Important Questions Class 12 Physics Chapter 5: Magnetism and Matter

Important Questions Class 12 Physics Chapter 5 focus on magnetism, which studies magnetic fields, dipoles, materials, and magnetic effects. Students use torque, magnetic flux, dipole fields, magnetisation, susceptibility, and permeability to solve numerical and conceptual problems.

Magnetism connects field patterns, dipole behaviour, material response, and Earth-like magnetic effects in one chapter. Important Questions Class 12 Physics Chapter 5 help students practise bar magnets, magnetic field lines, Gauss’s law for magnetism, torque, potential energy, magnetisation, magnetic intensity, susceptibility, and magnetic materials. The CBSE 2026 chapter also compares electric and magnetic dipoles and classifies materials as diamagnetic, paramagnetic, and ferromagnetic.

Key Takeaways

• Magnetic Monopoles: Isolated north and south magnetic poles do not exist.
• Torque on Dipole: A magnetic dipole in a uniform magnetic field experiences torque τ = mB sin θ.
• Gauss’s Law: The net magnetic flux through any closed surface is zero.
• Material Classification: Magnetic materials are diamagnetic, paramagnetic, or ferromagnetic.

Important Questions Class 12 Physics Chapter 5 Structure 2026

Important Questions Class 12 Physics Chapter 5 with Answers

Magnetism and Matter questions mix theory, diagrams, and numerical formulas.
Students must connect field direction, dipole alignment, flux, and material response.
These class 12 physics chapter 5 important questions follow the NCERT 2026 chapter sequence.

1. What does Important Questions Class 12 Physics Chapter 5 cover in Magnetism and Matter?

Important Questions Class 12 Physics Chapter 5 cover bar magnets, magnetic dipoles, Gauss’s law, magnetisation, and magnetic materials. The chapter also tests numerical problems on torque and magnetic fields.

1. Bar Magnet: Poles, field lines, and solenoid analogy.
2. Dipole: Torque and potential energy in a uniform magnetic field.
3. Gauss’s Law: Net magnetic flux through a closed surface.
4. Materials: Diamagnetic, paramagnetic, and ferromagnetic behaviour.
5. Final Result: The chapter tests magnetic fields and material response.

2. Why does a freely suspended bar magnet point north-south?

A freely suspended bar magnet points north-south because Earth behaves approximately like a magnet. Its north-seeking pole points towards geographic north.

1. Given Fact: Earth has a magnetic field.
2. Magnet Behaviour: A magnetic dipole aligns with the external magnetic field.
3. Observation: A suspended bar magnet settles along the north-south direction.
4. Final Result: The bar magnet aligns with Earth’s magnetic field.

3. Why do isolated magnetic poles not exist?

Isolated magnetic poles do not exist because every magnet has both north and south poles. Cutting a magnet creates smaller magnets, not single poles.

1. Given Fact: A bar magnet has two poles.
2. Action: Cut the magnet into two parts.
3. Observation: Each part has a north pole and a south pole.
4. Final Result: Magnetic monopoles do not exist in NCERT physics.

Class 12 Physics Chapter 5 Magnetism and Matter

The chapter begins with known magnetic facts and then builds mathematical laws.
Bar magnets, solenoids, and dipoles provide the bridge between field patterns and formulas.
This class 12 physics chapter 5 magnetism and matter section covers the core NCERT concepts.

4. What happens when a bar magnet is cut into two equal pieces?

Each piece becomes a smaller magnet with its own north and south poles. The magnetic strength becomes weaker than the original magnet.

1. Case 1: Cut transverse to its length.
2. Result 1: Two smaller magnets form.
3. Case 2: Cut along its length.
4. Result 2: Two smaller magnets form.
5. Final Result: Each piece has both north and south poles.

5. Why is a bar magnet compared with a solenoid?

A bar magnet is compared with a solenoid because both produce similar magnetic field lines. Their fields look alike at large distances.

1. Bar Magnet: Field lines emerge from north and enter south outside.
2. Solenoid: Field lines form closed loops like a bar magnet.
3. Analogy: A bar magnet acts like many circulating currents.
4. Final Result: A bar magnet can be treated as an equivalent solenoid.

6. What is Ampere’s hypothesis in magnetism?

Ampere’s hypothesis states that magnetic phenomena arise from circulating currents. A bar magnet behaves like many tiny current loops.

1. Basic Idea: Moving charges produce magnetic fields.
2. Atomic View: Circulating electrons act like current loops.
3. Bar Magnet View: Many loops produce a net magnetic moment.
4. Final Result: Magnetism can be explained through circulating currents.

Bar Magnet Class 12 Questions

A bar magnet behaves like a magnetic dipole with north and south poles.
Its field resembles a finite solenoid and helps derive axial and equatorial field formulas.
These bar magnet class 12 questions focus on poles, field strength, and dipole analogy.

7. What is the axial magnetic field of a short bar magnet?

The axial magnetic field of a short bar magnet is B = μ0(2m)/(4πr³). It acts along the dipole axis.

1. Given Condition: r is much greater than magnet size l.
2. Formula Used:
   B = μ0(2m)/(4πr³)
3. Direction: Along the axis of the magnet.
4. Final Result: Axial field = μ0(2m)/(4πr³).

8. What is the equatorial magnetic field of a short bar magnet?

The equatorial magnetic field of a short bar magnet is B = −μ0m/(4πr³). The negative sign shows opposite direction to m.

1. Given Condition: r is much greater than magnet size l.
2. Formula Used:
   B = −μ0m/(4πr³)
3. Direction: Opposite to the magnetic moment.
4. Final Result: Equatorial field = −μ0m/(4πr³).

9. How does the magnetic field change when distance from a short bar magnet doubles?

The magnetic field becomes one-eighth of its original value. The field varies as 1/r³ for a short bar magnet.

1. Formula Used: B ∝ 1/r³.
2. New Distance: r’ = 2r.
3. Calculation:
   B’/B = r³/(2r)³
   B’/B = 1/8
4. Final Result: The field becomes B/8.

Magnetic Field Lines Class 12 Questions

Magnetic field lines show the direction and relative strength of a magnetic field.
They form closed loops and never intersect, unlike electric field lines of isolated charges.
These magnetic field lines class 12 questions test diagrams and conceptual rules.

10. What are the main properties of magnetic field lines?

Magnetic field lines are continuous closed loops. The tangent at any point gives the direction of magnetic field B.

1. Closed Loop Rule: Magnetic field lines form closed loops.
2. Direction Rule: Tangent gives the direction of B.
3. Strength Rule: Crowded lines show stronger field.
4. Intersection Rule: Field lines never intersect.
5. Final Result: Magnetic field lines show direction and strength.

11. Why do magnetic field lines never intersect?

Magnetic field lines never intersect because the magnetic field cannot have two directions at one point. Intersection would create directional ambiguity.

1. Assumption: Two field lines intersect.
2. Problem: The point would have two tangents.
3. Meaning: Magnetic field would have two directions.
4. Final Result: Field lines cannot intersect.

12. Why do magnetic field lines form closed loops?

Magnetic field lines form closed loops because isolated magnetic monopoles do not exist. There are no starting or ending magnetic charges.

1. Electric Field: Lines start on positive charges and end on negative charges.
2. Magnetic Field: Lines have no source or sink.
3. Gauss Law Result: Net magnetic flux through a closed surface is zero.
4. Final Result: Magnetic field lines are continuous closed loops.

Magnetic Dipole Class 12 Questions

A bar magnet acts as a magnetic dipole at large distances.
Its magnetic moment determines torque, energy, and field strength.
These magnetic dipole class 12 questions cover formulas and physical meaning.

13. What is the magnetic moment of a current loop?

The magnetic moment of a current loop is m = IA. Here, I is current and A is loop area.

1. Given Quantity: Current = I.
2. Given Quantity: Area vector = A.
3. Formula Used: m = IA.
4. Unit: A m² or J T⁻¹.
5. Final Result: Magnetic moment = IA.

14. What is the magnetic moment of a solenoid with N turns?

The magnetic moment of a solenoid is m = NIA. Each turn contributes magnetic moment IA.

1. Given Quantity: Number of turns = N.
2. Given Quantity: Current = I.
3. Given Quantity: Cross-sectional area = A.
4. Formula Used: m = NIA.
5. Final Result: Solenoid magnetic moment = NIA.

15. Find the magnetic moment of a solenoid with 800 turns, area 2.5 × 10⁻⁴ m², and current 3 A.

The magnetic moment is 0.6 A m². Use m = NIA.

1. Given Data:
   N = 800
   A = 2.5 × 10⁻⁴ m²
   I = 3 A
2. Formula Used: m = NIA.
3. Calculation:
   m = 800 × 3 × 2.5 × 10⁻⁴
   m = 2400 × 2.5 × 10⁻⁴
   m = 0.6 A m²
4. Final Result: m = 0.6 A m².

Torque on Bar Magnet Class 12

A magnetic dipole in a uniform magnetic field experiences torque but no net force.
The torque tries to align the magnetic moment with the field.
These torque on bar magnet class 12 questions use τ = mB sin θ.

16. What is the torque on a magnetic dipole in a uniform magnetic field?

The torque on a magnetic dipole is τ = m × B. Its magnitude is τ = mB sin θ.

1. Vector Formula: τ = m × B.
2. Magnitude Formula: τ = mB sin θ.
3. Maximum Torque: θ = 90°.
4. Zero Torque: θ = 0° or 180°.
5. Final Result: Torque magnitude = mB sin θ.

17. A bar magnet at 30° in a 0.25 T field experiences torque 4.5 × 10⁻² J. Find its magnetic moment.

The magnetic moment is 0.36 J T⁻¹. Use τ = mB sin θ.

1. Given Data:
   τ = 4.5 × 10⁻² J
   B = 0.25 T
   θ = 30°
2. Formula Used: τ = mB sin θ.
3. Calculation:
   m = τ/(B sin θ)
   m = 4.5 × 10⁻²/(0.25 × 0.5)
   m = 0.045/0.125
   m = 0.36 J T⁻¹
4. Final Result: m = 0.36 J T⁻¹.

18. What is the torque when a magnetic dipole is parallel to the magnetic field?

The torque is zero when a magnetic dipole is parallel to the field. The angle θ equals 0°.

1. Given Data: θ = 0°.
2. Formula Used: τ = mB sin θ.
3. Calculation:
   τ = mB sin 0°
   τ = 0
4. Final Result: Torque = 0.

19. What is the torque when a magnetic dipole is antiparallel to the magnetic field?

The torque is zero when a magnetic dipole is antiparallel to the field. The angle θ equals 180°.

1. Given Data: θ = 180°.
2. Formula Used: τ = mB sin θ.
3. Calculation:
   τ = mB sin 180°
   τ = 0
4. Final Result: Torque = 0.

Class 12 Physics Magnetism Questions with Answers

Magnetism numericals often combine magnetic moment, torque, potential energy, and field formulas.
Students should identify the angle before choosing sin θ or cos θ.
These class 12 physics magnetism questions with answers focus on CBSE-style calculations.

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

The potential energy is U = −mB cos θ. It depends on the angle between m and B.

1. Formula Used: U = −mB cos θ.
2. Stable Case: θ = 0° gives U = −mB.
3. Unstable Case: θ = 180° gives U = +mB.
4. Final Result: Magnetic potential energy = −mB cos θ.

21. Find the stable and unstable potential energies for m = 0.32 J T⁻¹ and B = 0.15 T.

The stable energy is −0.048 J, and the unstable energy is +0.048 J. Use U = −mB cos θ.

1. Given Data:
   m = 0.32 J T⁻¹
   B = 0.15 T
2. Stable Orientation: θ = 0°.
   U = −0.32 × 0.15 × cos 0°
   U = −0.048 J
3. Unstable Orientation: θ = 180°.
   U = −0.32 × 0.15 × cos 180°
   U = +0.048 J
4. Final Result: Stable U = −0.048 J and unstable U = +0.048 J.

22. What work turns a magnet from field-aligned to perpendicular position?

The work required is mB. Potential energy changes from −mB to 0.

1. Initial Position: θ = 0°.
   Ui = −mB
2. Final Position: θ = 90°.
   Uf = 0
3. Work Required:
   W = Uf − Ui
   W = 0 − (−mB)
   W = mB
4. Final Result: Work required = mB.

23. What work turns a magnet from field-aligned to opposite field direction?

The work required is 2mB. Potential energy changes from −mB to +mB.

1. Initial Position: θ = 0°.
   Ui = −mB
2. Final Position: θ = 180°.
   Uf = +mB
3. Work Required:
   W = Uf − Ui
   W = mB − (−mB)
   W = 2mB
4. Final Result: Work required = 2mB.

Gauss Law Magnetism Class 12 Questions

Gauss’s law for magnetism states that net magnetic flux through any closed surface is zero.
The law expresses the absence of isolated magnetic poles.
These gauss law magnetism class 12 questions cover flux, monopoles, and closed field lines.

24. State Gauss’s law for magnetism.

Gauss’s law for magnetism states that net magnetic flux through any closed surface is zero. It means magnetic monopoles do not exist.

1. Mathematical Form:
   ∮ B · dS = 0
2. Physical Meaning: No isolated magnetic pole exists inside any closed surface.
3. Field Line Meaning: Lines entering equal lines leaving.
4. Final Result: Net magnetic flux is always zero.

25. How does Gauss’s law for magnetism differ from Gauss’s law in electrostatics?

Magnetic flux through a closed surface is always zero. Electric flux depends on the net charge enclosed.

1. Electrostatic Law:
   ∮ E · dS = q/ε0
2. Magnetic Law:
   ∮ B · dS = 0
3. Reason: Electric charges exist separately, but magnetic monopoles do not.
4. Final Result: Electric flux can be non-zero, but magnetic flux is zero.

26. What would Gauss’s law become if magnetic monopoles existed?

Gauss’s law would include magnetic charge on the right side. The flux would no longer always remain zero.

1. Current Law: ∮ B · dS = 0.
2. If Monopoles Existed:
   ∮ B · dS = μ0qm
3. Here: qm is magnetic charge enclosed.
4. Final Result: Magnetic flux would depend on enclosed magnetic charge.

27. Why is net magnetic flux through a closed surface around one pole of a magnet zero?

The net flux is zero because field lines are continuous loops. Lines leaving the surface also enter it.

1. Surface: Closed surface around one magnet end.
2. Observation: Field lines pass through the surface.
3. Gauss Law: Total outgoing flux equals incoming flux.
4. Final Result: Net magnetic flux remains zero.

Class 12 Physics Chapter 5 Questions and Answers on Magnetic Fields

Magnetic field questions test direction, magnitude, and physical interpretation.
The chapter compares magnetic dipoles with electric dipoles through parallel formulas.
These class 12 physics chapter 5 questions and answers focus on field values and directions.

28. Find the axial field of a bar magnet with m = 0.48 J T⁻¹ at r = 10 cm.

The axial field is 9.6 × 10⁻⁵ T. Use B = μ0(2m)/(4πr³).

1. Given Data:
   m = 0.48 J T⁻¹
   r = 10 cm = 0.10 m
   μ0/4π = 10⁻⁷ T m A⁻¹
2. Formula Used:
   B = μ0(2m)/(4πr³)
3. Calculation:
   B = 10⁻⁷ × (2 × 0.48)/(0.10)³
   B = 10⁻⁷ × 0.96/0.001
   B = 9.6 × 10⁻⁵ T
4. Final Result: Axial field = 9.6 × 10⁻⁵ T.

29. Find the equatorial field of a bar magnet with m = 0.48 J T⁻¹ at r = 10 cm.

The equatorial field has magnitude 4.8 × 10⁻⁵ T. Its direction is opposite to the magnetic moment.

1. Given Data:
   m = 0.48 J T⁻¹
   r = 0.10 m
   μ0/4π = 10⁻⁷ T m A⁻¹
2. Formula Used:
   B = μ0m/(4πr³)
3. Calculation:
   B = 10⁻⁷ × 0.48/(0.10)³
   B = 10⁻⁷ × 0.48/0.001
   B = 4.8 × 10⁻⁵ T
4. Final Result: Equatorial field = 4.8 × 10⁻⁵ T opposite to m.

30. Why is the axial field twice the equatorial field at the same distance?

The axial field is twice the equatorial field because the axial formula has factor 2. Both fields vary as 1/r³.

1. Axial Field: Baxis = μ0(2m)/(4πr³).
2. Equatorial Field: Beq = μ0m/(4πr³).
3. Ratio:
   Baxis/Beq = 2
4. Final Result: Axial field magnitude is twice equatorial field magnitude.

Magnetisation and Magnetic Intensity Class 12

Magnetisation describes the net magnetic moment per unit volume of a material.
Magnetic intensity separates the applied field effect from the material’s own magnetic response.
This magnetisation and magnetic intensity class 12 section covers definitions and relations.

31. What is magnetisation in Class 12 Physics?

Magnetisation is the net magnetic moment per unit volume of a material. It is a vector quantity.

1. Formula Used:
   M = mnet/V
2. Unit: A m⁻¹.
3. Meaning: It measures magnetic moment density.
4. Final Result: Magnetisation = net magnetic moment per unit volume.

32. What is magnetic intensity H?

Magnetic intensity H represents the external magnetising field. It relates B and M through H = B/μ0 − M.

1. Formula Used:
   H = B/μ0 − M
2. Unit: A m⁻¹.
3. Relation:
   B = μ0(H + M)
4. Final Result: H measures the magnetising field contribution.

33. What is the relation between B, H, and M?

The relation between B, H, and M is B = μ0(H + M). It separates applied field and material magnetisation.

1. B: Total magnetic field in the material.
2. H: Magnetic intensity.
3. M: Magnetisation.
4. Formula: B = μ0(H + M).
5. Final Result: B = μ0(H + M).

Magnetic Susceptibility Class 12 Questions

Magnetic susceptibility measures how strongly a material responds to an external magnetic field.
Its sign separates diamagnetic and paramagnetic substances.
These magnetic susceptibility class 12 questions focus on χ, μr, and μ.

34. What is magnetic susceptibility?

Magnetic susceptibility is the ratio of magnetisation to magnetic intensity. It shows the magnetic response of a material.

1. Formula Used:
   χ = M/H
2. Rearranged Form:
   M = χH
3. Nature: χ is dimensionless.
4. Final Result: Magnetic susceptibility = M/H.

35. What is the relation between relative permeability and susceptibility?

The relation is μr = 1 + χ. Relative permeability compares material permeability with free-space permeability.

1. Formula Used: μr = 1 + χ.
2. Also: μ = μ0μr.
3. Combined Form: μ = μ0(1 + χ).
4. Final Result: μr = 1 + χ.

36. A material has μr = 400 and H = 2 × 10³ A m⁻¹. Find M.

The magnetisation is 7.98 × 10⁵ A m⁻¹. Use M = (μr − 1)H.

1. Given Data:
   μr = 400
   H = 2 × 10³ A m⁻¹
2. Formula Used:
   M = (μr − 1)H
3. Calculation:
   M = (400 − 1) × 2 × 10³
   M = 399 × 2 × 10³
   M = 7.98 × 10⁵ A m⁻¹
4. Final Result: M = 7.98 × 10⁵ A m⁻¹.

Magnetic Properties of Materials Class 12 Questions

Magnetic materials respond differently to external magnetic fields.
NCERT classifies materials using susceptibility, relative permeability, and field-line behaviour.
These magnetic properties of materials class 12 questions compare dia-, para-, and ferromagnetism.

37. How are magnetic materials classified in Class 12 Physics?

Magnetic materials are classified as diamagnetic, paramagnetic, and ferromagnetic. The classification depends on magnetic susceptibility.

1. Diamagnetic: χ is small and negative.
2. Paramagnetic: χ is small and positive.
3. Ferromagnetic: χ is large and positive.
4. Final Result: Magnetic materials are grouped by susceptibility.

38. What are diamagnetic substances?

Diamagnetic substances move from stronger to weaker magnetic field regions. They develop magnetisation opposite to the applied field.

1. Susceptibility: −1 ≤ χ < 0.
2. Relative Permeability: 0 ≤ μr < 1.
3. Examples: Bismuth, copper, lead, water, sodium chloride.
4. Final Result: Diamagnetic materials are weakly repelled by magnets.

39. What are paramagnetic substances?

Paramagnetic substances move from weaker to stronger magnetic field regions. They develop weak magnetisation along the applied field.

1. Susceptibility: 0 < χ < ε.
2. Relative Permeability: 1 < μr < 1 + ε.
3. Examples: Aluminium, sodium, calcium, oxygen, copper chloride.
4. Final Result: Paramagnetic materials are weakly attracted by magnets.

40. What are ferromagnetic substances?

Ferromagnetic substances get strongly magnetised in an external magnetic field. They move strongly towards higher magnetic field regions.

1. Susceptibility: χ is large and positive.
2. Relative Permeability: μr >> 1.
3. Examples: Iron, cobalt, nickel, gadolinium.
4. Final Result: Ferromagnetic materials are strongly attracted by magnets.

Diamagnetic Paramagnetic Ferromagnetic Questions

The three material types differ by field response and microscopic dipole alignment.
Diamagnetic materials oppose the field, while para- and ferromagnetic materials support it.
These diamagnetic paramagnetic ferromagnetic questions focus on direct comparisons.

41. What is the difference between diamagnetic and paramagnetic materials?

Diamagnetic materials are weakly repelled, while paramagnetic materials are weakly attracted. Their susceptibilities have opposite signs.

1. Diamagnetic χ: Negative and small.
2. Paramagnetic χ: Positive and small.
3. Field Behaviour: Diamagnetic materials reduce internal field.
4. Paramagnetic Behaviour: Paramagnetic materials enhance internal field.
5. Final Result: χ sign separates diamagnetic and paramagnetic materials.

42. Why are ferromagnetic materials strongly magnetised?

Ferromagnetic materials are strongly magnetised because their atomic dipoles align in domains. External fields make domains grow and align.

1. Atomic Dipoles: Each atom has magnetic moment.
2. Domain Formation: Many dipoles align within a domain.
3. External Field Effect: Domains align along B0.
4. Final Result: Domain alignment gives strong magnetisation.

43. What is the difference between hard and soft ferromagnets?

Hard ferromagnets retain magnetisation, while soft ferromagnets lose magnetisation after field removal. Their applications differ.

1. Hard Ferromagnet: Magnetisation persists.
2. Example: Alnico.
3. Soft Ferromagnet: Magnetisation disappears after field removal.
4. Example: Soft iron.
5. Final Result: Hard ferromagnets make permanent magnets.

44. What happens to a ferromagnet at high temperature?

A ferromagnet becomes paramagnetic at high temperature. Its domain structure disintegrates with temperature.

1. Low Temperature: Domains support strong magnetisation.
2. High Temperature: Thermal agitation disturbs domain alignment.
3. Result: Ferromagnetic behaviour disappears gradually.
4. Final Result: A ferromagnet changes into a paramagnet at high temperature.

NCERT Class 12 Physics Chapter 5 Questions for Board Practice

NCERT questions often test formulas through direct numerical substitution.
They also ask conceptual reasons for magnetic field-line behaviour and material response.
These NCERT class 12 physics chapter 5 questions follow the 2026 exercise pattern.

45. A magnet has m = 1.5 J T⁻¹ and B = 0.22 T. Find work to turn it from parallel to perpendicular.

The work required is 0.33 J. Use W = mB for rotation from 0° to 90°.

1. Given Data:
   m = 1.5 J T⁻¹
   B = 0.22 T
2. Formula Used: W = mB.
3. Calculation:
   W = 1.5 × 0.22
   W = 0.33 J
4. Final Result: Work required = 0.33 J.

46. For the same magnet, find work to turn it from parallel to opposite direction.

The work required is 0.66 J. Use W = 2mB for rotation from 0° to 180°.

1. Given Data:
   m = 1.5 J T⁻¹
   B = 0.22 T
2. Formula Used: W = 2mB.
3. Calculation:
   W = 2 × 1.5 × 0.22
   W = 0.66 J
4. Final Result: Work required = 0.66 J.

47. What is the torque in the perpendicular and opposite cases for the same magnet?

The torque is 0.33 N m at 90° and 0 N m at 180°. Use τ = mB sin θ.

1. Given Data:
   m = 1.5 J T⁻¹
   B = 0.22 T
2. At 90°:
   τ = 1.5 × 0.22 × sin 90°
   τ = 0.33 N m
3. At 180°:
   τ = 1.5 × 0.22 × sin 180°
   τ = 0
4. Final Result: Torque = 0.33 N m at 90° and 0 at 180°.

48. A solenoid has 2000 turns, area 1.6 × 10⁻⁴ m², and current 4 A. Find magnetic moment.

The magnetic moment is 1.28 A m². Use m = NIA.

1. Given Data:
   N = 2000
   A = 1.6 × 10⁻⁴ m²
   I = 4 A
2. Formula Used: m = NIA.
3. Calculation:
   m = 2000 × 4 × 1.6 × 10⁻⁴
   m = 8000 × 1.6 × 10⁻⁴
   m = 1.28 A m²
4. Final Result: m = 1.28 A m².

49. Find torque on that solenoid in B = 7.5 × 10⁻² T at 30°.

The torque is 4.8 × 10⁻² N m. Use τ = mB sin θ.

1. Given Data:
   m = 1.28 A m²
   B = 7.5 × 10⁻² T
   θ = 30°
2. Formula Used: τ = mB sin θ.
3. Calculation:
   τ = 1.28 × 7.5 × 10⁻² × 0.5
   τ = 4.8 × 10⁻² N m
4. Final Result: Torque = 4.8 × 10⁻² N m.

50. What is the force on a solenoid in a uniform magnetic field?

The net force is zero in a uniform magnetic field. The solenoid can still experience torque.

1. Given Condition: Magnetic field is uniform.
2. Dipole Behaviour: Opposite forces balance each other.
3. Torque: A couple can rotate the solenoid.
4. Final Result: Net force = 0.

Class 12 Physics Chapter-Wise Important Questions