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.
SNP
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.
BBHBVδ
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.
HBRetentivityCoercivity
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?
BSN
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.
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