Class 12 Physics Assignment
Chapter: Magnetism and Matter
Q1. Define magnetic dipole moment of a bar magnet. Write its SI unit and direction.
Solution:
Magnetic dipole moment is defined as the product of pole strength and magnetic length of the magnet: M = m × 2l. Its SI unit is A m². Its direction is from south pole to north pole inside the magnet.Q2. A bar magnet has pole strength 20 A m and distance between its poles 10 cm. Find its magnetic dipole moment.
Answer: 2 A m²
M = pole strength × distance between poles = 20 × 0.10 = 2 A m².Q3. Derive the expression for magnetic field at an axial point of a short bar magnet.
Solution:
For a short magnet of magnetic moment M, at an axial point distance r from centre, B = μ₀/(4π) × 2M/r³. This field is along the magnetic moment direction.Q4. Derive the expression for magnetic field at an equatorial point of a short bar magnet.
Solution:
For a short magnet, the magnitude of magnetic field at an equatorial point distance r from centre is B = μ₀/(4π) × M/r³. Vector form includes a negative sign because the direction is opposite to the magnetic moment.Q5. A short bar magnet has magnetic moment 0.8 A m². Find magnetic field at an axial point 20 cm from its centre.
Answer: 2 × 10⁻⁵ T
B = 10⁻⁷ × 2M/r³ = 10⁻⁷ × 1.6/(0.2)³ = 1.6×10⁻⁷/0.008 = 2×10⁻⁵ T.Q6. For the same magnet of moment 0.8 A m², find magnetic field at an equatorial point 20 cm from the centre.
Answer: 1 × 10⁻⁵ T
B = 10⁻⁷ × M/r³ = 10⁻⁷ × 0.8/0.008 = 1×10⁻⁵ T.Q7. Compare magnetic field at axial and equatorial points of a short bar magnet at same distance.
Answer:
In magnitude, B_axial = 2B_equatorial at the same distance. Axial field is along magnetic moment; equatorial field is opposite to magnetic moment.Q8. A bar magnet is placed in uniform magnetic field B at angle θ. Write torque and potential energy.
Solution:
Torque τ = MB sinθ. Potential energy U = −MB cosθ.Q9. A magnet of moment 5 A m² is placed in field 0.2 T at 30°. Calculate torque.
Answer: 0.5 N m
τ = MB sinθ = 5 × 0.2 × sin30° = 1 × 0.5 = 0.5 N m.Q10. Calculate work done in rotating a magnetic dipole of moment 3 A m² from 0° to 90° in field 0.4 T.
Answer: 1.2 J
Work done = ΔU = MB(cosθ₁ − cosθ₂) = 3×0.4(cos0° − cos90°) = 1.2 J.Q11. A magnetic dipole is in stable equilibrium in a uniform magnetic field. What is the angle between M and B?
Answer: 0°
Potential energy U = −MB cosθ is minimum when cosθ = 1, so θ = 0°.Q12. A magnetic dipole is in unstable equilibrium in a uniform magnetic field. What is the angle between M and B?
Answer: 180°
Potential energy is maximum when θ = 180°, therefore equilibrium is unstable.Q13. Figure shows a bar magnet placed on its axis. Identify direction of magnetic field at point P.
Answer:
At an axial point on the north side, magnetic field is away from the north pole, i.e. towards right.Q14. Explain why isolated magnetic poles do not exist.
Solution:
Magnetic field lines always form closed loops and every magnet has both north and south poles. Cutting a magnet produces smaller dipoles, not isolated poles. Hence magnetic monopoles have not been observed.Q15. State Gauss's law in magnetism and explain its meaning.
Solution:
Gauss's law in magnetism: ∮B·dA = 0. It means net magnetic flux through any closed surface is zero because magnetic monopoles do not exist.Q16. A closed surface is placed in magnetic field. If magnetic flux entering it is 4 Wb, what is flux leaving it?
Answer: 4 Wb
Net magnetic flux through a closed surface is zero. Therefore flux entering = flux leaving.Q17. Define magnetic declination.
Solution:
Magnetic declination is the angle between geographic meridian and magnetic meridian at a given place.Q18. Define angle of dip or magnetic inclination.
Solution:
Angle of dip is the angle made by Earth's resultant magnetic field with the horizontal direction at a place.Q19. At a place, horizontal component of Earth's magnetic field is 3 × 10⁻⁵ T and angle of dip is 60°. Find total magnetic field.
Answer: 6 × 10⁻⁵ T
B_H = B cosδ. So B = B_H/cos60° = 3×10⁻⁵/0.5 = 6×10⁻⁵ T.Q20. At a place, total magnetic field is 5 × 10⁻⁵ T and dip is 37°. Find horizontal and vertical components. Use sin37° = 0.6, cos37° = 0.8.
Answer: B_H = 4 × 10⁻⁵ T, B_V = 3 × 10⁻⁵ T
B_H = B cosδ = 5×10⁻⁵×0.8 = 4×10⁻⁵ T. B_V = B sinδ = 5×10⁻⁵×0.6 = 3×10⁻⁵ T.Q21. Figure shows Earth's magnetic field components. Write relation between B, B_H, B_V and angle of dip δ.
Answer:
B_H = B cosδ, B_V = B sinδ, tanδ = B_V/B_H, and B = √(B_H² + B_V²).Q22. At magnetic equator, what is the value of angle of dip?
Answer: 0°
At magnetic equator, Earth's field is horizontal, hence angle of dip is zero.Q23. At magnetic poles, what is the value of angle of dip?
Answer: 90°
At magnetic poles, Earth's magnetic field is vertical, so dip angle is 90°.Q24. Define magnetization I of a magnetic material and write its SI unit.
Solution:
Magnetization I is magnetic dipole moment per unit volume: I = M/V. SI unit is A m⁻¹.Q25. A magnetic specimen of volume 5 × 10⁻⁶ m³ has magnetic moment 0.02 A m². Find magnetization.
Answer: 4 × 10³ A/m
I = magnetic moment/volume = 0.02/(5×10⁻⁶) = 4×10³ A/m.Q26. Define magnetic susceptibility and relative permeability.
Solution:
Magnetic susceptibility χ measures how easily material gets magnetized: I = χH. Relative permeability μᵣ = μ/μ₀. Relation: μᵣ = 1 + χ.Q27. If magnetic susceptibility of a material is 0.002, find its relative permeability.
Answer: 1.002
μᵣ = 1 + χ = 1 + 0.002 = 1.002.Q28. If relative permeability of a material is 800, find its susceptibility.
Answer: 799
μᵣ = 1 + χ, so χ = μᵣ − 1 = 800 − 1 = 799.Q29. Distinguish diamagnetic, paramagnetic and ferromagnetic materials on the basis of susceptibility.
Solution:
Diamagnetic: χ is small negative. Paramagnetic: χ is small positive. Ferromagnetic: χ is large positive and depends on temperature and magnetization history.Q30. Give two examples each of diamagnetic, paramagnetic and ferromagnetic substances.
Answer:
Diamagnetic: bismuth, copper. Paramagnetic: aluminium, oxygen. Ferromagnetic: iron, cobalt, nickel.Q31. Why are diamagnetic substances weakly repelled by a magnet?
Solution:
In diamagnetic substances, induced magnetic moment develops opposite to applied magnetic field. Hence they are weakly repelled by strong magnetic field.Q32. Why are paramagnetic substances weakly attracted by a magnet?
Solution:
Paramagnetic substances have permanent atomic magnetic dipoles. In external field, these dipoles partially align with the field, causing weak attraction.Q33. State Curie's law for paramagnetic substances.
Solution:
Curie's law: χ = C/T, where χ is magnetic susceptibility, C is Curie constant, and T is absolute temperature. Susceptibility decreases as temperature increases.Q34. A paramagnetic material has susceptibility 4 × 10⁻⁵ at 300 K. Find susceptibility at 600 K.
Answer: 2 × 10⁻⁵
By Curie's law, χ ∝ 1/T. χ₂ = χ₁T₁/T₂ = 4×10⁻⁵×300/600 = 2×10⁻⁵.Q35. Define Curie temperature.
Solution:
Curie temperature is the temperature above which a ferromagnetic substance loses ferromagnetic behaviour and becomes paramagnetic.Q36. Explain hysteresis in ferromagnetic materials.
Solution:
Hysteresis is the lagging of magnetization or magnetic induction behind the magnetizing field during cyclic magnetization. It is represented by B-H or I-H hysteresis loop.Q37. In a hysteresis loop, define retentivity and coercivity.
Solution:
Retentivity is residual magnetism left when magnetizing field is reduced to zero. Coercivity is the reverse magnetizing field required to reduce magnetization to zero.Q38. Figure shows a hysteresis loop. Identify retentivity and coercivity from the graph.
Answer:
Retentivity is the B-axis intercept when H = 0. Coercivity is the negative H-axis intercept where B becomes zero.Q39. Which material is suitable for making permanent magnets: high or low coercivity? Explain.
Answer:
Permanent magnets require high retentivity and high coercivity so they retain magnetism and are difficult to demagnetize.Q40. Which material is suitable for transformer cores: narrow or wide hysteresis loop? Explain.
Answer:
Transformer cores require soft ferromagnetic material with narrow hysteresis loop, high permeability and low hysteresis loss.Q41. Why is soft iron used as core of electromagnets?
Solution:
Soft iron has high permeability and low coercivity. It is easily magnetized and demagnetized, making it suitable for electromagnets.Q42. Why is steel used for permanent magnets?
Solution:
Steel has high coercivity and good retentivity. Once magnetized, it retains magnetism for long time and is difficult to demagnetize.Q43. A solenoid has 1000 turns per metre and carries current 2 A. Find magnetic field intensity H inside it.
Answer: 2000 A/m
For a solenoid, H = nI = 1000 × 2 = 2000 A/m.Q44. A solenoid has n = 800 turns/m, current 5 A and relative permeability 1000. Find B inside it.
Answer: approximately 5.03 T
B = μ₀μᵣnI = 4π×10⁻⁷ ×1000×800×5 = 1.256×10⁻³×4000 ≈ 5.03 T. This is a theoretical value for the given high relative permeability.Q45. A material has H = 1000 A/m and magnetization I = 500 A/m. Find susceptibility.
Answer: 0.5
I = χH, so χ = I/H = 500/1000 = 0.5.Q46. Figure shows a small magnetic needle placed in uniform field. In which orientation is potential energy minimum?
Answer:
Potential energy U = −MB cosθ is minimum when θ = 0°, so magnetic moment points along B. Since M is from S to N, the north end should point along field direction.Q47. A bar magnet of magnetic moment 4 A m² is kept perpendicular to a field of 0.25 T. Find torque and potential energy.
Answer: τ = 1 N m, U = 0
τ = MB sin90° = 4×0.25 = 1 N m. U = −MB cos90° = 0.Q48. A magnet of moment 2 A m² is turned from 30° to 150° in a uniform field 0.5 T. Find work done by external agent.
Answer: √3 J ≈ 1.73 J
Work done by external agent = ΔU = U₂ − U₁ = MB(cosθ₁ − cosθ₂). Thus W = 2×0.5(cos30° − cos150°) = 1(√3/2 + √3/2) = √3 J.Q49. Explain why magnetic field lines outside a bar magnet go from north to south but inside go from south to north.
Solution:
Magnetic field lines are continuous closed curves. Outside the magnet they emerge from north pole and enter south pole. Inside the magnet they complete the loop from south to north.Q50. A short bar magnet has axial field 8 × 10⁻⁵ T at 10 cm. Find its magnetic moment.
Answer: 0.4 A m²
Axial field B = 10⁻⁷ × 2M/r³. M = Br³/(2×10⁻⁷) = 8×10⁻⁵ × (0.1)³ /(2×10⁻⁷) = 8×10⁻⁸/(2×10⁻⁷) = 0.4 A m².