happens to the time period of oscillation at this new location, giving a physical reason.

(iv) At the original location, if the needle is held at 30° from the direction of B_H, find the restoring torque acting on it.

Show answer
(i) Relation between B_E and B_H:

By definition of angle of dip (δ), the horizontal component is:
B_H = B_E cos δ

∴ B_E = B_H / cos δ = (0.4 × 10⁻⁴) / cos 60° = (0.4 × 10⁻⁴) / 0.5

∴ B_E = 0.8 × 10⁻⁴ T

(ii) Time period of oscillation of a magnetic dipole:

For small angular displacement, the restoring torque on the needle is τ = −m B_H θ (since the needle aligns with B_H in the horizontal plane).

The time period is given by:
T = 2π √(I / m B_H)

Substituting values:
T = 2π √[(2 × 10⁻⁶) / (0.08 × 0.4 × 10⁻⁴)]

Denominator: m B_H = 0.08 × 0.4 × 10⁻⁴ = 3.2 × 10⁻⁶ N m rad⁻¹

T = 2π √[(2 × 10⁻⁶) / (3.2 × 10⁻⁶)]
= 2π √[0.625]
= 2π × 0.7906

∴ T ≈ 4.97 s ≈ 5.0 s

(iii) At the location where angle of dip δ = 90° (the magnetic poles of the Earth), the horizontal component of Earth's magnetic field is:
B_H = B_E cos 90° = 0

Since T = 2π √(I / m B_H) and B_H → 0, the denominator under the square root approaches zero.

∴ T → ∞ (the time period becomes infinitely large, i.e., the needle does not oscillate).

Physical reason: At the magnetic pole, there is no horizontal restoring component of the field to bring the needle back to equilibrium. The needle has no preferred orientation in the horizontal plane, so no restoring torque acts and oscillation ceases.

(iv) Torque on a magnetic dipole:

By the definition of torque on a magnetic dipole in a uniform field:
τ = m B_H sin θ

where θ = 30° is the angle between the needle and B_H.

τ = 0.08 × (0.4 × 10⁻⁴) × sin 30°
= 0.08 × 0.4 × 10⁻⁴ × 0.5
= 0.08 × 0.2 × 10⁻⁴
= 1.6 × 10⁻⁶ N m

∴ Restoring torque τ = 1.6 × 10⁻⁶ N m
Q29Short Answer3 marks

Distinguish between diamagnetic, paramagnetic, and ferromagnetic materials on the basis of (i) magnetic susceptibility and (ii) behaviour in a non-uniform magnetic field. Give one example of each.

Show answer
The magnetic susceptibility χ of a material is defined as the ratio of magnetisation M to the applied magnetic field intensity H, i.e., χ = M/H.

Comparison of the three classes of magnetic materials:

┌──────────────────┬──────────────────────────┬────────────────────────────────────────┬─────────────┐
│ Property │ Diamagnetic │ Paramagnetic │ Ferromagnetic│
├──────────────────┼──────────────────────────┼────────────────────────────────────────┼─────────────┤
│ (i) Susceptibility χ │ Small and negative (χ < 0) │ Small and positive (0 < χ < 1) │ Very large and positive (χ >> 1) │
├──────────────────┼──────────────────────────┼────────────────────────────────────────┼─────────────┤
│ (ii) In non-uniform magnetic field │ Moves from stronger to weaker field region (feebly repelled) │ Moves from weaker to stronger field region (feebly attracted) │ Moves strongly from weaker to stronger field region (strongly attracted) │
├──────────────────┼──────────────────────────┼────────────────────────────────────────┼─────────────┤
│ Example │ Bismuth (Bi) │ Aluminium (Al) │ Iron (Fe) │
└──────────────────┴──────────────────────────┴────────────────────────────────────────┴─────────────┘

∴ The three classes differ fundamentally in the sign and magnitude of χ and in the direction of their motion in a non-uniform field.
Q30Short Answer3 marks

A geophysics team is surveying a region and places a small bar magnet M of magnetic moment m = 0.50 A m² at the origin of the x–y plane, with its magnetic moment directed along the +x axis. A compass needle (treated as a tiny magnetic dipole of moment m′ = 0.10 A m²) is placed, one position at a time, at the two locations given below:

Position I : at point A(d, 0), i.e., on the axial line of M, with m′ pointing along the +x axis.
Position II : at point B(0, d), i.e., on the equatorial line of M, with m′ pointing along the +x axis.

Take d = 10 cm.

(a) Calculate the magnitude of the magnetic field produced by M at point A and at point B. (2 marks)
(b) Find the potential energy of the compass needle in each position and hence state in which position the system has lower potential energy. (1 mark)
(c) In which position is the compass needle in stable equilibrium, unstable equilibrium, or not in equilibrium at all? Justify briefly. (1 mark)

Diagram for question 30: Magnetism and Matter
Show answer
Part (a) — Magnetic field at axial and equatorial points

For a bar magnet of magnetic moment m, the magnetic field formulas (in SI, using μ₀/4π = 10⁻⁷ T m A⁻¹) are:

Axial field (along the axis of M): B⃗_axial = (μ₀/4π)(2m/d³), directed along +x axis.

Equatorial field (along the perpendicular bisector of M): B⃗_eq = (μ₀/4π)(m/d³), directed along −x axis (antiparallel to m).

Given: m = 0.50 A m², d = 10 cm = 0.10 m, so d³ = (0.10)³ = 1.0 × 10⁻³ m³.

At point A (axial):
B_A = (μ₀/4π)(2m/d³)
= 10⁻⁷ × (2 × 0.50)/(1.0 × 10⁻³)
= 10⁻⁷ × 1.0 × 10³

∴ B_A = 1.0 × 10⁻⁴ T, directed along +x axis.

At point B (equatorial):
B_B = (μ₀/4π)(m/d³)
= 10⁻⁷ × (0.50)/(1.0 × 10⁻³)
= 10⁻⁷ × 0.50 × 10³

∴ B_B = 0.5 × 10⁻⁴ T, directed along −x axis.

─────────────────────────────────────────
Part (b) — Potential energy in each position

The potential energy of a magnetic dipole m′ in a field B⃗ is:
U = −m⃗′ · B⃗ = −m′ B cosθ
where θ is the angle between m⃗′ and B⃗.

Position I (point A): m⃗′ is along +x; B⃗_A is along +x → θ = 0°.
U_I = −m′ B_A cos 0°
= −(0.10)(1.0 × 10⁻⁴)(1)

∴ U_I = −1.0 × 10⁻⁵ J

Position II (point B): m⃗′ is along +x; B⃗_B is along −x → θ = 180°.
U_II = −m′ B_B cos 180°
= −(0.10)(0.5 × 10⁻⁴)(−1)

∴ U_II = +0.5 × 10⁻⁵ J

Since U_I (= −1.0 × 10⁻⁵ J) < U_II (= +0.5 × 10⁻⁵ J),

∴ The system has lower (minimum) potential energy in Position I.

─────────────────────────────────────────
Part (c) — Equilibrium analysis

A magnetic dipole is in equilibrium when the torque τ = m′ B sinθ = 0, i.e., when θ = 0° (stable) or θ = 180° (unstable).

Position I (θ = 0°): The compass needle is aligned parallel to B⃗_A. Torque = 0. If slightly displaced, the restoring torque brings it back.
∴ Position I corresponds to STABLE EQUILIBRIUM.

Position II (θ = 180°): The compass needle is antiparallel to B⃗_B. Torque = 0. However, any slight displacement produces a torque that increases the deflection rather than restoring it.
∴ Position II corresponds to UNSTABLE EQUILIBRIUM.

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